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Ex 13.5

Ex 13.5, 1
Deleted for CBSE Board 2023 Exams
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Ex 13.5, 2 Deleted for CBSE Board 2023 Exams

Ex 13.5, 3 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 4 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 5 Deleted for CBSE Board 2023 Exams

Ex 13.5, 6 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 7 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 8 Deleted for CBSE Board 2023 Exams

Ex 13.5, 9 Deleted for CBSE Board 2023 Exams

Ex 13.5, 10 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 11 Deleted for CBSE Board 2023 Exams

Ex 13.5, 12 Deleted for CBSE Board 2023 Exams

Ex 13.5, 13 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at Dec. 30, 2021 by Teachoo

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 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of throws = 6 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)^𝑥 (1/2)^(6−𝑥) = 6Cx (1/2)^(6 − 𝑥 + 𝑥) = 6Cx (𝟏/𝟐)^𝟔 Probability 5 successes Probability 5 success = P(X = 5) Putting x = 5 in (1) P(X = 5) = 6C5 (1/2)^6 = 6 × 1/64 = 𝟑/𝟑𝟐 (ii) Probability appearing at least 5 successes 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 = 𝟕/𝟔𝟒 (iii) Probability appearing at most 5 success 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) = 𝟔𝟑/𝟔𝟒