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MATHS LECTURE # 14
Permutation and Combination,
Probability
Fundamental Counting Principles The Product Rule The Additive Rule If one thing can be done in m different ways and a second thing can be done in n different ways, then the two things in succession can be done in m × n different ways. This principle is known as Product Rule. Suppose we have 9 distinct red balls and 13 distinct blue balls. In how many ways can we choose a ball? Since we have 9 + 13 = 22 balls and we are going to choose exactly one ball, we can choose a ball in 9 + 13 = 22 ways. If in addition to 9 red balls and 13 blue balls, we have 8 distinct black balls then the number of ways in which we can choose a ball out of these is 9 + 13 + 8 = 30 ways. The above illustration explains a general principle, called the additive rule. Suppose that X and Y are two disjoint events, that is, they never occur together. Further, suppose that X occurs in m ways and Y occurs in n ways. Then X or Y can occur in m + n ways. This rule can be extended to more than two mutually exclusive events also. In the above illustration, X can be taken as choosing of a red ball and Y can be taken as choosing of a blue ball. The number of ways in which exactly one of X, Y can occur is 9 + 13 = 22. Fo r ex amp le, if the re are 3 ca ndid ate s f or g ove rno r po st and 5 for the mayor post, then the two offices may be filled in 3 × 5 = 15 way s. In general, if a 1 can be done in x 1 ways, a 2 can be done in x 2 ways, a 3 can be done in x 3 ways, ..., and a n can be done in x n w ay s, th en th e e ve n ts a 1 , a 2 , a 3 .. . a n ca n b e d on e i n x 1 × x 2 × x 3 ... x n ways. E2. A man has 3 jackets, 10 shirts and 5 pairs of slacks. If an outfit consists of a jacket, shirt and a pair of slacks, how many different outfits can the man make? Sol. x 1 × x 2 × x 3 = 3 × 10 × 5 = 150 outfits. Sol. If Ramesh is allowed to choose exactly one of the above mentioned toys in the shop, he will have to choose one toy out of 15 + 17 + 6 = 38 toys. There are 38 ways in which Ramesh can choose a toy. If there are m ways of doing a thing, n ways of doing a second thing and p ways of doing a third thing, then the total number of “distinct” ways of doing all these together is m × n × p. E3. Suppose, there are five routes for going from a place A to another place B and six routes for going from the place B to a third place C. Find the number of different ways through which a person can go from A to C via B.
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PT Education, All rights reserved. Important Result E1. Ramesh Kumar Joshi is taken to a toy shop “ToysRUs” containing 15 distinct toy cars, 17 distinct toy dolls and 6 distinct toy guns. Find the number of ways in which Ramesh can choose a toy. Regd. Office: Indore PT centres spread across India ~ Established 1993 Our motto “Kar Ke Dikhayenge” is delivered through our unique Technology Driven Process Engine (TDpro engine). Email: [email protected] Web: www.PTeducation.com , www.facebook.com/PTeducation
IC : PTtkmml14 (1) of (25) Sol. Since there are five different routes from A to B, person can go from A to B in five different ways. After reaching B, he has six different ways of finishing the second part of his journey (i.e. going from B to C). Thus for one way of going from A to B there are six different ways of completing the journey from A to C via B. Hence, the total number of different ways of finishing both parts of the journey (i.e. A to B and t h e n f r o m B t o C ) = 5 t i m e s s i x d i f f e r e n t w a y s = 5 × 6 = no. of ways from the first point to the second point × number of ways from the second point to the third point. E4. Suppose, there are 7 special buses running from Indore to Bhopal and 9 boats plying down the stream from Bhopal to Bairagarh and 4 boats intended for the return journey from Bairagarh to Bhopal. Find in how many different ways can a gentleman have a trip to Bairagarh from Indore via Bhopal and back to Bhopal again? Sol. A person can go from Indore to Bhopal through any one of the buses from Indore to Bhopal. Therefore he can finish the first part of his journey in 7 different ways. After he has completed the first part of the journey in any one way, he has 9 boats before him and through any one of them he can go to Bairagarh. Therefore, corresponding to 7 ways of going from Indore to Bhopal there are 7 × 9 ways of going from Indore to Bairagarh via Bhopal. Thus we conclude that one can go from Indore to Bairagarh via Bhopal in 7 × 9 different ways. Again corresponding to any one way of going to Bairagarh, he can return to Bhopal from Bairagarh in 4 different ways. Hence, there are altogether 7 × 9 × 4 = 252 different ways of going from Indore to Bairagarh via Bhopal and returning to Bhopal from Bairagarh. So the next time you intend to take such a journey, do remember this chapter on basic permutations. E5. Find the number of different ways in which four persons can be accommodated in three different chairs. Sol. Let 1 st chair 2 nd chair 3 rd chair be three chairs. F o r t h e 1 s t c h a i r a n y o f t h e f o u r p e r s o n s c a n b e accommodated, so it can be filled in 4 ways. For the 2 nd Chair we will have only 3 ways of filling it, Since one person is already accommodated. Similarly for 3 rd chair we have 2 ways of filling. Using the product rule. The total number of ways of accommodating 4 persons in 3 chairs will be 4 × 3 × 2 = 24 ways. E6. A shop has 8 doors and 12 windows. (a) In how many ways can a thief enter the shop, if he may enter through a door or a window? (b) In how many ways can the thief rob the shop by entering through a window and exiting through a door? Sol. Two events are involved: selecting a window and selecting a door. The first event can occur in 12 different ways and the second event can occur in 8 different ways. (a) Since the thief uses either a door or a window to enter the shop, exactly one of the two events occur. By the sum rule, the thief has 8 + 12 = 20 choices to enter the shop. (b) Since the thief enters through a window and exits through a door, both the events must occur. Therefore, by the product rule, the thief has 12 × 8 = 96 possible ways to rob the shop. E7. A department in an organisation contains 17 male employees and 15 female employees. (a) In how many ways, can an employee be chosen from the department to represent the department in the annual meeting? (b) In how many ways, can a male and a female employee be chosen from the department to represent the department in the employees union meeting? Sol. Two events are involved: selecting a male and selecting a female employee. The first event can occur in 17 different ways and the second event can occur in 15 different ways. (a) Since only one employee is to be selected to represent the department in the annual meeting, exactly one of the two events must occur. By the sum rule, there are 17 + 15 = 32 ways to choose an employee to represent the department in the annual meeting. (b) Since a male and a female are to be chosen from the department to represent the department on the employees union meeting, both the events must occur. Therefore, by the product rule, there are 17 × 15 = 255 ways to choose a male and a female employee to represent the department in the employees union meeting. E8. How many words (with or without meaning) of three distinct letters of the English alphabets are there? Sol. The number of letters in the English alphabet is 26. The number of ways of forming 3­letter words is equal to the number of ways of filling up three blocks with distinct alphabets. The first block can be filled up in 26 ways, the second in 25 ways and the third in 24 ways. Thus, the number of 3­letter words = 26 × 25 × 24 = 15600.
(2) of (25) IC : PTtkmml14 Factorial Function E9. Evaluate each factorial. (i) Here we introduce a very useful function, called the factorial function. This function is very convenient for calculations and formulae to be used in this lecture. (iii) 3! ¸ 4! (v) Sol. (i) 1, 2, 3, ... , n. (ii) 5! × 2! (iv) (3!)! 2 3! 7!/2! = (7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1) = 7 × 6 × 5 × 4 × 3 = 2520. (ii) Definition: If n is a natural number, then n factorial, denoted by n! or ën is defined to be the product 1 × 2 × 3 ... (n – 1) × n 7!/2! 5! × 2! = (5 × 4 × 3 × 2 × 1) × (2 × 1) = 120 × 2 = 240. (iii) 3! ¸ 4! = (3 × 2 × 1) ¸ (4 × 3 × 2 × 1) = 6/24 = 1/4. That is n! = 1 × 2 × 3 × 4 . . (n – 1) × n (iv) (3!)! = (3 × 2 × 1)! = 6! = 720 As a special case, we define 0! = 1. We now list the value of n! for some values of n. 0! = 1 (v) 2 3! = 2 3×2×1 = 2 6 = 64. Permutation 1! = 1 2! = 1 × 2 = 2 3! = 1 × 2 × 3 = 6 4! = 1 × 2 × 3 × 4 = 24 5! = 1 × 2 × 3 × 4 × 5 = 120 A permutation is an arrangement of all or part of a number of things in a definite order. For example, the permutations of the three letters a, b, c taken all at a time are abc, acb, bca, bac, cba, cab. The permutations of the three letters a, b, c taken two at a time are ab, ac, ba, bc, ca, cb. 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720. We can also define n! regressively as follows Number of Permutations of n different things taken r at a time n! = n (n – 1)! for n ³ 1 Thus 7! = 7 (6!) = 7 (720) = 5040. and 8! = 8 (7!) = 8 (5040) = 40320. Properties of Factorials 1 . It is a natural number. 2 . If n ³ 5, then n! ends in a zero or the units digit of n! always is zero. 3 . If n ³ 1 then n! is divisible by 2, 3, 4, ... , n. 4 . If n ³ 1, then n! is divisible by r! for 1 £ r £ n In fact n ! n ( n - 1 ) ( n - 2 ) ... ( r + 1 ) r × ( r - 1 ) ... 2 . 1 =
r ! r ( r - 1 ) ... 2 . 1 = n (n – 1) (n – 2) ... (r + 1). 5 . Product of r (³ 1) consecutive natural numbers can be written as quotient of two factorials. Let the r consecutive natural numbers be m, m + 1, m + 2, ..., m + r – 1. Then, m × (m + 1) × (m + 2) . ... × (m + r – 1)
( m - 1 ) ! m ( m + 1 ) ... ( m + r - 1 ) ( m + r - 1 ) ! =
=
( m - 1) ! ( m - 1 ) ! If n and r are positive integers such that 1 £ r £ n, then the number of all permutations of n distinct things, taken r at a time is denoted by the symbol P(n, r) or n P r n P r =
n ! = n n ( - 1 )( n - 2 )...( n - r + 1 ) ( n - r )! When r = n, n P r = n P n = n(n – 1) (n – 2) ... 1 = n!. Thus 8 P 3 denotes the numbers of permutations of 8 different things taken 3 at a time and 5 P 5 denotes the number of permutations of 5 different things taken 5 at a time. Important In permutations, the order of arrangement is taken into account; when the order is changed, a different permutation is obtained. E10. Find the values of 5 P 1 , 5 P 2 , 5 P 3 , 5 P 4 , 5 P 5 and 10 P 7 . Sol. 5 P 1 = 5, 5 P 2 = 5 × 4 = 20, 5 P 3 = 5 × 4 × 3 = 60, 5 P = 5 × 4 × 3 × 2 = 120, 4 5 P = 5! = 5 × 4 × 3 × 2 × 1 = 120 5 10 P = 10 × 9 × 8 × 7 × 6 × 5 × 4 = 604800. 7 E11. Find the total number of ways in which 4 persons can take their places in a cab having 6 seats. Sol. The number of ways in which 4 persons can take their places in a cab having 6 seats = 6 p 4 = 6 × 5 × 4 × 3 = 360 ways.
