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The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is
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The correct answer is:
$8$
The probability of getting exactly 1 head in $n$ toss $=\frac{1}{2^n}$ the probability of getting exactly 1 head in $\mathrm{n}$ toss $=\frac{{ }^n C_1}{2^n}$ The probability of getting at least 2 head $=1-\frac{n+1}{2^n}$
Now, $1-\frac{n+1}{2^n}>0.94 \Rightarrow 0.04>\frac{1+n}{2^n}$
So $\mathrm{n}=8$
Now, $1-\frac{n+1}{2^n}>0.94 \Rightarrow 0.04>\frac{1+n}{2^n}$
So $\mathrm{n}=8$
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