Question

How many times must a man toss a fair coin, so that the probability of having atleast one head is more than 80%?

Answer 1

Let the required coin toses be n

Pr(at least one head in n tosses)>0.8 

                  or

Pr(0 heads in n tosses)=<0.2

Probability of 0 heads in n tosses is (1/2)ⁿ

(1/2)ⁿ =< 0.2 take log on both sides

nlog(0.5)=<log(0.2)

n*(-0.3010)=<(-0.6989) or n=>2.3219

∴ approximately it's 3