IC : PTtkmml14 (3) of (25) Some Important Formulae E16. Find the number of permutations of the letters of the word ASSASSINATION. Ø The total number of permutations of n different Sol. We have 13 letters in all of which 3 are A’s, 4 are S’s, 2 are I’s and 2 are N’s
things taken all at a time is n! E12. In how many ways can 6 people can stand in a queue? Sol. The number of ways in which 6 people can stand in a queue = 6! = 720. Ø The total number of arrangements of n different things taken r at a time, in which a particular thing always occurs = r × n–1 P r–1 . E13. How many 4 digits number (repetition is not allowed) can be made by using digits 1­7 if 4 will always be there in the number? Sol. Total digits (n) = 7. Total ways of making the number if 4 is always there \ The no. of arrangements = 13 ! 3 ! 4 ! 2 ! 2 ! E17. Find the number of permutations of the letters of the words ‘DADDY DID A DEADLY DEED’. Sol. We have 19 letters in all of which 9 are Ds; 3 are As; 2 are Ys; 3 are Es and the rest are all distinct. Therefore, the number of arrangements
=
19 ! 9 ! 3 ! 2 ! 3 ! E18. How many different words can be formed with the letters of word ‘ORDINATE’? (1) so that the vowels occupy odd places. (2) beginning with ‘O’. Ø The total number of permutations of n different (3) beginning with ‘O’ and ending with ‘E’. things taken r at a time in which a particular thing never occurs = n – 1 P r Sol. (1) = r × n–1 P r–1 = 4 × 6 P 3 = 480. ORDINATE contains 8 letters 4 odd places, 4 vowels.
Þ number of arrangements of the vowels 4! E14. How many different 3 letter words can be made by 5 vowels, if vowel ‘A’ will never be included? Also number of arranging consonants is 4!
Þ Number of words = 4! × 4! Sol. Total letters (n) = 5. So total number of ways = n–1 P r = 5–1 P 3 = 4 P 3 = 24. = (4 × 3 × 2 × 1) 2 = 576. (2) Ø The total number of permutations of n dissimilar things taken r at a time with repetitions = n r . E15. How many 3 digits number can be made by using digits 1 to 7 if repetition is allowed? Sol. Total digits (n) = 7. When O is fixed we have only seven letters at our disposal.
Þ Number of words = 7! = 5040. (3) When we have only six letters at our disposal, leaving ‘O’ and ‘E’ which are fixed. Number of permutations = 6! = 720. Circular Permutation So total ways = 7 3 = 343. Ø The number of permutations of n things taken all at a time when p of them are alike and of one kind, q of them are alike and of second kind, all other being different, is n !
p ! q ! Remark: The above theorem can be extended if in addition to the above r things are alike and of third kind and so on, then total permutation =
n ! p ! q ! r ! .... So far we have discussed permutation of objects (or things) in a row. This type of permutations are generally known as linear permutations. If we arrange the objects along a closed curve v i z . a c i r c l e , t h e p e r mu t a t i o n s a r e k n o w n a s c ir c u la r permutations. As we have seen in the earlier sections of this chapter that every linear arrangement has a beginning and an end, but there is nothing like a beginning or an end in a circular permutation. Thus, in a circular permutation, we consider one object as fixed and the remaining objects are arranged as in case of linear arrangements. The number of ways of arranging n distinct objects around a circle is (n – 1)! & the total number when taken n r at a time will be (4) of (25) P r r .
IC : PTtkmml14 Note: In the above, anti­clockwise and clockwise order of arrangements are considered as distinct permutations. E19. In how many ways can 7 Australians sit down at a round table? Difference Between Permutation & Combination In a combination only a group is made and the order in which the objects are arranged is immaterial. Sol. 7 Aust ral ian s ca n t ake the ir seat at th e ro und ta ble in (7 – 1) ! = 6! ways. On the other hand, in a permutation, not only a group is formed, but also an arrangement in a definite order is considered. E20. In how many ways can 12 persons among whom two are brothers be arranged along a circle so that there is exactly one person between the two brothers? Examples Sol. One person between the two brothers can be chosen in 10 P = 10 ways. The remaining 9 persons can be arranged in 1 9 P = 9!. 9 1 . ab and ba are two different permutations, but each represents the same combination. 2 . abc, acb, bac, bca, cab, cba are six different permutations but each one of them represents the same combination, namely a group of three objects a, b and c. The two brothers can be arranged in 2! ways. Therefore, the total number of ways = (9!) (10) (2!) = (10!) (2!). Remark: We use the word ‘arrangements’, for permutations and ‘selections’, for combinations. Important Number of combinations of n different things taken r at a time Ø If there be no difference between clockwise and n C r =
anticlockwise arrangements, the total no. of circular permutations of n things taken all at a time is b n - 1 g ! and the total number when taken r at a time 2
n will be Pr 2r P r n ! n n ( - 1 )( n - 2). ..( n - r + 1 ) =
r ! r !( n - r )! =
n ! n For example, the number of handshakes that may be exchanged among a party of 12 students if each student shakes hands once with each other student is . 12 C2 =
12 ! 12 ! 12 ´ 11 =
=
= 66 . 2 !( 12 - 2 )! 2 10 ! ! 1 ´ 2 E21. In how many ways can a garland of 10 different flowers be made? The following formula is very useful in simplifying calculations Sol. S ince, there is no difference between clockwise and anticlockwise arrangements, the total number of ways. Ø n C = n C r n–r (complementary combination) = (10 – 1)!/2 = 9!/2. Combination A combination is a grouping or selection of all or part of a number of things without reference to the arrangement of the things selected. Thus, the combinations of the three letters a, b, c taken 2 at a time are ab, ac, bc. Note that ab and ba are 1 combination but 2 permutations of the letters a, b. The symbol n C r represents the number of combinations (selections, groups) of n different things taken r at a time. Thus 9 C 4 denotes the number of combinations of 9 different things taken 4 at a time. Note. The symbol C(n, r) having the same meanings as n C r is sometimes used. This formula indicates that the number of selections of r out of n things is the same as the number of selections of n – r out of n things. Like given in the following cases 5 C1 =
9 C7 =
25 5 5 ´ 4 5 ! = 5 , 5 C2 =
= 10 , 5 C5 =
= 1 1 1 ´ 2 5 ! 9 ´ 8 = 36 , 1 ´ 2 C22 =
25 ´ 24 ´ 23 = 2300 . 1 ´ 2 ´ 3 Note that in each case the numerator and denominator have the same number of factors. E22. In how many ways a hockey team of eleven can be elected from 16 players? Sol. Total number of ways = 16 C11 =
IC : PTtkmml14 16 ! = 4368.
11 ! ´ 5 ! (5) of (25) Some important formulae For example, Ø Number of combinations of n different things taken r at a time in which p particular things will always occur is n–p C r–p Ø No. of combinations of n dissimilar things taken ‘r’ at a time in which ‘p’ particular things will never occur is n–p C r E23. In a class of 25 students, find the total number of ways to select two representative, if a particular person will be never selected. Sol. Total students (n) = 25. A particular students will not be selected (P) = 1, So total number of ways = 25–1 C 2 = 24 C 2 = 276. (i) 25 C 23 (ii) 75 C 74 25 C = 25 C 25 23 25–23 = C 2 = 300. (ii) Groups with 3 alphabets Groups with 2 alphabets ABC DE ABD CE ACD BE BCD AE ABE DC ACE BD BCE AD ADE BC BDE AC CDE AB E26. In how many ways we can make two groups of 8 and 3 students out of total 11 students. Sol. Total students (m + n) = 11. So total number of ways = 11 C 8 =
E24. Evaluate Sol. (i) The number of ways of dividing of 3 boys & 2 girls respectively, into groups of 3 & 2 are, as follows Ø If 2m things are to be divided into two groups, each containing m things, the number of ways
75 C = 75 C 75 74 75–74 = C 1 = 75. =
Ø ( 2 m)! 2 ( m !) 2 n C + n C n + 1 C r r – 1 = r For example, if we divide 4 alphabets A, B, C and D into two groups containing 2 alphabets, the number of ways are 3. The arrangements are shown as below. E25. Evaluate (i) 5 C + 5 C . 3 2 (ii) 7 C + 7 C . 5 4 5 C + 5 C = 5 C + 5 C 6 3 2 3 3–1 = C 3 = 20. I II AB CD 7 C + 7 C = 7 C + 7 C 8 5 4 5 5–1 = C 5 = 56. AC BD AD BC Sol. (i) (ii) Ø 11 ! = 165 . 8 ! ´ 3 ! n C + n C + n C + .... n C = 2 n 0 1 2 n Ø The number of ways in which (m + n) things can be divided into two groups containing m & n things respectively
= ( m + n ) C n =
b m + n g ! =
b m ! n ! g Ø The number of ways to divide n things into different groups, one containing p things, another q things and so on =
b p + q + r + ... ! g b p ! q ! r ! ... g where {n = p + q + r + ...} ( m + n ) C m Ø Total number of combinations of n dissimilar things taken some or all at a time = 2 n – 1.
