Question

How many times do you need to toss a fair coin to get 100 heads with probability 90%?

Answer 1

Let we toss a coin n times-

Probability of getting 1 heads $= ^nC_1(\dfrac{1}{2})^1)\dfrac{1}{2})^{n-1}$

$= n\dfrac{1}{2^n}$

So Probability of getting 100 head $=\dfrac{100n}{2^n}$

As per Question-

$\dfrac{100n}{2^n}=0.9\\[9pt] 100n=9.2^n$

At n=9, The condition is satisfied.

So, The number of times we have to toss is 9.