**Misc 10**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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 𝒒𝒏−𝒙 𝒑𝒙 n = number of coins tosses p = Probability of head = 12 q = 1 – p = 1 – 12 = 12 Hence, ⇒ P(X = x) = nCx 12𝑥 12𝑛−𝑥 P(X = x) = nCx 12𝑛 − 𝑥 + 𝑥 P(X = x) = nCx 𝟏𝟐𝒏 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 12𝑛> 90 % 1 − 1 2𝑛 > 90100 1 − 1 2𝑛 > 910 1 − 910 > 1 2𝑛 10 − 910 > 1 2𝑛 110 > 1 2𝑛 𝟐𝒏 > 10 We know that 24 = 16 > 10 So n ≥ 4.

About the Author

Davneet Singh

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.