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Ex 13.5
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
Last updated at March 22, 2023 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