# Example 24 - Chapter 13 Class 12 Probability

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 24 (Method 1) Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces. Picking a card is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx Here, n = number of cards drawn = 2 p = Probability of getting ace = 4 52 = 1 13 q = 1 p = 1 1 13 = 12 13 Hence, P(X = x) = 2Cx Since 2 cards are drawn, We can get 0 ace, 1 ace or 2 ace So, X can be 0, 1 , 2 Putting values in (1) P(X = 0) = 2C0 1 13 0 12 13 2 0 = 1 1 12 13 2 = 144 169 P(X = 1) = 2C1 1 13 1 12 13 2 1 = 2 1 13 12 13 = 24 169 P(X = 2) = 2C2 1 13 2 12 13 2 2 = 1 1 13 2 = 1 169 The probability distribution is Example 24 (Method 2) Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces. Let X : Number of aces We select two cards, So, we can select 2 Aces or 1 Aces or 0 Aces So, value of X can be 0, 1 & 2 There are 4 aces out of 52 So, P(ace) = 4 52 P(not ace) = 1 P(ace) = 1 4 52 = 48 52 For X = 0 Out of two cards, no ace is selected P(X = 0) = P(no ace) P(no ace) = 48 52 48 52 = 144 169 For X = 1 Out of two Cards, One ace is selected There can be two cases Aces is selected first, then no ace No ace is selected first, then ace P(X = 1) = P(ace) P(no ace) + P(no ace) P(ace) = 4 52 48 52 + 4 52 48 52 = 12 169 + 12 169 = 24 169 X = 2 We select two aces P(X = 2) = P(ace) P(ace) = 4 52 4 52 = 1 169 Thus Probability distribution is

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Example 24 You are here

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Chapter 13 Class 12 Probability

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.