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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
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B
>3
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C
<3
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D
none of these
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Solution

The correct option is B $$>3$$
In any fair coin toss, P (getting a head) = P (getting a tail) i.e., p=q=$$\dfrac 12$$
We need to find n such that the probability of getting at least one head is more than $$80\%$$
$$P(X≥1)=1−P(X<1)>80\%$$
$$\implies 1−P(X=0)>\dfrac 8{10}\\\implies P(X=0)<1−\dfrac 8{10}\\\implies P(X=0)<\dfrac 2{10} or P(X=0)<\dfrac 15$$
For a bionomial distribution, $$P(X=0)=^nC_0\left(\dfrac 12\right)^0\left(\dfrac 12\right)^{n−0}=\left(\dfrac 12\right)^n$$
$$\implies \left(\dfrac 12\right)^n<\dfrac 1{5}\\\implies 2^n>5$$
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.

Maths

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