Misc 5 - An urn contains 25 balls of which 10 balls bear mark 'X' - Binomial Distribution

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  1. Chapter 13 Class 12 Probability
  2. Serial order wise
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Misc 5 An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear 'X' mark. (ii) not more than 2 will bear 'Y' mark. (iii) at least one ball will bear 'Y' mark. (iv) the number of balls with 'X' mark and 'Y' mark will be equal. Let X : Number of balls with mark ‘X’ Drawing a ball is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ n = number of balls drawn = 6 p = Probability of getting ball with ‘X’ mark = 10﷮25﷯ = 2﷮5﷯ q = 1 – p = 1 – 2﷮5﷯ = 3﷮5﷯ Hence, ⇒ P(X = x) = 6Cx 𝟐﷮𝟓﷯﷯﷮𝒙﷯ 𝟑﷮𝟓﷯﷯﷮𝟔 − 𝒙﷯ • Probability that all will bear 'X' mark. Probability all balls has ‘X’ mark = P(X = 6) Putting x = 6 in (1) P(X = 6) = 6C6 2﷮5﷯﷯﷮6﷯ 3﷮5﷯﷯﷮6 −6﷯ = 6C6 2﷮5﷯﷯﷮6﷯ 3﷮5﷯﷯﷮0﷯ = 1 × 2﷮5﷯﷯﷮6﷯ × 1 = 2﷮5﷯﷯﷮6﷯ • Probability that not more than 2 will bear 'Y' mark. P(not more than 2 bear Y) = P(6X, 0Y) + P(5X, 1Y) + P(4X, 2Y) = P(X = 6) + P(X = 5) + P(X = 4) = 6C6 2﷮5﷯﷯﷮6﷯ 3﷮5﷯﷯﷮6−6﷯+ 6C5 2﷮5﷯﷯﷮5﷯ 3﷮5﷯﷯﷮6−5﷯+6C4 2﷮5﷯﷯﷮4﷯ 3﷮5﷯﷯﷮6−4﷯ = 6C6 2﷮5﷯﷯﷮6﷯ 3﷮5﷯﷯﷮0﷯+ 6C5 2﷮5﷯﷯﷮5﷯ 3﷮5﷯﷯﷮1﷯+6C4 2﷮5﷯﷯﷮4﷯ 3﷮5﷯﷯﷮2﷯ = 1 × 2﷮5﷯﷯﷮6﷯× 3﷮5﷯﷯﷮0﷯+ 6 × 2﷮5﷯﷯﷮5﷯ 3﷮5﷯﷯+15 2﷮5﷯﷯﷮4﷯ 3﷮5﷯﷯﷮2﷯ = 2﷮4﷯﷯﷮4﷯ 2﷮5﷯﷯﷮2﷯+6 × 2﷮5﷯﷯ 3﷮5﷯﷯+15 3﷮5﷯﷯﷮2﷯﷯ = 2﷮4﷯﷯﷮4﷯ 4﷮25﷯+ 36﷮25﷯+ 135﷮25﷯﷯= 2﷮5﷯﷯﷮4﷯ 175﷮25﷯﷯ =7 2﷮5﷯﷯﷮4﷯ (iii) Probability that at least one ball will bear 'Y' mark. P(atleast one bears ‘Y’) = 1 – P(no balls bear ‘Y’) = 1 – P(all ball bears ‘X’) = 1 – P(X = 6) = 1 – 6C6 2﷮5﷯﷯﷮6﷯ 3﷮5﷯﷯﷮6−6﷯ = 1 – 6C6 2﷮5﷯﷯﷮6﷯ 3﷮5﷯﷯﷮0﷯ = 1 – 1 × 2﷮5﷯﷯﷮6﷯ × 1 = 1 – 2﷮5﷯﷯﷮6﷯ (iv) Probability that number of balls with 'X' mark and 'Y' mark will be equal. So, we will have 3X & 3Y balls P(X & Y marked balls are equal) = P(X = 3) = 6C3 2﷮5﷯﷯﷮3﷯ 3﷮5﷯﷯﷮6−3﷯ = 6C3 2﷮5﷯﷯﷮3﷯ 3﷮5﷯﷯﷮3﷯ = 6 × 5 × 4 × 3 !﷮3 ! × 3 × 2 × 1﷯ × 8﷮125﷯ × 27﷮125﷯ = 5 × 4 × 8﷮125﷯ × 27﷮125﷯ = 864﷮3125﷯

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