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Question

(C) Let the coin be tossed $n$ times. Probability of getting a head in a single toss is $p = \frac{1}{2}$. Probability of not getting a head (tail) is

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let the coin be tossed $n$ times. Probability of getting a head in a single toss is $p = \frac{1}{2}$. Probability of not getting a head (tail) is $q = 1 - p = \frac{1}{2}$. The probability of getting at least one head in $n$ tosses is $P(X \geq 1) = 1 - P(X = 0)$. Given that $P(X \geq 1) > \frac{99}{100}$. Therefore,$1 - P(X = 0) > \frac{99}{100}$. Since $P(X = 0) = (\frac{1}{2})^n$,we have $1 - (\frac{1}{2})^n > \frac{99}{100}$. This simplifies to $1 - \frac{99}{100} > (\frac{1}{2})^n$,which means $\frac{1}{100} > (\frac{1}{2})^n$. Taking the reciprocal,we get $100 For $n = 6$,$2^6 = 64$,which is less than $100$. For $n = 7$,$2^7 = 128$,which is greater than $100$. Thus,the minimum number of tosses required is $7$.

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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