The minimum number of times a fair coin needs | vedclass

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Question

(D) Let the coin be tossed $n$ times. The probability of getting at least two heads is given by $P(X \geq 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]$.

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let the coin be tossed $n$ times. The probability of getting at least two heads is given by $P(X \geq 2) = 1 - P(X Since the coin is fair,$P(X=k) = \binom{n}{k} (\frac{1}{2})^n$. Thus,$P(X=0) = \frac{1}{2^n}$ and $P(X=1) = \frac{n}{2^n}$. The condition is $1 - [\frac{1}{2^n} + \frac{n}{2^n}] \geq 0.96$. $1 - \frac{n+1}{2^n} \geq 0.96$. $\frac{n+1}{2^n} \leq 1 - 0.96 = 0.04 = \frac{4}{100} = \frac{1}{25}$. $\frac{2^n}{n+1} \geq 25$. Testing values for $n$: For $n=7$: $\frac{2^7}{7+1} = \frac{128}{8} = 16 For $n=8$: $\frac{2^8}{8+1} = \frac{256}{9} \approx 28.44 \geq 25$. 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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