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

Misc 10 - Chapter 13 Class 12 Probability - Part 2
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Question 5 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 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, 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)^𝑥 (1/2)^(𝑛−𝑥) P(X = x) = nCx (1/2)^(𝑛 − 𝑥 + 𝑥) P(X = x) = nCx (𝟏/𝟐)^𝒏 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)^𝑛> 90 % 1 − 1/2^𝑛 > 90/100 1 − 1/2^𝑛 > 9/10 1 − 9/10 > 1/2^𝑛 (10 − 9)/10 > 1/2^𝑛 1/10 > 1/2^𝑛 𝟐^𝒏 > 10 We know that 24 = 16 > 10 So, n ≥ 4.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.