Question

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

• A. $3$
• B. $>3$
• C. $<3$
• D. none of these

Answer 1

The correct option is B $>3$

In any fair coin toss, P (getting a head) = P (getting a tail) i.e., p=q=$\frac{1}{2}$

We need to find n such that the probability of getting at least one head is more than $80\%$

$P(X\ge 1)=1-P(X<1)>80\%$

$⟹1-P(X=0)>\frac{8}{10}⟹P(X=0)<1-\frac{8}{10}⟹P(X=0)<\frac{2}{10}orP(X=0)<\frac{1}{5}$

For a bionomial distribution, $P(X=0){=}^n{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{n-0}={\left(\frac{1}{2}\right)}^n$

$⟹{\left(\frac{1}{2}\right)}^n<\frac{1}{5}⟹2^n>5$

Since $2^1=2,2^2=4,2^3=8,2^4=16$, the minimum value for n that satisfies the inequality is $n=3$, i.e, the coin should be tossed $3$ or more times.