    1. Chapter 13 Class 12 Probability
2. Serial order wise
3. Ex 13.5

Transcript

Ex 13.5, 1 A die is thrown 6 times. If getting an odd number is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes? Let X : Number of times we get odd numbers in 6 throws of die Throwing a die is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx n = number of throws = 10 p = Probability of getting an odd number = 3 6 = 1 2 q = 1 p = 1 1 2 = 1 2 Hence, P(X = x) = 6Cx 1 2 1 2 6 P(X = x) = 6Cx 1 2 6 + P(X = x) = 6Cx Probability 5 successes Probability 5 success = P(X = 5) Putting x = 5 in (1) P(X = 5) = 6C5 1 2 6 = 6 1 64 = (ii) Probability appearing at least 5 succeses i.e. P(X 5) P(X 5) = P(X = 5) + P(X = 6) = 6C5 1 2 6 + 6C6 1 2 6 = 6 1 2 6 + 1 2 6 = 7 1 2 6 = (iii) Probability appearing at most 5 succeses i.e. P(X 5) P(X 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 6C0 1 2 6 + 6C1 1 2 6 + 6C2 1 2 6 + 6C3 1 2 6 + 6C4 1 2 6 + 6C5 1 2 6 = 1 2 6 (6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5) = 1 2 10 (1 + 6 + 15 + 20 + 15 + 6) =

Ex 13.5 