(6) of (25) IC : PTtkmml14 E27. In a city no persons have identical set of teeth and there is no person without a tooth. Also no person has more than 32 teeth. If we disregard the shape and size of tooth and consider only the positioning of the teeth, then find the maximum population of the city. (Assume no two persons have similar configuration regarding positioning of teeth) Sol. We have 32 places for teeth. For each place we have two choices either there is a tooth or there is no tooth. Therefore the number of ways to fill up these places is 2 32 . As there is no person without a tooth, then the maximum population is 2 32 – 1. E30. Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the seating arrangement can be made. Sol. Since four particular guests want to sit on a particular side A (say) and three others on the other side B (say). So, we are left with 11 guests out of which we choose 5 for side A in 11 C 5 ways and the remaining 6 for side B in 6 C 6 ways. Hence, the number of selections for the two sides is 11 C 5 × 6 C . Now 9 persons on each side of the table can be arranged 6 among themselves in 9! ways. Hence, the total number of arrangements Ø Total number of combinations of n things, taking some or all at a time, when p of them are alike of one kind, q of them are alike of another kind and so on is = 11 C 5 ´ 6 C6 ´ 9 ! ´ 9 ! =
11 ! ´ 9 ! ´ 9 ! . 6 ! 5 ! {(p + 1)(q + 1)(r + 1) ...} – 1, where n = p + q + r + ... E28. Find the total number of combinations of 5 alphabets A, B, A, B, B, taking some or all at a time. Sol. Here A is twice and B is thrice, so by formula, total combinations = (2 + 1) (3 + 1) – 1 = 11. In our daily life, we often come across events whi c h are concern ed wi t h the i dea of the likelihood or the chance of occurrence of future events. Consider the following events The combinations are as follows 1 at a time 2 at a time 3 at a time 4 at a time 5 at a time Total Combinations A B AA AB BB AAB ABB BBB AABB ABBB AABBB Total 2 3 3 2 1 11 Miscellaneous Problems on Permutations and Combinations A . Amit is a Class X student in a certain school. There is a test in Mathematics on the coming Monday. We may ask “Will Amit pass the test?” Before we answer this question, we have to consider some past records. If Amit failed in all his past Mathematics tests, we say that he has a very slim chance of passing the test. But, if Amit passed all his past Mathematics tests, we say that he has a very great chance of passing the test. B. Ø In many problems the concepts of both permutation and combination are used. Following examples will illustrate the same. E29. Out of 6 consonants and 5 vowels, how many words of 3 consonants and 2 vowels can be formed? Sol. Three consonants out of 6 and 2 vowels out of 5 can be chosen in 6 C 3 × 5 C 2 ways. Since each of these words contains 5 letters, which can be arranged among themselves in 5! ways. So, the required number of words = 6 C 3 × 5 C 2 × 5! = 24000. Prakash is a student of standard X in a certain school. If a student is chosen at random in standard X, will Prakash be chosen? Before we answer this question, we have to know about the present statistics, the number of students in standard X. If there are 45 students in the class, then Prakash’s chance of being chosen is 1/45. C . If we toss a fair coin, is it more likely for a ‘head’ or a ‘tail’ to come up? In this event, we can answer the question immediately without considering any past or present statistics. We all know that there i is an equal chance for a ‘head’ or a ‘tail’ to come up. Thus the chance for a ‘head’ (or a ‘tail’) to come up is 1/2. An alternative word used for ‘chance’ is ‘probability’ and it is generally represented by “P”. Random Experiments In most branches of knowledge, experiments are the ways of life. In probability and statistics, too, we concern ourselves with s p e ci a l t y pe o f e x pe r i m e n ts . W e ca l l t h e se a s r an d o m experiments.
IC : PTtkmml14 (7) of (25) A random experiment is an experiment in which (i) S all the possible outcomes of the experiment are known in advance and (ii) the exact outcome of any specific performance of the experiment is unpredictable, i.e., depends upon chance. For example, Suppose a fair coin is tossed. There are two possible outcomes of the experiment; heads and tails. For any performance of the experiment one does not know what the exact outcome will be. Therefore, the tossing of a fair coin is a random experiment. A Elementary and compound events Single element subsets of sample space associated with a random experiment are known as elementary events. Sample space The set of all possible outcomes of a random experiment is called the sample space. If x Î S, then x is called as a sample point. For instance, in illustration, the sample space is S 1 = {H, T} where H stands for heads and T for tails. If the same coin is tossed twice, then S 2 = {HH, HT, TH, TT}. In case the coin is tossed three times, the sample space will be S 3 = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. In case of a single toss, the sample space consists of 2 points. In case of two tosses, it consists of 2 × 2 = 2 2 = 4 points whereas in case of three tosses, it consists of 2 × 2 × 2 = 2 3 = 8 points. In case of n tosses, the sample space contains 2 n points. To get all the points in the sample space S n , corresponding to n tosses of a fair coin we first write S 1 = {H, T} corresponding to a single toss. To obtain S 2 we postfix H to each element in S 1 to obtain HH, TH and then postfix T to each element in S 1 to obtain HT, TT. These four elements constitute S 2 . That is, S 2 = {HH, TH, HT, TT}. Now, to obtain S 3 we postfix H to each element in S 2 to obtain HHH, THH, HTH, TTH and then T to each element in S 2 to obtain HHT, THT, HTT, TTT. These eight elements constitute S 3 . That is, S 3 = {HHH, THH, HTH, TTH, HHT, THT, HTT, TTT}. In a similar way we can obtain S 4 , S 5 , etc. Alternatively we may use a tree diagram as shown below. Those subsets of sample space S associated to an experiment which are disjoint union of single element subsets of sample space S are known as compound events. e.g.: Consider the experiment of throwing a die. The sample space associated with this experiment is S = {1, 2, 3, 4, 5, 6}. Events E 1 = {1}, E 2 = {2}, E 3 = {3}, E 4 = {4}, E 5 = {5}, E 6 = {6} are elementary events whereas A 1 = {2, 4, 6}, A 2 = {1, 3, 5}, A 3 = {3, 6} etc. are compound events. Equally Likely Events All possible results of a random experiment are called equally likely outcomes and we have no reason to expect any one rather than the other. For example, as the result of drawing a card from a well shuffled pack, any card may appear in draw, so that the 52 cards become 52 different events which are equally likely. Mutually Exclusive Events and S 3 = {HHH, HHT, HTH, HTT, THH, THT, THH, TTT}. Events are called mutually exclusive or disjoint or incompatible if the occurrence of one of them precludes the occurrence of all the others. For example in tossing a coin, there are two mutually exclusive events viz turning up a head and turning up of a tail. Since both these events cannot happen simultaneously. But note that events are compatible if it is possible for them to happen simultaneously. For instance in rolling of two dice, the cases of the face marked 5 appearing on one dice and face 5 appearing on the other, are compatible. Event Exhaustive Events Thus, S 2 = {HH, HT, TH, TT} Any subset of a sample space is called an event. A sample space S serves as the universal set for all questions related to an experiment 'S' and an event A w.r.t it is a set of all possible outcomes favourable to the event A. For example, Events are exhaustive when they include all the possibilities associated with the same trial. In throwing a coin, the turning up of head and of a tail are exhaustive events assuming of course that the coin cannot rest on its edge.
A random experiment: flipping a coin twice Sample space: S = {(HH), (HT), (TH), (TT)} The question: "both the flips show same face" Therefore, the event A: {(HH), (TT)}. See figure. (8) of (25) IC : PTtkmml14 Independent Events S Two events are said to be independent if the occurrence of any event does not affect the occurrence of the other event. For example in tossing of a coin, the events corresponding to the two successive tosses of it are independent. The flip of one penny does not affect in any way the flip of a nickel. B A A Ç B
Dependent Events Important If the occurrence or non­occurrence of any event affects the happening of the other, then the events are said to be dependent events. For example, in drawing a card from a pack of cards, let the event A be the occurrence of a king in the 1st draw and B be the occurrence of a king in the second draw. If the card drawn at the first trial is not replaced then events A and B are independent events. Note that A Ç B occurs if and only if the outcome of the experiment is an element of A and B both. In other words A Ç B occurs if and only if both A and B occur simultaneously. Thus, A Ç B also denotes the simultaneous occurrence of A and B. We often write A Ç B as “A and B”. Note Definitions of Probability We shall now consider two definitions of probability (1) Mathematical or a priori or classical. (1) If an event contains a single sample point i.e. it is a singleton set, then this event is called an elementary or a simple event. (2) An event corresponding to the empty set is an ‘impossible event’. (3) An event corresponding to the entire sample space is called a 'certain event'. Complement of an event The complement of an event A with respect to S is the set of all elements of S that are not in A. We denote the complement of A by the symbol A’. In the following figure, the shaded portion of S represents A’ or A . We also call A’ as “not A”. . A’ S (2) Statistical or empirical. Classical Definition of Probability If a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes out of which m are favourable to the occurrence of an event A, then the probability of the occurrence of A, usually denoted by P(A) is given by Thus, P ( A ) =
m the number of favourable outcomes . =
n the total number of outcomes In the application of this definition, the term favourable is used rather loosely ­ favourable may mean that a patient may have a viral fever or a brand new television does not work or a book published by TMH company has more than 120 printing errors. A Thus probability is a concept which measures numerically the degree of certainty or uncertainty of the occurrence of an event. Intersection of events The intersection of two events A and B, denoted by A Ç B, is the event containing all elements that are common to A and B. In the following figure, A Ç B is given by the shaded portion. For example, the probability of randomly drawing king from a well­shuffled deck of cards is 4/52. Since 4 is the number of favourable outcomes (i.e. 4 kings of diamond, spade, club and heart) and 52 is the number of total outcomes (the number of cards in a deck). If A is any event of sample space having probability P, then clearly, P is a positive number (expressed as a fraction or usually as a decimal) not greater than unity.
