The minimum number of times a fair coin must | vedclass

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(C) Let $n$ be the number of tosses. The probability of getting at least one head is given by $P(	ext{at least one head}) = 1 - P(	ext{no head})$.Si

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let $n$ be the number of tosses. The probability of getting at least one head is given by $P(	ext{at least one head}) = 1 - P(	ext{no head})$.Since the coin is fair,the probability of getting no head in $n$ tosses is $\left(\frac{1}{2}ight)^n$.We want $1 - \left(\frac{1}{2}ight)^n > \frac{99}{100}$.This simplifies to $1 - \frac{99}{100} > \left(\frac{1}{2}ight)^n$,which means $\frac{1}{100} > \left(\frac{1}{2}ight)^n$.This is equivalent to $2^n > 100$.We know that $2^6 = 64$ and $2^7 = 128$.Since $128 > 100$,the minimum integer value for $n$ is $7$.

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than $99\%$ is:

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