Misc 10 - How many times must a man toss a fair coin - Miscellaneous

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  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

Misc 10 How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%? 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 p = Probability of head = 1 2 q = 1 p = 1 1 2 = 1 2 Hence, P(X = x) = nCx 1 2 1 2 P(X = x) = nCx 1 2 + P(X = x) = nCx We need to find How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%? So, given P(X 1) > 90%, we need to find n Now, P(X 1) > 90 % 1 P(X = 0) > 90 % 1 P(X = 0) > 90 % 1 nC0 1 2 > 90 % 1 1 2 > 90 100 1 1 2 > 9 10 1 9 10 > 1 2 10 9 10 > 1 2 1 10 > 1 2 > 10 We know that 24 = 16 > 10 So n 4.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.