Bernoulli Trial

Chapter 13 Class 12 Probability
Serial order 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.