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Example 31 - If a fair coin is tossed 10 times, find probability - Binomial Distribution

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  1. Class 12
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Example 31 If a fair coin is tossed 10 times, find the probability of (i) exactly six heads (ii) at least six heads (iii) at most six heads Let X : Number of heads appearing Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ n = number of coins tosses = 10 p = Probability of head = 1﷮2﷯ q = 1 – p = 1 – 1﷮2﷯ = 1﷮2﷯ Hence, ⇒ P(X = x) = 10Cx 1﷮2﷯﷯﷮𝑥﷯ 1﷮2﷯﷯﷮10−𝑥﷯ P(X = x) = 10Cx 1﷮2﷯﷯﷮10 − 𝑥 + 𝑥﷯ P(X = x) = 10Cx 𝟏﷮𝟐﷯﷯﷮𝟏𝟎﷯ • Probability exactly six heads Probability exactly six heads = P(X = 6) Putting x = 6 in (1) P(X = 6) = 10C6 1﷮2﷯﷯﷮10﷯ = 10 !﷮ 10 − 6﷯ ! ×6 !﷯ × 1﷮2﷯﷯﷮10﷯= 10 !﷮4 ! × 6 !﷯ × 1﷮ 2﷮10﷯﷯= 105﷮512﷯ (ii) Probability appearing at least six heads i.e. P(X ≥ 6) P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 10C6 1﷮2﷯﷯﷮10﷯ + 10C7 1﷮2﷯﷯﷮10﷯ + 10C8 1﷮2﷯﷯﷮10﷯ + 10C9 1﷮2﷯﷯﷮10﷯ + 10C10 1﷮2﷯﷯﷮10﷯ = 1﷮2﷯﷯﷮10﷯(10C6 + 10C7 + 10C8 + 10C9 + 10C10) = 1﷮2﷯﷯﷮10﷯(210 + 120 + 45 + 10 + 1) = 1﷮2﷯﷯﷮10﷯(386) = 193﷮512﷯ (ii) Probability appearing at most six heads i.e. P(X ≤ 6) P(X ≤ 6) = P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 10C6 1﷮2﷯﷯﷮10﷯ + 10C5 1﷮2﷯﷯﷮10﷯ + 10C4 1﷮2﷯﷯﷮10﷯ + 10C3 1﷮2﷯﷯﷮10﷯ + 10C2 1﷮2﷯﷯﷮10﷯ + 10C1 1﷮2﷯﷯﷮10﷯+ 10C0 1﷮2﷯﷯﷮10﷯ = 1﷮2﷯﷯﷮10﷯(10C6 + 10C5 + 10C4 + 10C3 + 10C2 + 10C1 + 10C0) = 1﷮2﷯﷯﷮10﷯(210 + 252 + 210 + 120 + 45 + 10 + 1) = 1﷮2﷯﷯﷮10﷯(848) = 53﷮64﷯

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