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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
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B
6
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C
5
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D
3
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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.

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