Question

1. How many times must a man toss a fair coin so that the probability of having atleast one one head is more than 90%

Answer 1

In any fair coin toss, P (getting a head) = P (getting a tail) →p=q=12→p=q=12
We need to find nn such that the probability of getting at least one head is more than 90%
P(X≥1)=1−P(X<1)>90P(X≥1)=1−P(X<1)>90%
⇒1−P(X=0)>910⇒1−P(X=0)>910→P(X=0)<1−910→P(X=0)<1−910→P(X=0)<110→P(X=0)<110
For a bionomial distribution,P(X=0)=nC012012n−0P(X=0)=nC012012n−0=(12)n=(12)n
⇒(12)n<110⇒(12)n<110→2n>10→2n>10
Since 21=2,23=8,24=1621=2,23=8,24=16, the minimum value for nn that satifies the inequality is n=4n=4, i.e, the coin should be tossed 4 or more times.