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(B) Let the coin be tossed $n$ times. Let getting a head be considered a success. $	herefore p = \frac{1}{2}, q = 1 - p = 1 - \frac{1}{2} = \frac{1}{

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$1 / 32$

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(B) Let the coin be tossed $n$ times. Let getting a head be considered a success. $	herefore p = \frac{1}{2}, q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$. It is given that $P(X = 3) = P(X = 5)$. Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$: ${}^{n}C_{3} (\frac{1}{2})^{3} (\frac{1}{2})^{n-3} = {}^{n}C_{5} (\frac{1}{2})^{5} (\frac{1}{2})^{n-5}$. Since the powers of $(\frac{1}{2})$ sum to $n$ on both sides, we have ${}^{n}C_{3} = {}^{n}C_{5}$. Using the property ${}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x + y = n$ (where $x 
eq y$), we get $n = 3 + 5 = 8$. Now, we need to find the probability of getting exactly one head, $P(X = 1)$: $P(X = 1) = {}^{8}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{8-1} = 8 	imes (\frac{1}{2})^{8} = \frac{8}{256} = \frac{1}{32}$.

$A$ fair coin is tossed a fixed number of times. If the probability of getting exactly $3$ heads equals the probability of getting exactly $5$ heads, then the probability of getting exactly one head is

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