maihuna wrote:How many times must a man toss a fair coin so that the probability of having
at least one head is more than 90%?

With a fair coin, probability of getting heads = Prob (tails) = .5

Prob (at least 1 head in n tries) = 1 - Prob ( all tails in n tries)
= 1 - (.5)^n

According to question

=> 1- (.5)^n >= .9
=> .5^n =< .1

=> (1/2)^n =< .1
=> 1 =< .1* (2^n)
=> 10 =< 2^n

So, n = 4 ... minimum integral value for which the above equation is satisfied.