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Example 24 - Two cards are drawn successively with replacement from 52

Example 24 - Chapter 13 Class 12 Probability - Part 2
Example 24 - Chapter 13 Class 12 Probability - Part 3
Example 24 - Chapter 13 Class 12 Probability - Part 4

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Example 24 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 Finding probabilities separately 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 There are 4 aces out of 52 So, P(ace) = 4/52 P(not ace) = 1 – P(ace) = 1 – 4/52 = 48/52 Finding probabilities separately 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 Ace 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.