IC : PTtkmml14 (9) of (25) Tips to remember E33. What is the chance that a leap year selected at random will contain 53 Sundays? Ø If an event E is sure to occur, we say that the probability of the event E is equal to 1 and we write P(E) = 1. Sol. A leap year has 52 weeks and 2 more days. The two days can be Ø If an event E is sure not to occur, we say that the probability o f t h e e v e n t E i s e q u a l t o 0 a n d w e w r i t e P (E) = 0. Therefore for any event E, 0 £ P(E) £ 1. Monday ­ Tuesday Tuesday ­ Wednesday Wednesday ­ Thursday Thursday ­ Friday E31. A coin is tossed three times. Find the probability of getting at least one head. Sol. Sample space: S = {(H H H), (H H T), (HTH), (HTT), (THT), (TTH), (THH), (TTT) } Þ n (S) = 8 Event A: getting at least one head so that A Friday ­ Saturday Saturday ­ Sunday and Sunday ­ Monday. There are 7 outcomes and 2 are favourable to the 53rd Sunday. Now for 53 Sundays in a leap year, P(A) = 2/7. getting no head at all = { (TTT) Þ n ( A ) = 1 Compound Probability P ( A ) = 1/8 Therefore, P(A) = 1 – P ( A ) = 1 – 1/8 = 7/8. E32. A ball is drawn at random from a box containing 6 red balls, 4 white balls and 5 blue balls. Determine the probability that the ball drawn is (i) red (ii) white (iii) blue (iv) not red (v) red or white. Two or more events are said to be independent if the occurrence or non­occurrence of any one of them does not affect the probabilities of occurrence of any of the others. Thus if a coin is tossed four times and it turns up a head each time, the fifth toss may be head or tail and is not influenced by the previous tosses. Sol. Let R, W and B denote the events of drawing a red ball, a white ball and a blue ball respectively. The probability that two or more independent events will happen is equal to the product of their individual probabilities. (i) The probability that the ball drawn is red = P ( R ) =
n R ( ) the number of favourable outcomes =
n S ( )
the total number of outcomes Thus the probability of getting a head on both the fifth and sixth tosses is = (ii) 6 6 2 =
= . 6 + 4 + 5 15 5
Similarly, the probability that the ball drawn is white = P(W) = 4/15. (iii) The probability that the ball drawn is black F G I J
H K 1 1 1 = . 2 2 4 Two or more events are said to be dependent if the occurrence or non­occurrence of one of the events affects the probability of occurrence of any of the others. = P(B) = 5/15 = 1/3. (iv) The probability that the ball drawn is not red ( ) = 1 - P R ( ) = 1 = P R (v) 2 3 = . 5 5 Consider that two or more events are dependent. If P 1 is the probability of a first event, P 2 the probability that after the first has happened the second will occur, P 3 the probability that after the first and second have happened the third will occur etc., then The probability that the ball drawn is red or white
6 4 n R ( ) n w ( )
10 2 =
+
=
+
=
= . 15 15
n S ( ) ns 15 3 the Probability that all events will happen in the given order is the product P 1 × P 2 × P 3 .... For example, a box contains 3 white balls and 2 black balls. If a ball is drawn at random, the probability that it is black is 2 2 = . 3 + 2 5
If this ball is not replaced and a second ball is drawn, the probability that it also is black is that both will be black is (10) of (25) 1 1 = . Thus the probability 3 + 1 4
F G I J
H K 2 1 1 =
.
5 4 10 IC : PTtkmml14 Mutually Exclusive Events E34. A single card is selected from a deck of 52 bridge cards. What is the probability that Two or more events are said to be mutually exclusive if the occurrence of any one of them guarantees the non occurrence of the others. The probability of occurrence of one or two or more mutually exclusive events is the sum of the probabilities of the individual events. 1 . it is not a heart? (ii) it is an ace or a spade? Sol. A deck of bridge cards has 4 suits – spade, heart, diamond and club. Each suit has 13 cards.– Ace, two, three, .... , ten, jack, Queen, King. (i) P (not a heart) = 1 – P (a heart) = 1 – 13/52 = 39/52 = 3/4. (ii) F : The event that the number is even. There are 4 aces and 12 spades besides the ace of spades
G : The event that the number is a multiple of three. Þ P (an ace or a spade) = 16/52 = 4/13. In rolling a die E : The event that the number is odd. 2 . (i) In drawing a card from a deck of 52 cards E35. If a die is thrown, what is the probability of getting a 5 or a 6? E : The event that it is a spade. F : The event that it is a club. Sol. Getting a 5 and getting a 6 are mutually exclusive G : The event that it is a king. so, P(5 or 6) = P(5) + P(6) =
1 1 2 1 + = = . 6 6 6 3 In the above 2 cases, events E & F are mutually exclusive but the events E & G or F & G are not mutually exclusive or disjoint since they may have common outcomes. Two events are said to be overlapping if the events have at least one outcome in common, hence they can happen at the same time. Addition Law of Probability The probability of occurrence of some one of two overlapping events is the sum of the probabilities of the two individual events minus the probability of their common outcomes. If E & F are two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given by
+ P(E or F) = P(E) + P(F) If the events are not mutually exclusive, then
E36. If a die is thrown, what is the probability of getting a number less than 4 or an even number? Sol. The numbers less than 4 on a die are 1, 2 and 3. The even numbers on a die are 2, 4 and 6. Since these two events have a common outcome, 2, they are overlapping events. P (less than 4 or even) + P(E or F) = P(E) + P(F) – P(E & F together) = P(less than 4) + P(even) – P(less than 4 and even)
=
Note: Compare this with n(A È B) = n(A) + n(B) – n(A Ç B) of set theory. 3 3 1 5 + - =
. 6 6 6 6 Independent Events Similarly
+ P (neither E nor F) = 1 – P(E or F) Two events are independent if the happening of one has no effect on the happening of the other. For example 1 . 2 . On rolling a die & tossing a coin together E : The event that number 6 turns up. F : The event that head turns up. In shooting a target E : Event that the first trial is missed. F : Event that the second trial is missed. In both these cases events E & F are independent.
IC : PTtkmml14 (11) of (25) 3 . In drawing a card from a well shuffled pack E : Event that first card is drawn. F : Event that second card is drawn without replacing the first. G: Event that second card is drawn after replacing the first. E39. A bag contains 3 red and 4 black balls and another bag has 4 red and 2 black balls. One bag is selected at random and from the selected bag a ball is drawn. Let E be the event that the first bag is selected, F be the event that the second bag is selected, G be the event that ball drawn is red and H be the event that ball drawn is black. Then, find In this case E & F are not independent but E & G are independent. Multiplication Law of Probability ( i ) P(E) (ii) (iii) P(G/E) (iv) P(G/F) (v) (vi) P(H/F) P(H/E) P(F) Sol. If the events E & F are independent then
+ P(E & F) = P (E) × P (F) & P (not E & F) = 1 – P(E & F together). ( i ) P (E) = (iii) Odds In Favour And Against If P is the probability that an event will occur and q (= 1 – p) is the probability of the non­occurrence of the event; then we say that the odds in favour of the event occurring are p : q (p to q) and the odds against its occurring are q : p. For example, if the event consists of drawing a card of club from the deck of 52 cards, then the odds in favour are 13 39 1 3 : = : = 1 : 3. 52 52 4 4 Similarly the odds against a card of diamond would be 3 : 1. The odds in favour of 4 or 6 in a single toss of a fair die are 1 2 : ; i.e. 1 : 2 and the odds against are 2 : 1. 3 3 P P (F) = 1 2 3 7 F G G I J = Probability of drawing a red ball from second H F K bag = (v) (ii) F G G I J = Probability of drawing a red ball from first H E K bag = (iv) P 1 2 4 2 =
6 3
F G H I J = Probability of drawing a black ball from first H E K P bag = 4 7 F G H I J =Probability of drawing a black ball from second H F K (vi) P bag = 2 1 = 6 3
E37. The odds in favour of an event are 2:7. Find the probability of occurrence of this event. Sol. Total number of outcomes = 2 + 7 = 9 Favourable number of cases = 2
\
P (E) = 2 9 E38.The odds against of an event are 5:7, find the probability of occurrence of this event. Sol. Total number of outcomes = 5 + 7 = 12 Number of cases against the occurrence of event = 5
\
Number of cases in favour of event = 12 – 5 = 7
\
P (E) = (12) of (25) 7 12 IC : PTtkmml14 SOLVED EXAMPLES
The following questions will help you get a thorough hold on the basic fundamentals of the chapter and will also help you develop your concepts. It is advised that a s incere student should solve each of the given questions at least twice to develop the required expertise.
DIRECTIONS: For each of the following questions, please give the complete solution. 1 . 2 . 3 . 4 . In how many ways 5 Indians and 4 Americans be seated around the table, if no two Americans sit together? A gentleman has 6 friends to invite for a party. In how many ways can he send invitation cards to them if he has three servants to carry the cards? In how many ways can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? Three villages A, B and C lie on a highway. There are 10 more routes connecting A and B 8 more routes connecting B and C, 5 more routes connecting A and C. If B is in between A and C, in how many ways can one go from B to C? 5 . A man has six friends, in how many ways can he invite one or more of them to a party? 6 . Sheela can choose to go for a party with Yogesh, Nikhilesh, Shashank, Amol, Shainu, Vivek, Girish. She has 10 pairs of sandals and 17 different dresses. In how many different ways she will go to the party if she has a choice of taking 5 of her friends. Also she likes her 7 pair of sandals and 15 dresses which she wants to wear for the party? 7. 8 . 9 . In how many ways can the 7 letters M, N, O, P, Q, R, S be arranged so that P and Q occupy continuous positions? In how many ways can 4 Indians and 4 Nepalese people be seated around a round table such that no two Indians are in adjacent positions? There are 150 members in a table­tennis club. They are playing a tournament such that a member is out of the tournament if he loses a game. If we know that there were no ties, then the number of games to determine the champion will be IC : PTtkmml14 10. In how many different ways can 8 examination papers be arranged in a row, so that the best and the worst papers may never come together? 11. Three letters are to be sent to different persons and addresses on the three envelopes are also written. Without looking at the addresses, what is the probability that the letters go into the right envelope? 12. A bag contains 3 red, 7 white and 4 black balls. If three balls are drawn from the bag, then what is the probability that all of them are of the same colour? 13. If four persons are chosen at random from a group of 3 men, 2 women and 4 children. Then, what is the probability that exactly two of them are children? 14. From 10,000 lottery tickets numbered from 1 to 10,000, one ticket is drawn at random. What is the probability that the number marked on the drawn ticked is divisible by 20? 15. The chances to fail in physics are 20% and the chances to fail in mathematics are 10%. What are the chances to fail in at least one subject? 16. A problem of mathematics is given to three students whose chances of solving the problem are 1/3, 1/4 and 1/5 respectively. What is the probability that the question will be solved? 17. In a lottery 50 tickets are sold in which 14 are of prize. A man bought 2 tickets, then what is the probability that the man win the prize? 18. A three digit number is formed by using numbers 1, 2, 3 and 4 without repetition. What is the probability that the number is divisible by 3? 19. Word ‘UNIVERSITY’ is arranged randomly. Then, what is the probability that both T does not come together? 20. A bag contains 3 red. 4 white and 5 blue balls. All balls are different. Two balls are drawn at random. What is the probability that they are of different colour?
