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Example 31 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
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Example 31 Important You are here
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Chapter 13 Class 12 Probability
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Transcript
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 headsIf a trial is Bernoulli, then There is finite number of trials They are independent Trial has 2 outcomes i.e. Probability success = P then Probability failure = q = 1 β P (4) Probability of success (P) is same for all trials Let X : Number of heads appearing Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx π^(πβπ) π^π n = number of coins tosses = 10 p = Probability of head = 1/2 q = 1 β p = 1 β 1/2 = 1/2 Hence, P(X = x) = 10Cx (1/2)^π₯ (1/2)^(10βπ₯) P(X = x) = 10Cx (1/2)^(10 β π₯ + π₯) P(X = x) = 10Cx (π/π)^ππ 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 β 6) ! Γ6 !) Γ (1/2)^10= (10 !)/(4 ! Γ 6 !) Γ 1/2^10 = 105/512 (ii) Probability appearing at least six heads i.e. P(X β₯ 6) P(X β₯ 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) = πππ/πππ (iii) Probability appearing at most six heads i.e. P(X β€ 6) P(X β€ 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) = ππ/ππ
