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Ex 13.5, 12 - Find probability of throwing at most 2 sixes in - Ex 13.5

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  1. Chapter 13 Class 12 Probability
  2. Serial order wise
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Ex 13.5, 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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ 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 𝟏﷮𝟔﷯﷯﷮𝒙﷯ 𝟓﷮𝟔﷯﷯﷮𝟔 − 𝒙﷯ 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﷯ ﷯ = 𝟑𝟓﷮𝟏𝟖﷯ 𝟓﷮𝟔﷯﷯﷮𝟒﷯ So, the required Probability is 35﷮18﷯ 5﷮6﷯﷯﷮4﷯

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