(13) of (25) PRACTICE EXERCISE # 01
The following exercise is to be taken by the students as a homework assignment, only after going through the theory provided. After having solved this exercise, a sincere student shall stand to gain a lot of speed and also will reinforce the concepts studied earlier. For all questions, mark the answers in the score sheet that is provided AT THE END. Please fill all ovals properly with an HB pencil. Scratch work (if any) has to be done on test paper itself.
Sug gested Ti me : 25 mi n. max.
DIRECTIONS: Two dice are rolled simultaneously. Find the probability of 1 . 2 . 3 . 8 . The letters of the word ENTRANCE are arranged in all possible ways. The number of arrangements having the E's and N's together is (1) 7! (2) 6! (3) 8! (4) 9! A picnic party of four persons is to be selected from 8 Engineers and 3 Doctors so as to include at least one Doctor. The possible number of ways are (1) 168 (2) 84 (3) 8 (4) 260 How many numbers, each consisting of four different digits, can be formed with the digits 0, 1, 2, 3? (1) 15 (2) 19 (3) 18 (4) 24 4. How many numbers lying between 3000 and 4000 and which are divisible by 5 can be made with the digits 3, 4, 5, 6, 7 and 8? (Digits are not to be repeated in any number). (1) 11 (2) 12 (3) 13 (4) 14 5. The total number of ways of answering 5 objective type questions, each question having 4 choices is (1) 256 (2) 512 (3) 1024 (4) 4096 6. A “necklace” is a circular string with several beads on it. It is allowed to rotate a necklace but not to turn it over. How many different necklaces can be made using 13 different beads? (1) 13! (2) 13!/2 (3) 12! (4) 12!/2 7. The number of ways of distributing 10 different books among 4 students (S 1 , S 2 , S 3 , S 4 ) such that S 1 and S 2 get 2 books each and S 3 and S 4 get 3 books each is (1) 12600 (2) 15200 (3) 10 C 4 (4) 10 ! 2 ! 2 3 3 ! ! ! (14) of (25) 9 . Each of two friends has 20 stamps and 10 postcards. We call an exchange fair if they exchange a stamp for a stamp or a postcard for a postcard. How many ways are there to carry out one fair exchange between these two friends? (1) 350 (2) 400 (3) 600 (4) 500 4 men and 3 women are to be seated in a row so that the w o m e n o c c u p y t h e e v e n p l a c e s . H o w m a n y s u c h arrangements are possible? (1) 7! (2) 144 (3) 30 (4) 6! 10. In how many ways can 4 men and 4 women be seated at a round table if each woman is to be between two men? (1) 288 (2) 144 (3) 1440 (4) 2880 11. By stringing together 9 different coloured beads, how many different bracelets can be made? (1) 20160 (2) 40320 (3) 80640 (4) 10080 12. How many parallelograms are formed by a set of 4 parallel lines intersecting another set of 7 parallel lines? (1) 121 (2) 139 (3) 115 (4) 126 13. In how many ways can 3 women be selected out of 15 women; if one particular woman is always included and two particular women are always excluded? (1) 66 (2) 77 (3) 88 (4) 99 14. In how many ways can a person choose 1 or more of 4 electrical appliances? (1) 10 (2) 12 (3) 14 (4) 15 15. A father has 2 apples and 3 pears. Each weekday (Monday through Friday) he gives one of the fruits to his daughter. In how many ways can this be done? (1) 120 (2) 10 (3) 24 (4) 12
IC : PTtkmml14
16. Getting more heads than the number of tails. (1) 2 (3) 5 8 (2) 7/8 (4) 1 2 DIRECTIONS: One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that 17. The card drawn is black. 1 4 23. The card drawn is neither a spade nor a king. (1) 0 (2) 9 13 (3) 1 2 (4) 4 13 24. The card drawn is neither an ace nor a king. (1) 11 13 (2) 1 2 (3) 2 13 (4) 11
26 (1) 1 2 (2) (3) 8 13 (4) can’t be determine 1 12 (2) 1 13 (1) 2 3 (2) 1 3 1 4 (4) 3 4 (3) 1 6 (4) 5 6 DIRECTIONS : A bag contains 8 red and 4 green balls. Find the probability that 18. The card drawn is a queen. 25. The ball drawn is red when one ball is selected at random. (1) (3) 19. The card drawn is black and a queen. (1) 1 13 (3) 1 26 (2) 1 52 (4) 5 6 26. All the 4 balls drawn are red when four balls are drawn at random. (1) 17
32 (2) 14
99 (3) 7 12 (4) None of these 20. The card drawn is either black or a queen. (1) 15
26 (2) 13 17 (3) 7 13 (4) 15
26 21. The card drawn is either king or a queen. (1) 5 26 (2) 1 13 (3) 2 13 (4) 12 13 22. The card drawn is either a heart, a queen or a king. (1) 17
52 (2) 21 52 (3) 19
52 (4) 9 26 IC : PTtkmml14 27. All the 4 balls drawn are green when four balls are drawn at random. (1) 1
495 (2) 7 99 (3) 5 12 (4) 2 3 28. Two balls are red and one ball is green when three balls are drawn at random. (1) 56
99 (2) 112 495 (3) 78 495 (4) None of these
(15) of (25) 29. Three balls are drawn and none of them is red. (1) (3) 68 99 (2) 4
495 (4) 30. At least one balls are green when two balls are drawn at random 7 99 (1) 1 4 (2) 19
33 (3) 2 3 (4) None of these
None of these Please make sure that you mark the answers in this score­sheet with an HB pencil. The marking of answers must be done in the stipulated time for the test. Do not take extra time over and above the time limit. S C O R E S H E E T 1 2 3 4 5 6 (16) of (25) 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 7 8 9 10 11 12 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 13 14 15 16 17 18 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 19 20 21 22 23 24 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 25 26 27 28 29 30 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 IC : PTtkmml14 PRACTICE EXERCISE # 02
The following exercise is to be taken by the students as a homework assignment, only after going through the theory provided. After having solved this exercise, a sincere student shall stand to gain a lot of speed and also will reinforce the concepts studied earlier. For all questions, mark the answers in the score sheet that is provided AT THE END. Please fill all ovals properly with an HB pencil. Scratch work (if any) has to be done on test paper itself.
Sug gested Ti me : 25 mi n. max.
1 . 2 . 3 . 4 . 5 . 6 . 7. 8 . How many even numbers less than 10,000 can be formed with the digits 3, 5,7,8,9 without any repetition? 9 . Wh at a re the pos sib le v alu es o f n whi ch wil l ma ke 13 C < 13 C n n+2 ? (1) 32 (2) 16 (1) 4 (2) 3 (3) 44 (4) 41 (3) 5 (4) 2 Three men have 4 coats, 5 waist coats and 6 caps. In how many ways can they wear them? (1) 17280 (2) 172800 (3) 1728000 (4) 1728 How many different words can be formed of the letters of the word “MATHEMATICS ”, so that no two vowels are together? (1) 1058400 (2) 105840 (3) 10584000 (4) None of these In the above question if the relative position of the vowels and consonant remains unaltered then how many different words can be formed? (1) 15120 (2) 151200 (3) 15200 (4) 15210 How many 6 digit number can be formed from the digits 1, 2, 3, 4, 5, 6 which are divisible by 4 and digits are not repeated? (1) 192 (2) 122 (3) 140 (4) 242 There are three coplanar parallel lines. If any p points are taken on each of the lines, then find the maximum number of triangles with the vertices of these points. 10. A man has 3 daughters and 4 identical doughnuts to distribute among them. In how many ways can he distribute these doughnuts without making parts of any one of them and each daughter getting atleast one? (1) 3! (2) 3! × 4! (3) 3 (4) 4 11. On a plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B, and no two are parallel, then the number of intersection points the lines have is equal to (1) 535 (2) 601 (3) 728 (4) 963 12. How many nine digit numbers have an even sum of their digits? (1) 9! – 6! (2) 14 × 10 7 (3) 45 × 10 7 (4) 10 9 2 13. The letters of the word SURITI are written in all possible orders and these words are written in alphabetical order. Then the rank of the word SURITI is (1) p 2 (4p – 3) (2) p 3 (4p – 3) (1) 236 (2) 245 (3) p (4p – 3) (4) p 3 (3) 307 (4) 315 How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together? (1) 216 (2) 45360 (3) 1260 (4) 43200 If all permutations of the letters of the word AGAIN are arranged as per dictionary, then find the fiftieth word of the arrangement. 14. The number of five digit telephone numbers having at least one of their digits repeated is (1) 90000 (2) 100000 (3) 30240 (4) 62784 15. Find the number of words of 5 letters such that each can be formed with the letters of the word “CHROMATE”. (These words need not have meaning) if each letter may be repeated in any arrangement. (1) NAAGI (2) NAIAG (1) 262144 (2) 4096 (3) NAAIG (4) NAGAI (3) 1024 (4) 32768
IC : PTtkmml14
(17) of (25) 16. A natural number is chosen at random from amongst the first 300. What is the probability that the number, so chosen is divisible by 3 or 5? (1) 48 515 (2) 4 150 (3) 1/2 (4) None of these 17. A natural number is chosen at random from the first 100 natural numbers. What is the probability that the number chosen is a multiple of 2 or 3 or 5? (1) (3) 30 100 74 100 (2) (4) 1 33 7 10 18. A box contains 5 red balls, 8 green balls and 10 pink balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green ? (1) 13
23 (2) 10
23 (3) 11
23 (4) 13 529 19. A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective? (1) 136 345 (3) 316 435 (2) 17
87 (4) 158 435 20. In a class 40% of the students offered Physics 20% offered Chemistry and 5% offered both. If a student is selected at random, find the probability that he has offered Physics or Chemistry only. (1) 45% (2) 55% (3) 36% (4) None of these 21. The probability that an MBA aspirant will join IIM is he will join XLRI is 1 . Find the probability that he will join 3 IIM or XLRI. (1) 4 15 (2) 7 15 (3) 11 15 (4) 8 15 (18) of (25) 2 and 5 22. In a given race, the odds in favour of horses H 1 , H 2 , H 3 and H 4 are 1:2, 1:3, 1:4, 1:5 respectively. Find the probability that one of them wins the race. (1) 57
60 (2) 1 20 (3) 2 7 (4) 7 60 23. Two balls are drawn from a bag containing 2 white, 3 red and 4 black balls one by one without replacement. What is the probability that atleast one ball is red? (1) 7 12 (2) 5 12 (3) 3 10 (4) None of these 24. A bag contains 6 red and 9 blue balls. Two successive drawing of four balls are made such that the balls are not replaced before the second draw. Find the probability that the first draw gives 4 red balls and second draw gives 4 blue balls. (1) 3
715 (2) 7
715 (3) 15 233 (4) None of these 25. An urn contains 4 white 6 black and 8 red balls. If 3 balls are drawn one by one without replacement, find the probability of getting all white balls. (1) 5 204 (2) 1 204 (3) 13/204 (4) None of these 26. Two numbers are selected at random from the integers 1 through 9. If the sum is even, find the probability that both numbers are odd. (1) 5/ 8 (2) 3/8 (3) 3/10 (4) None of these 27. A box contains 25 tickets, numbered 1, 2, 3, ..... 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show odd numbers. (1) 37
50 (2) 13
50 (3) 13 / 25 (4) None of these
IC : PTtkmml14 28. The probability that A can solve a problem is 2 and the probability 3 30. The probabilities of A, B, C solving a problem are 3 . Find the probability that 5 atleast one of A and B will be able to solve the problem. 3 respectively. If all the three try to solve the problem 8 simultaneously, find the probability that exactly one of them will solve it. that B can solve the same problem is (1) (3) 12 15 (2) 19
45 (4) 1 2 , and 3 7 13 15 none of these (1) 25 52 (2) 25 56 (3) 13/42 (4) None of these
29. In the previous question (no. 28), Find the probability that none of the two will be able to solve the problem. (1) 13 15 (2) 4 15 (3) 2 15 (4) 23 30 Please make sure that you mark the answers in this score­sheet with an HB pencil. The marking of answers must be done in the stipulated time for the test. Do not take extra time over and above the time limit. S C O R E S H E E T 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 IC : PTtkmml14 7 8 9 10 11 12 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 13 14 15 16 17 18 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 4 5 1 2 3 19 20 21 22 23 24 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 25 26 27 28 29 30 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 (19) of (25) Solutions Success is where
preparation and
opportunity meet...
