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

Solution

The correct option is D $$3$$Let p denote the probability that we will get the head, then $$p=\dfrac 12, q=1-p=\dfrac 12$$Suppose the coin is tossed n times. Let X denotes the number of times he get head in n-trials.$$\therefore P(X=r)=^nC_rp^rq^{n-r}\\\implies = ^nC_r \left(\dfrac 12\right)^r\left(\dfrac 12\right)^{n-r}=^nC_r\left(\dfrac 12\right)^n, r=0, 1, 2, ..., n$$Now,$$P(X\ge 1)>0.8\\\implies P(X=1)+P(X=2)+...+P(X=n)>\dfrac 8{10}\\\implies [P(X=0)+P(X=1)+...+P(X=n)]-P(X=0)>\dfrac 8{10}\\\implies 1-P(X=0)>\dfrac 8{10}\quad [\because \text{sum of probability is 1}]\\\implies P(X=0)<1-\dfrac 8{10}\\\implies P(X=0)<\dfrac 2{10}\\\implies ^nC_0\left(\dfrac 12\right)^n<\dfrac 15\\\implies \left(\dfrac 12\right)^n<\dfrac 15$$This is true only when the value of n is equal to or greater than $$3$$$$\therefore ^nC_0\left(\dfrac 12\right)^n<\dfrac 15\implies n=3, 4, 5, ...$$Hence the coin must be tossed at least $$3$$ times.Maths

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