# Ex 13.5, 1 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

Transcript

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 succeses 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

Ex 13.5

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Chapter 13 Class 12 Probability (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.