Solved Examples 1 . Total number of persons = 5 + 4 = 9. 14. Since there is no restriction on Indians, we arrange the 5 Indian around a table in (5 – 1)! = 4!, ways. If the American sit at the places indicated by ‘x’ then no two Americans will sit together. No. of tickets numbered such that it is divisible by 20 are Now there are 5 places for 4 Americans. 10000 = 500. 20 \ required number = 4! × 5 P 4 = 4! × 5!. 15. 2 . Total number of ways = 3 6 = 729. 3 . First prize can be given away to 4 boys in 4 ways Let P(A) = Since events are independent and we have to find] P(AÈ B) = P(A) + P(B) – P(A) . P(B) Hence the required number of way in which all the 5 prizes can be given away to 4 boys = 16. There are 8 indirect roads from B to C and 1 direct road i.e.,, highway from B to C. Also, one can go from B to A and then to C. There are 10 + 1 indirect and direct roads from B to A and then 5 roads from A to C. This can be done in 11 × 5 = 55 ways. 1 1 1 1 3 1 14 14 +
- ´
=
=
=
´ 100 = 28% . 5 10 5 10 10 50 50 50 T h e pr ob ab i l i t i e s of s tu de nts n ot s ol v i ng t he p ro bl em a re 1 - 1 1 2 1 3 1 4 = 1 - = , 1 - = and 1 - = 3 3 3 4 4 5 5 Therefore the probability that the problem is not solved by any one of them \ In all one can go 55 + 8 + 1 = 64 ways. 5 . 20 1 10 1 = , P(B) = = 100 5 100 10 Similarly second, third, fourth and fifth prizes can also be given away to four boys in 4 ways. = 4 × 4 × 4 × 4 × 4 = 1024 4 . 500 1 = . 10000 20 Hence required probability = \ 4 Americans can be seated in 5 P 4 ways. = The man has to select some or all of his 6 friends 2 3 4 2 ´ ´ = 3 4 5 5 \ The number of ways to do the same is given by 6 C + 6 C + 6 C + 6 C + 6 C + 6 C = 2 6 – 1 = 63. 1 2 3 4 5 6 6 . Number of combinations are 7 C 5 × 15 C 1 × 7 C 1 = 2205. 7 . For arranging the 7 letters keeping P and Q always together, we have to view P and Q as one letter. Let this be denoted by PQ. Number of ways both the tickets are blank = 36 C 2 Then, we have to arrange the letters M, N, O, P, Q, R and S in a linear arrangement. Here, it is like arranging 6 letters in 6 places (since 2 letters are counted as one). This can be done in 6! ways. Thus the probability of not winning the prize = However, the solution is not complete at this point of time since in the count of 6!, the internal arrangement between P and Q is neglected. This can be done in 2! ways. Hence the required answer is 6! × 2! = 1440. Hence the probability of winning the prize = 1 – 8 . If we first put 4 Indians around the round table. We can do this in 3! ways. Once the 4 Indians are placed around the round table, we have to place the four Nepalese around the same round table. Now, since the Indians are already placed we can do this in 4! ways (as the starting point is defined when we put the Indians. Try to visualise this around a circle for placing 2 Indians and 2 Nepalese). Hence, the answer is 3! × 4! = 144. 9 . 150 – 1 = 149. This is so because all members except one lose a game. 10. Total number of arrangements = 8! = 40320 17. 18. C 3 + 7 C 3 + 4 C 3 1 + 35 + 4 40 10 =
= = . 14 × 13 × 2 14 × 26 91 12. Required probability = 13. Total number of ways = 9 C 4 , 2 children are chosen in 4 C 2 ways and other 2 persons are chosen in 5 C 2 ways. 14 C 3 4 Hence required probability = 5 C 2 ´ C 2 9 C 4 10 = . 21 C 2 50 C 2 = 18
35 18 17
= . 35 35 Total number of ways to form the numbers of three digit with 1, 2, 3 and 4 are 4 P = 4! = 24 3 If the numbers are divisible by three then their sum of digits must be 3, 6 or 9 But sum 3 is impossible. Then for sum 6, digits are 1, 2, 3 Number of ways = 3! Similarly for sum 9, digits are 2, 3, 4. Number of ways = 3! Thus number of favourable ways = 3! + 3! 1 3 ! + 3 ! 12
= = . 24 2 4 ! Hence required probability = 19. Total no. of ways = 10 ! 2 ! Favourable number of ways for T come together is 9! 1 1 Hence the required probability = 3 ! = 6 . 3 In 50 tickets 14 are of prize and 36 are blank. 36 Nu mbe r of arr ang eme nt s w hen be s t an d w ors t pap ers ar e t og eth er = 7! × 2! = 10080 Number of arrangements in which best and worst papers are not together = 40320 – 10080 = 3024011. Total no. of ways placing 3 letters in three envelops = 3 !, out of these ways only one way is correct. 2 3 = . 5 5 Hence the probability that problem is solved = 1 -
Thus probability that T come together is = 20. 9 ! ´ 2 ! 2 1 1 4 =
= Hence required probability = 1 – = . 10 ! 10 5 5 5 We have the following pattern ––– ( 1 ) red, white P(A) = ( 3 ) blue, white P(C) = 3 ´ 4 12 C 2 ( 2 ) red, blue P(B) = 3 ´ 5 12 C 2 4 ´ 5 12 C 2 Since all these cases are exclusive, so the required probability = (20) of (25) c 12 + 15 + 20 h = c 47 ´ 2 h = 47 . 12 C 2 12 ´ 11 66 IC : PTtkmml14 Solutions Practice
makes perfect! 1 . Practice Exercise # 01 There are two E's and two N's and they are bund l ed. They could be arranged among themselves in 4!/(2! 2!) = 6 ways. Besides, the remaining 4 letters and the bundle of E's and N's can be arranged in 5! ways. \ Required number of arrangements = 6 (5!) = 6!. Ans.(2) 2 . ( 1 ) Suppose we take only one Doctor. 10. Consider that the men are seated first. Then the men can be arranged in 3! ways and the women in 4! ways. Required number of circular arrangements = 3!4! = 144. Ans.(2) 11. There are 8! arrangements of the beads on the bracelet, but half of these can be obtained from the other half simply by turning the bracelet over. Hence there are ½(8!) = 20160 different bracelets. Ans.(1) 12. Number of selections of Doctor Each combination of 2 lines out of 4 can intersect each combination of 2 lines out of 7 to form a parallelogram. Number of parallelograms = 4 C 2 × 7 C 2 = 6 × 21 = 126. Ans.(4) = 3 C 1 = 3. Also number of selections of Engineers = 8 C 3 = 8! / (3! × 5!) Number of ways = 15 -1 - 2 C 3 - 1 =
14. number of ways of selecting Doctors = 3 C 2 = 3. Number of ways of selecting Engineers= 8 C 2 = (8 × 7)/2 = 28. Each appliance may be dealt with in 2 ways, as it can be chosen or not chosen. Since each of the 2 ways of dealing with an appliance is associated with 2 ways of dealing with each of the other appliances, the number of ways of dealing with the 4 appliances Number of parties = 28 × 3 = 84 = 2 × 2 × 2 × 2 = 2 4 ways. When we take 3 Doctors and one Engineer, But 2 4 ways included the case in which no appliance is chosen. Þ Number of ways of selecting Doctors = 3 C 3 = 1. Þ Number of ways of selecting Engineer = 8 C 1 = 8 Hence the required number of ways = 2 4 – 1 = 16 – 1 = 15 ways.
