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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
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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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ 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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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