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ID : ww-10-Probability [1] Grade 10 Probability For more such worksheets visit www.edugain.com Answer t he quest ions (1) T wo dice are rolled. What is the probability that the two numbers add up to a prime number? (2) From a deck of cards, Elizabeth took out one card at random. What is the probability that she got a prime number? (3) William is participating in a race. T he probability that he will come f irst in the race is 0.1. T he probability that he will come second in the race is 0.1. T he probability that he will come in 3rd is 0.6, and the probability that he will be 4th is 0.2. What is the probability that he will come in 2nd position or better in the race? (4) From among a group of 3 men, 3 women and 7 children, 2 individuals are selected randomly. What is the probability that exactly 1 among the chosen are children? (5) What is the probability that an integer in the set 1, 2, 3, ...100 is divisible by 2 and not divisible by 3? Choose correct answer(s) f rom given choice (6) T here are 3 pens belonging to 3 students. T he pens were put into a box, and each student pulls out an pen one af ter the other. What is the probability that each student gets his or her own pen? a. 3 b. 3 c. 1 1 6 d. 2 2 6 (7) Coin A is f lipped 3 times and coin B is f lipped 4 times. What is the probability that the number of heads obtained f rom f lipping the two coins is the same? a. 37 b. 130 c. 33 128 (C) 2016 Edugain (www.Edugain.com) 35 128 d. 39 128 Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [2] (8) What is the probability that a leap year will contain 53 Wednesdays and 52 T hursdays? a. 1 b. 7 366 1 c. 2 d. 366 1 52 (9) T he letters of the word PROBABILIT Y are rearranged in a random order. What is the probability that the letters have exactly 4 letters between them? a. 1 b. 7 110 c. 12 d. 110 8 105 (10) Richard selects 3 numbers randomly f rom the f ollowing set of 5 numbers 2, 11, 9, 3 and 7. He puts them in the f orm of a proper f raction of the type a b . c What is the probability that you will get a f raction greater than 3 a. 24 b. 30 c. 12 20 ? 21 24 36 d. 36 18 30 (11) A card is drawn f rom a shuf f led deck of 52 cards. What is the probability of getting a card greater than 4 but lesser than 9? a. 16 b. 52 c. 1 8 52 d. 52 4 52 (12) Patricia draws 3 cards out of a deck of 52 cards. What is the probability that she draws a Jack, a 4, and a 2? a. 4 b. 52 c. 3 52 (C) 2016 Edugain (www.Edugain.com) 64 22100 d. 256 22100 Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [3] (13) A bag contains 6 red balls, 6 blue balls, and 4 green balls. Kevin draws 2 balls out of the bag. What is the probability that he gets a green ball and a blue ball? a. 24 b. 120 c. 24 136 26 d. 120 24 127.5 (14) A tyre manuf acturer keeps the record of how much distance the tyres manuf actured by the company travel bef ore f ailing. T hey f ind the f ollowing data Distance traveled in kilometers Number of failing tyres Less than 5000 20 5000 to 10000 100 10001 to 20000 440 20001 to 40000 576 More than 40000 864 If Sarah buys a tyre f rom them, what is the probability that it will last more than 10000 kilometers? a. 1880 b. 2000 c. 1865 1842 2000 d. 2000 1854 2000 (15) In 1992, the meteorological of f ice predicted the weathers completely right f or the months of f ebruary and april, and completely wrong f or all the other months. What is the probability that the f orecast was wrong f or a given day that year? a. 308 b. 366 c. 59 365 307 d. 366 306 365 © 2016 Edugain (www.edugain.com). All Rights Reserved (C) 2016 Edugain (www.Edugain.com) Many more such worksheets can be generated at www.edugain.com Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [4] Answers (1) 15 36 Step 1 T he two dice that are rolled can show any of these values Dice 1 : 1, 2, 3, 4, 5, 6 Dice 2 : 1, 2, 3, 4, 5, 6 So we can get a total of 36 combinations between them (6 x 6) Step 2 If we take one value f rom the list of possible values f rom each Dice, we get numbers ranging f rom 2 (when both Dice show 1) to 12 (when both dice show 6). Let's enumerate the prime numbers between 2 and 12. T hey are 2, 3, 5, 7 and 11 We need to see in how many ways we can get each of these values Let's put the value rolled by the dice as (x,y), where x is the value rolled by Dice 1, and y the value rolled by Dice 2 - 2: T he only way to get this is when we roll (1,1). 1 possibility - 3: We can get this by (1,2) or (2,1). 2 possibilities. - 5: We can get this by (2,3), (3,2), (1,4) or (4,1). 4 possibilities. - 7: We can get this by (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6 possibilities. - 11: We can get this by (5,6), or (6,5). 