Question

The number of times a fair coin must be tossed so that the probability of getting at least one head is $0.8$ is-

• A. $7$
• B. $6$
• C. $5$
• D. $3$

Answer 1

The correct option is D $3$

Let p denote the probability that we will get the head, then $p=\frac{1}{2},q=1-p=\frac{1}{2}$

Suppose the coin is tossed n times. Let X denotes the number of times he get head in n-trials.

$\therefore P(X=r){=}^n{C}_r{p}^r{q}^{n-r}⟹{=}^n{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^n,r=0,1,2,...,n$

Now,

$P(X\ge 1)>0.8⟹P(X=1)+P(X=2)+...+P(X=n)>\frac{8}{10}⟹\left[P\right(X=0)+P(X=1)+...+P(X=n\left)\right]-P(X=0)>\frac{8}{10}⟹1-P(X=0)>\frac{8}{10}[\because \text{sum of probability is 1}]⟹P(X=0)<1-\frac{8}{10}⟹P(X=0)<\frac{2}{10}⟹{}^n{C}_0{\left(\frac{1}{2}\right)}^n<\frac{1}{5}⟹{\left(\frac{1}{2}\right)}^n<\frac{1}{5}$

This is true only when the value of n is equal to or greater than $3$

${\therefore }^n{C}_0{\left(\frac{1}{2}\right)}^n<\frac{1}{5}⟹n=3,4,5,...$

Hence the coin must be tossed at least $3$ times.