Þ Number of parties = 8. Hence the required number of ways = 4 C 1 + 4 C 2 + 4 C 3 + 4 C 4 Total number of parties = 168 + 84 + 8 = 260. Ans.(4)
= 4 + 6 + 4 + 1 = 15 ways. Ans.(4) Þ Number of parties = 56 × 3 = 168 ( 2 ) ( 3 ) When we select 2 Doctors and 2 Engineers, Short­cut: The appliances may be chosen singly, in twos, etc. Short­cut: Total number of ways having at least one Doctor = Total number of ways – all are Engineers = 11 C 4 – 8 C 4 = 260. 3 . Required number of ways = 16. E = {HHH, HHT, HTH, THH} Þ n (E) = 4 Such numbers will be 3! = 6. Þ (24 – 6) = 18 numbers can be formed which will be of four digits. Ans.(3) Every number between 3000 and 4000, which is divisible by five and which can be formed by the given digits, must contains 5 in unit’s place and 3 in thousand’s plac e. Thus we are left with four digits out of which we are to place two digits between 3 and 5, which can be done in 4 P 2 = 12 ways. Hence 12 numbers can be formed. Ans.(2) 5 . 17. n (E) = 26 18. n (E) = 4 \ P (E) = 19. Henc e Then it is clear that we have 13! different necklaces. n (E) = 2 However, any arrangement of beads must be considered identical to those \ P (E) = 20. Short­cut: Total ways = (n – 1)! = (13 – 1)! = 12!. Ans.(3) 7 . Required number of ways = 8 . Since any two stamps can be exc hanged so there are 20 × 20 ways to exchange the stamps and any two postcards can be exchanged so there are 10 × 10 ways to exchange the postcard. Hence total ways to exchange stamps and postcards = 20 × 20 + 10 × 10 = 500. Ans.(4) 9 . Each arrangement of the men may be associated with each arrangement of the women. 2 1 = . Ans.(3) 52 26 There are 26 black cards (including two queens). Besides it there are two more queens (i n red colours) T hus n (E) = 26 + 2 = 28 10 ! . 2 2 3 3 ! ! ! ! (\ S 1 & S 2 get 2 books each and S 3 & S 4 get 3 books each). Ans.(4) 4 1 = . Ans.(2) 52 13 Since drawn card must be black so there are only two queens. It is first assumed that it is prohibited to rotate the necklace. 13 ! 12 that can be obtained from it by relation = 12 ! 13 26 1 = . Ans.(1) 52 2 \ P (E) = Since each question can be answered in 4 ways. = 4 × 4 × 4 × 4 × 4 = 4 5 = 1024. Ans.(3) 4 1 = . Ans.(4) 8 2 \ P (E) = So, the total number of ways of answering 5 questions 6 . 5 ! = 10 . Ans.(2) 2 3 ! ! 15. With 0, 1, 2, 3 we can form 4! = 24 numbers. But it includes the numbers in which 0 occurs in thousand’s place and they will be numbers only of 3 digits. 4 . 12 . 11 = 66 ways . Ans.(1) 2 13. = 8 × 7 × 6 /(3 × 2 × 1) = 56 \ P (E) = 21. 28
7 = . Ans.(3) 52 13 There are 4 kings and 4 queens \ E = K È Q \ n(E) = 4 + 4 = 8 \ P (E) = 8 2 = . Ans.(3)
52 13 The required number of arrangement = 4! × 3! = 144. Ans.(2)
IC : PTtkmml14 (21) of (25) 22. There are 13 hearts (including one queen and one king). Besides it there are 3 queens and 3 kings in remaining 3 suits each. 27. n (E) = 4 C 4 = 1 T hus n (E) = 13 + 3 + 3 = 19 \ P (E) = 23. n (S) = 12 C 4 = 495 \ P (E) = 19
. Ans.(3) 52 28. n (S) = 12 C 4 = 495 n (E) = 8 C 2 × 4 C 1 = 112 There are 13 spades (including one king). Besides there are 3 more kings in remaining 3 suits). \ P (E) = T hus n (E) = 13 + 3 = 16 29. 24. \ P (E) = 30. 4
. Ans. (3) 495 If no ball is green. n (S) = 12 C 2 = 66 \ n (E) = 52 – 8 = 44 n (E) = 8 C 2 = 28 44 11
= . Ans.(1) 52 13 n (S) = 8 + 4 = 12 \ P (E) = \ P (E) = 14
33 P ( E ) = 1 -
n (E) = 8 26. n (S) = 495 \ n (E) = 4 + 4 = 8 \ P (E) = 25. 36
9 = . Ans.(2) 52 13 There are 4 aces and 4 kings 112 . Ans.(2) 495 n (E) = 4 C 3 = 4 Henc e n (E) = 52 – 16 = 36 \ P (E) = 1
. Ans.(1) 495 14 19
= . Ans.(2)
33 33 8 2 = . Ans.(1) 12 3 n (S) = 12 C 4 = 495 n (E) = 8 C 4 = 70 \ P (E) = 70 14
= . Ans.(2) 495 99 Objective key – Practice Exercise # 01 1.( 2) 2.( 4) 3.( 3) 4.( 2) 5.( 3) 6.( 3) 7.(4) 8.( 4) 9.( 2) 1 0. (2 ) 1 1. (1 ) 1 2. (4 ) 1 3. (1 ) 1 4. (4 ) 1 5. (2 ) 1 6. (4 ) 17.( 1) 1 8. (2 ) 1 9. (3 ) 2 0. (3 ) 2 1. (3 ) 2 2. (3 ) 2 3. (2 ) 2 4. (1 ) 2 5. (1 ) 2 6. (2 ) 27.( 1) 2 8. (2 ) 2 9. (3 ) 3 0. (2 )
(22) of (25) IC : PTtkmml14 Solutions Success is where
preparation and
opportunity meet...
Practice Exercise # 02
1.
Case 1. There is only one even number of one digit.
Case 2. 4 7 . 13 ! 3 !. 4 !. 2 !. 2 ! in which there are 2M’s, 2T’s, 2E’s. 1 There are only 4 even numbers of two digit T he number of ways i n whi c h 9 l etters c an be arranged. (Numbers are 38, 58, 78 and 98)
Case 3. 4 3 1 9 ! = 2 !. 2 !. 2 ! = 45360 Þ 4 × 3× 1 = 12 There are only 13 even numbers of 3 digits.
Case 4. 4 3 2 There are 4 vowels O, I, E, E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters 1 6 !
which include 2Ms and 2Ts and this be done in 2 ! ´ 2 ! = 180 ways. There are 24 even numbers of 4 digits Thus there are total 1 + 4 + 12 + 24 = 41 even numbers. Ans.(4) 2 . 3 . The total number of ways in which three men can wear 4 coats is the number of arrangements of 4 different coats taken 3 at a time. So, three men can wear 4 coats in 4 P 3 ways. Similarly, 5 waist coats and 6 caps can be worn by three men in 5 P 3 and 6 P 3 ways respectively. Hence, by the fundamental principle of counting, the required number of ways. In which of 180 ways, the 4 vowels O, I, E, E remaining together can be = 4 P 3 × 5 P 3 × 6 P 3 = 180 × 12 = 2160. = (4!) × (5 × 4 × 3) × (6 × 5 × 4) = 172800. Ans.(2) Hence, the required number of ways in which the four vowels do not come together = 45360 – 2160 = 43200. Ans.(4) The word MATHEMATICS contains 11 letters of which 4 are vowels A, A, E, I and 7 consonants. Let us first arrange the 7 consonants. This can be done arranged in in 4 ! = 12 ways 2 ! The number of ways in which the four vowels always come together 8 . 7 P 7 7 ! = ways. 2 ! ´ 2 ! 4 If all the arrangements are arranged as per dictionary then the first letter will start from A. Total words starting from A = 4! = 24. 25th word will start by G. Total words starting with G = 4!/2 = 12. If the v owels are placed at the places marked x , no two v owels will be together. 37th word will start by I. x M x T x T x M x H x C x S x Now 49th word will be NAAGI and the 50th word will be NAAIG. Ans.(3) 8 So the four vowels can be arranged at these eight places in 4 . There are total 9 letters in the word COMMITTEE P 4 ways. 2 ! Total word starting from I = 4!/2! = 12. 9 . 13 C < 13 C n n + 2 Þ
13 ! 13 ! 13 ! c 13 - n h ! n ! < c 13 - n - 2 hc ! n + 2 h! 13 ! 7 ! 8 P 4 ´ Total number of arrangements = = 1058400. Ans.(1) 4 2 ! Þ
c 13 - n h ! n ! < c 11 - n hc ! n + 2 h! Si nc e th e re l a ti v e p os i ti on o f th e c o ns on ant s an d v o we l s r em ai ns unchanged, we can arrange vowels at four places (2 nd , 5 th , 7 th and 9 th ) in Þ
c 13 - n hc 12 - n hc 11 - n h ! n ! < c 11 - n hc ! n + 2 hc n + 1 hn ! Þ
c 13 - n hc 12 - n h < c n + 2 hc n + 1 h
4 P 4 = 12 w ay s an d th e c o ns on an ts a t re ma i n i n g s e v e n pl ac es i n 2 ! 1 ! 1 1 ! 1
Þ (n + 2) (n + 1) < (13 – n) (12 – n) Þ n 2 + 3n + 2 < n 2 – 25n + 156 7 P 7 = 1260 ways. 2 ! ´ 2 ! Þ 3n + 2 < – 25n + 156 Þ 28n < 154 Þ n < 154 / 28 or n < 5.5. Hence the required number of ways =1260 × 12 = 15120. Ans.(1) 5 . 6 . Since we need integer value, so n £ 5. Ans.(3) For the numbers to be divisible by 4, the last two digits must be any of 12, 24, 16, 64, 32, 36, 56 and 52. The last two digit places can be filled in 8 ways. Remaining 3 places in 4C 3 × 3! ways. Hence no. of 5 digit nos. which are divisible by 4 are 24 × 8 = 192. Ans.(1) 10. The number of triangles with vertices on different lines p C 1 × p C 1 × p C 1 = p 3 11. Since each daughter has to get at least one doughnut, there will be only one doughnut left after distributing one to each. This doughnut can be given to any of the three daughter. So, number of ways of distributing = 3. Ans.(3) The number of triangles with 2 vertices on one line and the third vertex on any one of the other two lines = 3 C 1 ( p C 2 × 2p C 1 ) = 6p. c h = 3p (p – 1) p p -1 A 13 pass through A 2 11 Pass through B \ the required number of triangles = p 3 + 3p 2 (p – 1) = 4p 3 – 3p 2 = p 2 (4p – 3) \ number of intersection points 37 C 2 – 13 C 2 – 11 C 2 + 2 (The work “maximum” shows that no selection of points from each of the three li nes are collinear). Ans.(1)
