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(D) Let the probability of getting a head be $p$. Then the probability of not getting a head is $1-p$. The person wins in an odd number of tosses if t

Vedclass · Accepted Answer

$\frac{32}{243}$

Vedclass · Answer

(D) Let the probability of getting a head be $p$. Then the probability of not getting a head is $1-p$. The person wins in an odd number of tosses if the first head appears on the $1^{st}, 3^{rd}, 5^{th}, \dots$ toss. This forms a geometric series: $p + (1-p)^2 p + (1-p)^4 p + \dots = 3/4$. Using the sum formula for an infinite geometric series $S = \frac{a}{1-r}$, where $a = p$ and $r = (1-p)^2$: $\frac{p}{1-(1-p)^2} = \frac{3}{4}$ $\frac{p}{1-(1-2p+p^2)} = \frac{3}{4}$ $\frac{p}{2p-p^2} = \frac{3}{4}$ $\frac{1}{2-p} = \frac{3}{4}$ $4 = 6 - 3p \implies 3p = 2 \implies p = 2/3$. If $5$ such coins are tossed at a time, the probability that a head appears on all $5$ coins is $p^5 = (2/3)^5 = 32/243$.

$A$ person tossing a biased coin indefinitely wins the game by getting head for the first time. The probability that he wins the game in an odd number of tosses is $3/4$. If $5$ such coins are tossed at a time, then the probability that head appears on all the coins is:

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