Subscribe to our Youtube Channel - https://you.tube/teachoo

Last updated at Feb. 15, 2020 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 ๐^(๐โ๐) ๐^๐ Here, n = number of coins tosses p = Probability of head = 1/2 q = 1 โ p = 1 โ 1/2 = 1/2 Hence, P(X = x) = nCx (1/2)^๐ฅ (1/2)^(๐โ๐ฅ) P(X = x) = nCx (1/2)^(๐ โ ๐ฅ + ๐ฅ) P(X = x) = nCx (๐/๐)^๐ 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 โฅ 1) > 90%, we need to find n Now, P(X โฅ 1) > 90 % 1 โ P(X = 0) > 90 % 1 โ P(X = 0) > 90 % 1 โ nC0 (1/2)^๐> 90 % 1 โ 1/2^๐ > 90/100 1 โ 1/2^๐ > 9/10 1 โ 9/10 > 1/2^๐ (10 โ 9)/10 > 1/2^๐ 1/10 > 1/2^๐ ๐^๐ > 10 We know that 24 = 16 > 10 So, n โฅ 4.

Miscellaneous

Misc 1

Misc 2 Important

Misc 3

Misc 4 Not in Syllabus - CBSE Exams 2021

Misc 5 Important Not in Syllabus - CBSE Exams 2021

Misc 6 Important Not in Syllabus - CBSE Exams 2021

Misc 7 Important

Misc 8 Important

Misc 9 Important Not in Syllabus - CBSE Exams 2021

Misc 10 Important You are here

Misc 11 Important

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18

Misc 19

Chapter 13 Class 12 Probability

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.