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

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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 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 1 2 10 P(X = x) = 10Cx 1 2 10 + P(X = x) = 10Cx 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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