The minimum number of times a fair coin needs | vedclass

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(D) Let $n$ be the number of tosses. The probability of getting $k$ heads in $n$ tosses is given by the binomial distribution $P(X=k) = {^nC_k} (\frac

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $n$ be the number of tosses. The probability of getting $k$ heads in $n$ tosses is given by the binomial distribution $P(X=k) = {^nC_k} (\frac{1}{2})^n$.The probability of getting at least two heads is $P(X \ge 2) = 1 - P(X=0) - P(X=1)$.$P(X=0) = {^nC_0} (\frac{1}{2})^n = \frac{1}{2^n}$.$P(X=1) = {^nC_1} (\frac{1}{2})^n = \frac{n}{2^n}$.So,$P(X \ge 2) = 1 - \frac{1+n}{2^n}$.We want $1 - \frac{1+n}{2^n} \ge 0.96$,which implies $\frac{1+n}{2^n} \le 0.04 = \frac{1}{25}$.Testing values for $n$:For $n=7$: $\frac{1+7}{2^7} = \frac{8}{128} = \frac{1}{16} = 0.0625 > 0.04$.For $n=8$: $\frac{1+8}{2^8} = \frac{9}{256} \approx 0.03515 Thus,the minimum number of tosses required is $8$.

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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