Binomial Distribution

Chapter 13 Class 12 Probability
Concept wise

### Transcript

Question 1 A die is thrown 6 times. If βgetting an odd numberβ 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 π^(πβπ) π^π Here, n = number of throws = 6 p = Probability of getting an odd number = 3/6 = 1/2 q = 1 β p = 1 β 1/2 = 1/2 Hence, P(X = x) = 6Cx (1/2)^π₯ (1/2)^(6βπ₯) = 6Cx (1/2)^(6 β π₯ + π₯) = 6Cx (π/π)^π Probability 5 successes Probability 5 success = P(X = 5) Putting x = 5 in (1) P(X = 5) = 6C5 (1/2)^6 = 6 Γ 1/64 = π/ππ (ii) Probability appearing at least 5 successes i.e. P(X β₯ 5) P(X β₯ 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 = π/ππ (iii) Probability appearing at most 5 success i.e. P(X β€ 5) P(X β€ 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) = ππ/ππ