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TutorBreeze.com 13. Probability (Random Variables and Binomial Distributions) QUESTION 1: Let X denote a random variable. The probability that X can take the values x, has the following form, where k is some unknown constant. Find the value of k. SOLUTION: x=0 0.2, kx, x = 1, 2 P(X = x) = x = 3, 4 k (6 − x), 0, otherwise P(X = 0) = 0.2 , P( X=1) = k , P ( X= 2) = 2k P (X = 3) =3k , P ( X= 4) =2 k , So 0.2 + k + 2k + 2k + 3k =1 i.e.8 k+0.2 = 1 8k = 0.8 k= 0.1 QUESTION 2: From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. SOLUTION: Let X = number of defective bulbs So, X = 0,1,2,3,4 Then p = 1/5 , q = 4/5 4 4 24 4 256 P( X = 0) = C0 p q = = = 30 5 625 4 0 4 Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com 3 256 1 4 P ( X = 1) = 4C1 p1 q 3 = 4 = 625 5 5 2 2 96 1 4 P ( X = 2) = 4C2 p 2 q 2 = 4 = 5 5 625 3 1 16 1 4 P ( X = 3) = C3 p q = 4 = 5 5 625 4 3 1 4 0 1 1 4 P ( X = 4) = 4C4 p 4 q 0 = 4 = 625 5 5 X 0 1 P(X) 256 256 625 625 2 3 4 96 625 16 625 1 625 QUESTION 3: Suppose X follows the Binomial distribution B(6,1/2)Which is the most likely outcome. SOLUTION: Here , n = 6 , p = ½ , q = ½ 6 1 1 1 P ( X = 0) = 6C0 p 0 q 6 = 6C0 = 6 = 64 2 2 1 5 1 6 1 1 P ( X = 1) = 6C1 p1q 5 = 6C1 = 6 6 = 2 64 2 2 2 4 2 4 1 1 6 × 5 1 1 15 P ( X = 2) = 6C2 p 2 q 4 = 6C2 = × = 2 2 1× 2 2 2 64 3 3 3 3 1 1 6 × 5 × 4 1 1 20 P ( X = 3) = 6C3 p 3q 3 = 6C3 = × = 2 2 1× 2 × 3 2 2 64 4 2 4 2 1 1 6 × 5 1 1 15 P ( X = 4) = 6C4 p 4 q 2 = 6C4 = × = 2 2 1× 2 2 2 64 5 1 5 1 6 1 1 1 1 P ( X = 5) = 6C5 p 5 q1 = 6C5 = 6 × = 2 2 2 2 64 6 6 1 1 1 P ( X = 6) = 6C6 p 6 q 0 = 6C6 = 1× = 2 2 64 20 Maximum Value = 64 ∴X = 3 is the most likely outcome. Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com QUESTION 4: A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes. SOLUTION: S = “throwing a Doublet in a single throw of a pair of dice”. F= “not throwing a Doublet in a single throw of a pair of dice”. p= 6 1 = 36 6 q = 1− p = 1− 1 5 = 6 6 3 125 5 P ( X = 0) = 3C0 p 0 q 3 = 1.q 3 = = 216 6 2 75 1 5 P ( X = 1) = 3C1 p1 q 2 = 3 p.q 2 = 3 = 216 6 6 2 1 5 15 P ( X = 2) = 3C2 p 2 q1 = 3 p 2 .q = 3 = 6 6 216 3 1 1 P ( X = 3) = 3C3 p 3 q 0 = .1. p 3 = = 216 6 X 0 1 2 3 P(X ) 125 216 75 216 15 216 1 216 QUESTION 5: The mean and variance of a Binomial distribution for a random variable are 10 and 5/3 respectively. Find P ( X≥1) SOLUTION: (q + p )n. np = 10 , …(1) npq = 5/3 ….(2) Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com q= 5 1 = 3 × 10 6 p = 1− q = 1− n= 1 5 = 6 6 10 10 × 6 = = 12 p 5 12 1 P ( X ≥ 1) = 1 − P ( X = 0) = 1 − 12C0 p 0 q12 = 1 − q12 = 1 − 6 QUESTION 6(i): You are taking a 10 question multiple choice test. If each question has four choices and you guess on each question, what is the probability of getting exactly 7 questions correct? SOLUTION: QUESTION 6(ii): In a box containing 100 bulbs, 10 are defective. The what is the probability that out of a sample of 5 bulbs, none is defective . SOLUTION: Here n= 5, p = 10/100 = 1/10, q = 9/10 P(0) = q 5 = (9/10)5 QUESTION 7: Find the mean of the Binomial distribution B(3, ¼) SOLUTION: : Let X be the random variable that follows the given binomial distribution. X can take the values 0,1,2,3 Here n = 3 , p = ¼ and q = 1- p = 1- ¼ = ¾ .P(x) = nCx px qn-x Substituting in this X = 0,1,2,3, we can get the respective probabilities. They work out to be P(X=0) = 27/64 , P(X=1) also equal to the same 27/64, P(X=2) =9/64 and P(X=3) = 1/64. Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com In the tabular form the Binomial distribution is : X P(X) 0 1 2 3 27/64 27/64 9/64 1/64 We know that the mean of a distribution is given by : µ = Σxi pi, Let us add one more row to our table and find the product of each x with its corresponding probability. X P(X) XP(X) 0 1 2 3 27/64 27/64 27/64 9/64 18/64 1/64 3/64 0 µ = Σxi pi, =48/ 64 =3/4,which is the mean QUESTION 8: State True or False with justification : The above distribution can not be the probability distribution of a random variable SOLUTION:False, because , the sum of all probabilities is one . QUESTION 9: Find the mean of the variable Y whose probability distribution is given graphically SOLUTION: Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com QUESTION 10: Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and standard deviation of the number of kings. SOLUTION: the probability distribution of X is MEAN VARIANCE Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com STANDARD DEVIATION QUESTION 11: . The chances of a patient having a heart attack is 40%. According to latest research Drug A reduces the risk of heart attack by 30% and drug B reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient had been prescribed Drug A SOLUTION: Let E1 be the event that a patient used Drug A∴P(E1) = ½ Let E2 be the event that a patient used Drug B∴P(E2 ) = ½ Let E be the event that a patient had a heart attack Required probability :P(E1/E) Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com P ( E / E1 ) = 40 30 28 1 − = 100 100 100 40 25 30 1 − = 100 100 100 P ( E / E1 ) P ( E1 ) P ( E1 / E ) = P ( E / E1 ) P ( E1 ) + P ( E / E2 ) P ( E2 ) P ( E / E1 ) = 28 1 × 14 100 2 = = 28 1 30 1 29 × + × 100 2 100 2 QUESTION 12: A random variable X has the following probability distribution , find (i) The value of K (ii) P( X ≤ 1) X P(X) 0 0.1 1 K 0 0.1 1 K 2 0.2 3 2K 4 0.3 5 K SOLUTION: X P(X) i) 2 0.2 3 2K The value of K:0.1 + K + 0.2 +2K + 0.3 + K = 1 ⇒ 4K = 0.4 ⇒K = 0.1 (ii) P( X≤ 1) = P(0) + P1) = 0.1 + 0. 1 = 0.2 Contact us for Online Tutoring at http://www.TutorBreeze.com 4 0.3 5 K TutorBreeze.com QUESTION 13: 27.A bag contains 25 balls of which 10 are purple and the remaining are pink . A ball is drawn at random , its colour is noted and it is replaced . 6 balls are drawn in this way, find the probability that (i) All balls were purple (ii) Not more than 2 were Pink (iii) Atleast one ball was pink (iv) An equal number of purple and pink balls were drawn. SOLUTION: This is a case of Bernoulli’s trials . Let Success : Getting a purple ball on a draw Failure : Getting a pink ball on a draw p = P( success ) = 10 2 3 = ⇒q= 25 5 5 6 6 2 2 (i ) P(6 success ) = 6C6 p 6 q 0 = 1× ×1 = 5 5 (ii ) P( not more than 2 failures ) = P( not less than 4 success) = P(4) + P(5) + P(6) 4 2 5 2 3 2 3 2 = 6C4 p 4 q 2 + 6C5 p 5 q1 + 6C6 p 6 q 0 = 15 × + 6 × + 1× 5 5 5 5 5 4 2 2 2 3 2 3 2 = 15 × + 6 × + 5 5 5 5 5 6 4 2 135 36 4 = + + 5 25 25 25 4 2 175 2 = = 7× 5 25 5 4 3 3 864 2 3 (iii )P(3success7failures) = P(3) = 6C3 p 3 q 3 = 15 = 5 5 3125 (iv) P (atleast 1 failure)=P( at most 5 success) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 1 - P(6)=1- 6C6 p 6 q 0 6 2 2 = 1 − 1× × 1 = 1 − 5 5 6 Contact us for Online Tutoring at http://www.TutorBreeze.com TutorBreeze.com Four cards are drawn successively with replacement from a well shuffled deck of 52 cards . What is the probability that (i) All the four cards are spades ? (ii) Only 3 cards are spades (iii) None is a spade QUESTION 14: SOLUTION: This is a case of bernoullis trial. p = P(Success) = P(getting a spade in a single draw ) = q = P( Failure) = 1 − p = 1 − 13 1 = 52 4 1 3 = 4 4 4 1 1 All the four cards are spades =P(X =4) = C4 p q = = 256 4 12 3 Only 3 cards are spades=P(X=3)= 4C3 p 3 q1 = = 256 64 4 (i) (ii) 4 0 4 81 3 None is a spade=P(X=0) = C0 p q = = 256 4 4 (iii) 0 4 QUESTION 15: Find the probability distribution of doublets in three throws of a pair of dice SOLUTION:. S = “throwing a Doublet in a single throw of a pair of dice”. F= “not throwing a Doublet in a single throw of a pair of dice” p= 6 1 = 36 6 q = 1− p = 1− 1 5 = 6 6 3 5 125 P ( X = 0) = 3C0 p 0 q 3 = 1.q 3 = = 6 216 2 75 1 5 P ( X = 1) = C1 p q = 3 p.q = 3 = 216 6 6 3 1 2 2 2 1 5 15 P ( X = 2) = C2 p q = 3 p .q = 3 = 6 6 216 3 2 1 2 3 1 1 P ( X = 3) = 3C3 p 3 q 0 = .1. p 3 = = 6 216 X 0 1 2 3 P(X ) 125 216 75 216 15 216 1 216 Contact us for Online Tutoring at http://www.TutorBreeze.com

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