Example 31
Last updated at March 11, 2017 by Teachoo
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 heads Let X : Number of heads appearing Coin toss is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 n = number of coins tosses = 10 p = Probability of head = 12 q = 1 – p = 1 – 12 = 12 Hence, ⇒ P(X = x) = 10Cx 12𝑥 1210−𝑥 P(X = x) = 10Cx 1210 − 𝑥 + 𝑥 P(X = x) = 10Cx 𝟏𝟐𝟏𝟎 • Probability exactly six heads Probability exactly six heads = P(X = 6) Putting x = 6 in (1) P(X = 6) = 10C6 1210 = 10 ! 10 − 6 ! ×6 ! × 1210= 10 !4 ! × 6 ! × 1 210= 105512 (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 1210 + 10C7 1210 + 10C8 1210 + 10C9 1210 + 10C10 1210 = 1210(10C6 + 10C7 + 10C8 + 10C9 + 10C10) = 1210(210 + 120 + 45 + 10 + 1) = 1210(386) = 193512 (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 1210 + 10C5 1210 + 10C4 1210 + 10C3 1210 + 10C2 1210 + 10C1 1210+ 10C0 1210 = 1210(10C6 + 10C5 + 10C4 + 10C3 + 10C2 + 10C1 + 10C0) = 1210(210 + 252 + 210 + 120 + 45 + 10 + 1) = 1210(848) = 5364
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Chapter 13 Class 12 Probability
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.