# Question 12 - Bernoulli Trial - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Bernoulli Trial

Question 1
Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Question 7 Important Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Question 9 Deleted for CBSE Board 2025 Exams

Question 10 Important Deleted for CBSE Board 2025 Exams

Question 11 Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams You are here

Question 13 Important Deleted for CBSE Board 2025 Exams

Question 14 (MCQ) Important Deleted for CBSE Board 2025 Exams

Question 15 (MCQ) Important Deleted for CBSE Board 2025 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 12 Find the probability of throwing at most 2 sixes in 6 throws of a single die. Let X : be the number six we get on 5 throws Throwing a pair of die is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx Where n = number of times die is thrown = 6 p = Probability of getting a six = 1 6 q = 1 1 6 = 5 6 Hence, P(X = x) = 6Cx We need to find probability of throwing at most 2 sixes in 6 throws of a single die. i.e. P(X 2) P(X 2) = P(X = 0) + P(X = 1) + P(X = 2) = 6C0 1 6 0 5 6 6 +6C1 1 6 1 5 6 5 +6C2 1 6 2 5 6 4 = 1 1 5 6 6 + 6 1 6 5 6 5 + 15 1 6 2 5 6 4 = 5 6 6 + 5 6 5 + 15 1 36 5 6 4 = 5 6 6 + 5 6 5 + 5 12 5 6 4 = 5 6 4 5 6 2 + 5 6 + 5 12 = 5 6 4 25 36 + 5 6 + 5 12 = 5 6 4 25 + 30 + 15 36 = 5 6 4 70 36 = So, the required Probability is 35 18 5 6 4