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Example 24 - Find probability distribution of number of aces - Probability distribution

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  1. Chapter 13 Class 12th Probability
  2. Serial order wise
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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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ 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 𝟏﷮𝟏𝟑﷯﷯﷮𝒙﷯ 𝟏𝟐﷮𝟏𝟑﷯﷯﷮𝟐 − 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 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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