2 Possibilities T his gives us a total of 1 + 2 + 4 + 6 + 2 = 15 possible ways to get a prime number Step 3 So the probability of getting the two numbers add up to a prime is 15 36 (2) 4 13 Step 1 T here are 52 cards in a deck, with 13 of each suits (diamonds, clubs, hearts and spades Step 2 Each of the 13 cards of a suits are the numbers 1 to 9, then 4 f ace cards, including the ace Step 3 In the numbers 1 to 9, only 2,3,5,7 are primes. So f our primes per suit, which gives a total of 16 prime cards Step 4 So there are 16 possible prime cards out of the total 52. T he probability that we pick a prime card is theref ore 16 52 (C) 2016 Edugain (www.Edugain.com) , which can be reduced to 4 13 Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [5] (3) 0.2 Step 1 We are looking f or the probability that he will come 2nd or better in the race T his is the probability that he will either win the race or be 2nd Since he can be f irst or second, the probabilities can be added Step 2 T he probability that he will be f irst is 0.1 T he probability that he will be in second place is 0.1 Step 3 Adding them, we get 0.1 + 0.1 = 0.2 (4) 42 78 Step 1 T he total number of people in the group is 13 Step 2 We can select 2 individuals f rom among them in 13 C 2 ways = 13 x (13-1) = 78 2x1 Step 3 Here we want the probability that exactly 1 children among these chosen T his means 1 of the chosen 2 are children, and 1 are either men or women Step 4 Now we need to f igure out the number of ways of choosing 1 children f rom among 7 children. T his is 7 C 1 = 7 Step 5 We also need to f ind out the ways of choosing 1 men/women f rom among 6 men/women T his is 6 C 1 = 6 Step 6 T he probability that exactly 1 among the chosen 2 are children is theref ore 6x7 78 (C) 2016 Edugain (www.Edugain.com) = 42 78 Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [6] (5) 17 50 Step 1 We know that, the integers divisible by 2 and 3 are also divisible by 6. T heref ore, the number of integers divisible by 2 and not divisible by 3 = T he number of integers divisible by 2 - T he number of integers divisible by 6 Step 2 T he number of integers in the set 1, 2, 3, ...100, divisible by 2 = quotient of 100/2 = 50 T he number of integers in the set 1, 2, 3, ...100, divisible by 6 = quotient of 100/6 = 16 Now, the number of integers in the set 1, 2, 3, ...100, divisible by 2 and not divisible by 3 = 50 - 16 = 34 Step 3 T he probability that an integer in the set 1, 2, 3, ...100 is divisible by 2 and not divisible by 3 T he integers in the set 1, 2, 3, ...100 is divisible by 2 and not divisible by 3 = T he total number of integers in the set 1, 2, 3, ...100 34 = 100 17 = 50 (6) b. 1 6 Step 1 We have 3 pens and 3 students. Let's f irst see in how many dif f erent ways can the pens be distributed among the students Step 2 We know that 3 objects can be distributed in 3 ! = 3 x 2 x...x 1 = 6 ways Step 3 Out of these 6 ways, there is only one distribution where each student got his or her own pen Step 4 So the probability of each student getting his or her own pen = 1 6 (7) b. 35 128 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [7] (8) a. 1 7 Step 1 T here are 366 days in a leap year Step 2 If we divide 366 by 7 (since there are seven days in a week), we will get an answer of 52, with a remainder of 2 T his means that a leap year will have 52 Sundays, 52 Mondays, 52 T uesdays, 52 Wednesdays, 52 T hursdays, 52 Fridays and 52 Saturdays. Apart f rom these there will be two other days. T his means that there will be two weekdays that occur 53 times. Step 3 T he two days could be (Sunday, Monday), (Monday, T uesday), (T uesday, Wednesday), (Wednesday, T hursday), (T hursday, Friday), (Friday, Saturday), or (Saturday, Sunday) - a total of seven combinations Step 4 Out of these seven combinations, only one of them has Wednesday but no T hursday Step 5 So the probability of either of those two days being a Wednesday is 1 7 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [8] (9) c. 12 110 Step 1 T he f irst step is f inding out all the possible arrangements of the letters of the word PROBABILIT Y. Note that we are not interested in "unique" arrangements (in some kind of problems we would be), but just the total possible arrangement Step 2 T he answer to this is that there are 11 letters in this word, and theref ore the letters can be arranged in 11! ways i.e. 11x10x9x...2x1 ways T o see this, take any of the letters - it can be in any of the possible 11 positions. For each of these position, a second letter can be in any of the other 10 positions (11 minus the one taken up by the f irst letter) So the two letters can appear in 11x10 combinations For each of these 110 combinations, a third letter can be in any of the remaining 9 places, and so on Step 3 Now we know the total number of arrangements (11!) possible, and need to look f or the possible arrangements where the letters P and L have exactly 4 letters between them) Step 4 T his can be seen by inspection If P is in the f irst position (f irst letter of the word), then L would need to be in position 6 (since there are 4 letters between them Similarly, if P is in the second position , then L would need to be in position 7 T here are 6 such positions, the last one having P in the 6th position and L being in the last position Step 5 T he same thing can be seen with L being bef ore P. T here are 6 such positions So the total number of positions f or the two letters where this condition is met is 12 Now we have f illed in 2 of the 11 letters with P and L in 12 ways T he remaining 9 letters can take any of the remaining 9 positions f or each of these Since there are no restriction on the remaining 9 letters, the number of possible arrangements of 9 letters in 9 positions is 9! So the total ways to rearrange all the letters so that P and L have exactly 4 letters between them is 12 x 9! Step 6 Now we can work out the probability of rearranging the letters of PROBABILIT Y so that P and L have exactly 4 letters between them P (arrangement) = Arrangements where the two letters have 4 letters between them = T otal possible arrangements of the letters of PROBABILIT Y 12 x 9! 