     1. Chapter 13 Class 12 Probability
2. Serial order wise
3. Miscellaneous

Transcript

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 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 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

Miscellaneous 