(\ two points A & B) = 535. Ans.(1) IC : PTtkmml14 B 2 (23) of (25) 12. The first eight digit can be chosen arbitrarily. n (A Ç B Ç C) = There are 9 × 10 7 ways to do this. Then the last digit can always be chosen in exactly 5 ways. Required probability = P (A È B È C) (If the sum of the previous eight digits is odd, then we must choose an odd digit. Otherwise the last digit must be even) = P (A) + P (B) + P (C) – [P (A Ç B) + P (B Ç C) + P (A Ç C) + P (A Ç B Ç C) Hence the total ways = 9 × 10 7 × 5 = 45 × 10 7 . Ans.(3)
= Short­cut: Total nine digits number 9 × 10 8 . Sum of the digits will be either even or odd and the probability of getting even and odd sum are same. So, total numbers which has even sum of their 18. n (S) = 5 + 8 + 10 = 23 [n (A Ç B) = 0] \ P (A) = 5 8 and P (B) = 23 23 \ P (A È B) = P (A) + P (B) = The alphabetic order is I, R, S, T and U. 19. 5 ! R is = 60. 2 ! and (A Ç B) be the event of getting two non defective oranges. 20 4 ! 4 ! = 12 and ST is = 12. 2 ! 2 ! \ P (A) = C 2 30 C 2 22 , P (B) = Total arrangements of telephone numbers in which no digit is repeated = = Total ways – ways in which no digit is repeated. 16. n (S) = 300 P (A Ç B) = 20 So the telephone numbers having at least one digit repeated Number of words = 8 × 8 × 8 × 8 × 8 = 8 5 = 32768 words. Ans.(4) 20. C 2 30 C 2 21. C 2 = 316 . Ans.(3) 435 P (A) = 55 = 55%. Ans.(2) 100 2 1 , P (B) = 5 3 \ P (A È B) = 2 1 11
+ = 5 3 15 ( Q A and B are mutually exclusive events) . Ans.(3) 22. \ (A Ç B) be the event of getting a number divisible by both 2 and 3. P (H 1 ) = 1 1 1 1 , P (H 2 ) = , P (H 3 ) = , P (H 4 ) = 3 4 5 6 \ P (H 1 È H 2 È H 3 È H 4 ) = P (H 1 ) + P (H 2 ) + P (H 3 ) + P (H 4 ) \ (B Ç C) be the event of getting a number divisible by both by 3 and 5. \ (A Ç C) be the event of getting a number divisible by both 2 and 5. \ (A Ç B Ç C) be the event of getting a number divisible by A, B and C. C 2 30 40 20 5
+
- 100 100 100 P (A È B) = n (S) = 100 Now, n (A) = 50, n (B) = 33, n (C) = 20, n (A Ç B) = 16 C 2 15 – n (S) = 100 = 1 1 1 1 + = . Ans.(3) 3 5 30 2 Let A be the event of getting a number divisible by 2 and B be the event of getting a number divisible by 3 and C be the event of getting a number divisible by both 2 and 3. C 2 30 \ P (A È B) = P (A) + P (B) – P (A Ç B) = 5 100 1 60 1 20 1 = , P (B) = = , P (A Ç B) = =
\ P (A) = 300 3 300 5 600 30 17. C 2 22 + \ n (A) = 100, n (B) = 60, n (A Ç B) = 20 = C 2 30 n (A) = 40, n (B) = 20, n (A Ç B) = 5 Let A be the event of getting a number divisible by 3 and B be the event of getting a number divisible by 5 and (A Ç B) be event of getting a number divisible by 3 and 5 both \ P (A È B) = P (A) + P (B) – P (A Ç B) C 2 \ P (A È B) = P (A) + P (B) – P (A Ç B) = 9 × 9 × 8 × 7 × 6. = 90000 – 9 × 9 × 8 × 7 × 6 = 62784. Ans.(4) C 2 30 15 a n d Total ways of telephone numbers = 90000. 15. 5 8 13 +
= . Ans.(1) 23 23 23 n (S) = 20 C 2 We get 3 P 3 = 3! = 6 words beginning with SUI, after which come the words SURIIT and SURITI. Thus the rank of SURITI is 236. Ans.(1) 14. P (A Ç B) = 0 Let A be the event of getting two oranges and B be the event of getting two non defective fruits. The words beginning with SI is 4! = 24 and the number of words beginning with SR is I J
K n (B) = 8 6 ! The letters of SURITI can be arranged in = 360 ways. 2 ! The number of words beginning with I is 5! = 120 and those beginning with F G H 50 33 20 16 6
10 3
74 +
+
+
+ +
= . Ans.(3) 100 100 100 100 100 100 100 100 n (A) = 5 9 ´ 10 8 = 45 ´ 10 7 digit = 2 13. 3
100 = 23. 1 1 1 1 57
+ + + = . Ans.(1) 3 4 5 6 60 Let A be the event of not getting a red ball in first draw and B be the event of not getting a red ball in second draw. Then Required probability n (B Ç C) = 6, n (A Ç C) = 10, n (A Ç B Ç C) = 3 = Probability that atleast one ball is red 50 1 33 = , P (B) = \ P (A) = 100 2 100 = 1 – Probability that none is red P (C) = 20 1 16 = , n (A Ç B) = 100 5 100 n (B Ç C) = 6
10 , n (A Ç C) = 100 100 = 1 – P (A and B) = 1 – P (A Ç B) F G B I J = 1 – F G 2 ´ 5 I J = 7 H A K H 3 8 K 12 F B I J = 5 6 2 Here P (A) = = and P G H A K 8 9 3 ( ) × P = 1 - P A [There are 5 balls (excluding 3 red balls) after the selection of one non­red ball]. Ans.(1)
(24) of (25) IC : PTtkmml14 24. Let A be the event drawing 4 red ball in first draw and B be the event of drawing 4 blue balls in the second draw. 6 ( ) =
T he n P A C 4 15 F G B I J =
H A K P C 4 9 11 C 4 =
Required probability = P(A Ç B) = P(A) P 126 21
= 330 55 P(A) = F G B I J = 1 ´ 21 = 3 .
H A K 91 55 715 25. F G B I J = 12 = 1 H A K 24 2 Ans.(1) \ P Let A, B, C be the events of getting a white ball in first, second and third draw respectivel y then Hence, Required probability = Required probability = P (A Ç B Ç C) F G I J F G H K H I J K B C ( ) P P = P A A A Ç B 28. P (A) = 2 3 , P (B) = 3 5 be the events of solving the problem by A, B, C 4 2 = . 18 9 = P(A) + P(B) – P(A) P(B) When a white ball is drawing the first draw there are 17 balls left in the urn, out of which 3 are white. F G I J
H K = 29. B 3 \ P A = 17 2 3 2 3 13
+ - × = . Ans.(2) 3 5 3 5 15 P(none) = 1 – P (atleast one) = 1 – P (A È B) = 1 -
Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white. F C I = 2 = 1 G H A Ç B JK 16 8 30. 13 2 = . Ans.(3) 15 15 Let A, B, C respectively. , , C are the respective events of not solving the problem by then Then A B P Now P (A) = 2 3 1 1
Hence the required probability = ´ ´ =
Ans.(2) 9 17 8 204 26. 13 1 13
´ = . Ans.(2) 25 2 50 Required probability = P (A or B) = P(A È B) Now, P(A) = Probability of drawing a white ball in first draw = 13
, since there 13 odd numbers 1, 3, 5......25. 25 Since the ticket drawn in the first draw is not replaced, therefore second ti ck et drawn is from the remai ni ng 24 ti ckets , out of whic h 12 are odd numbered. Hence, the required probability = P (A Ç B) ( ) = P A P Let A be the event of drawing an odd numbered ticket in the first draw and B be the event of drawing an odd numbered ticket in the second draw. Then F G B I J H A K 15 1 = 1365 91 =
C 4 27. There are 4 even numbers and 5 odd numbers ( ) = \ P A 1 2 3 , P (B) = and P(C) = 3 7 8 2 5 5 ( ) = and P C ( ) = , P B 3 7 8 Let A = the event of choosing odd numbers \ The probability that exactly one of them will solve it B = the event of getting the sum of an even number. Then A Ç B = The event of choosing odd numbers whose sum is even. = {[A Ç (not B) Ç (not C)] or [(not A) Ç B Ç (not C)] or [(not A) Ç (not B) Ç C]} \ n (B) = 4 C 2 + 5 C 2 = 16 = P {( A Ç B Ç C ) È ( A Ç B Ç C ) È ( A Ç B Ç C )} a n d n(A Ç B) = 5 C 2 = 10 F A I ( ) G J \ Required probability = P A H B K = n A ( Ç B ) 10 5 =
= . Ans.(1) n( B ) 16 8 = P A ( Ç B Ç C ) + P A ( Ç B Ç C ) + P A ( Ç B Ç C ) [ Q A ( Ç B Ç C ) Ç ( A Ç B Ç C ) Ç P A ( Ç B Ç C ) = Æ = P A ( ) P B ( ) P C ( ) + P A ( ) × P B ( ) × P C ( ) + P A ( ) × P B ( ) × P C ( ) = = F G 1 ´ 5 ´ 5 I J + F G 2 ´ 2 ´ 5 I J + F G 2 ´ 5 ´ 3 I J H 3 7 8 K H 3 7 8 K H 3 7 8 K 25 5 5 25
+
+
= . Ans.(2)
168 42 28 56 Objective key – Practice Exercise # 02 1.( 4) 2.( 2) 3.( 1) 4.( 1) 5.( 1) 6.( 1) 7.(4) 8.( 3) 9.( 3) 1 0. (3 ) 1 1. (1 ) 1 2. (3 ) 1 3. (1 ) 1 4. (4 ) 1 5. (4 ) 1 6. (3 ) 17.( 3) 1 8. (1 ) 1 9. (3 ) 2 0. (2 ) 2 1. (3 ) 2 2. (1 ) 2 3. (1 ) 2 4. (1 ) 2 5. (2 ) 2 6. (1 ) 27.( 2) 2 8. (2 ) 2 9. (3 ) 3 0. (2 )
IC : PTtkmml14 (25) of (25) (26) of (25) IC : PTtkmml14