11! = 12 110 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [9] (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [10] (10) d. 18 30 Step 1 We need to select 3 numbers out of the 5 given in order to get a f raction of the f orm a b c We are also told it is a proper f raction, so b should be greater than c Step 2 Let's put the integers in a sorted manner. We get 2, 3, 7, 9 and 11 Step 3 Now we need to see how many proper f ractions can be f ormed f rom them A proper f raction of the type a b has three integers, a whole number a, a numerator b, c and a denominator c Let's assume we use one of the integers in the list above as the whole number a We now can select b and c f rom the remaining 4 integers We can select 2 integers f rom 4 in 4C2 = 4x3 = 6 ways 2x1 For any pair we select, one will be greater than the other, and the smaller integer will f orm the numerator and the larger one the denominator - Note: that the other way won't work - if the numerator is larger than the denominator it is not a proper f raction So f or each of the 5 integers, if we select one as the whole number, we get 6 possible combinations of numerator and denominator that can f orm a proper f raction T his means there are 5x6 = 30 possible proper f ractions of the f orm a b that can be c f ormed f rom these 5 integers Step 4 Now we need to f igure out how many of these 30 f ractions are greater than 3 20 21 In gt, we see that the whole number is 3, the numerator is 20 and the denominator is 21 We see that the numerator 20 is larger than the largest number in the set of numbers given to us, and the denominator 21 is one larger than 20 T he implication of this is that 3 20 will be larger than any f raction that can be f ormed 21 f rom the set of numbers 2, 3, 7, 9 and 11 where the whole number of the f raction is 3 So we only need to count the f ractions that have the whole number greater than 3 T hese are the f ractions that will have the whole numbers as 7, 9 and 11 Step 5 Now, remember there are 30 proper f ractions you can f orm f rom this list of number (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [11] From our analysis above, we also saw that f or each number selected as the whole number, we can f orm 6 f ractions f rom this list of numbers Step 6 So using each of the numbers f rom 7, 9 and 11 as the whole number, we can f orm 6 proper f ractions T he total number of f ractions that can be f ormed using 7, 9 and 11 as the whole number = 6 x 3 = 18 Step 7 Out of the 30 f ractions, 18 will be greater than 3 20 21 Step 8 T he probability is theref ore = 18 30 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [12] (12) b. 64 22100 Step 1 T here are 4 Jacks in a deck of cards. At this point there are also 52 cards in the deck Step 2 T here are 4 4s in a deck of cards. Now there are 51 cards remaining in the deck Step 3 T here are 4 2s in a deck of cards. Now there are 50 cards remaining in the deck Step 4 You can select 3 cards without replacement f rom a deck in 52 C 3 ways = 52 x 51 x 50 3x2x1 Step 5 4x4x4 So the answer is 52 x 51 x 50 3x2x1 Step 6 Simplif ying we get 64 22100 (13) a. 24 120 Step 1 T here are a total of 6+6+4 = 16 balls in the bag Step 2 T he number of ways to pick out 2 balls f rom a set of 16 balls is 16 C2 = 16 x (16-1)/2 = 120 Step 3 T he number of ways to pick a green ball and a blue ball is obtained by multiplying the number of the balls of each of these colors. T his is 4 x 6 = 24 Step 4 T he probability that he gets a green ball and a blue ball is theref ore 24 120 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ww-10-Probability [13] (14) a. 1880 2000 Step 1 First we need to f ind the total number of tyres that are given here. Step 2 We add the number of tyres 20 + 100 + 440 + 576 + 864 = 2000. Step 3 T o f ind the probability that the tyre Sarah purchased would last more than 10000 kilometers, we need to add the number of tyres that lasted more than 10000 kilometers. T his is 440 + 576 + 864 = 1880. Step 4 T he probability that the tyre lasts more than 10000 km = 1880 . 2000 (15) c. 307 366 Step 1 T he key thing to note here is that 1992 is a leap year A leap year has 366 days Step 2 Now we need to f igure out the number of days in f ebruary and april f ebruary has 29 days and april has 30 days Step 3 Adding them together we get 29 + 30 = 59 days Step 4 So the f orecast was right f or 59 days and wrong f or 307 days Step 5 T he probability that the f orecast was wrong on a given day would theref ore be 307 366 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited