Ex 13.5, 1 - A die is thrown 6 times. If 'getting odd number' - Ex 13.5

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  1. Chapter 13 Class 12 Probability
  2. Serial order wise
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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﷯ n = number of throws = 10 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﷯ P(X = x) = 6Cx 1﷮2﷯﷯﷮6 − 𝑥 + 𝑥﷯ P(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 succeses 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﷯

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