# Question 5 - Miscellaneous - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 2 (i) Important

Misc 2 (ii)

Misc 3

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9 Important

Misc 10 Important

Misc 11 (MCQ) Important

Misc 12 (MCQ)

Misc 13 (MCQ)

Question 1 Important

Question 2 Important

Question 3 Important

Question 4

Question 5 Important You are here

Question 6 Important

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 by Teachoo

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.