**Example 24 **

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 𝒒𝒏−𝒙 𝒑𝒙 Here, n = number of cards drawn = 2 p = Probability of getting ace = 452 = 113 q = 1 – p = 1 – 113 = 1213 Hence, ⇒ P(X = x) = 2Cx 𝟏𝟏𝟑𝒙 𝟏𝟐𝟏𝟑𝟐 − 𝒙 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 1130 12132 −0 = 1 × 1 × 12132 = 144169 P(X = 1) = 2C1 1131 12132 − 1 = 2 × 113 × 1213 = 24169 P(X = 2) = 2C2 1132 12132 − 2 = 1 × 1132 = 1169 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) = 452 P(not ace) = 1 – P(ace) = 1 – 452 = 4852 For X = 0 Out of two cards, no ace is selected P(X = 0) = P(no ace) × P(no ace) = 4852 × 4852 = 144169 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) = 452 × 4852+ 452 × 4852 = 12169+ 12169 = 24169 X = 2 We select two aces P(X = 2) = P(ace) × P(ace) = 452 × 452 = 1169 Thus Probability distribution is

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.