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(B) Let $p$ be the probability of getting a head,so $p = \frac{1}{2}$.Let $q$ be the probability of not getting a head,so $q = 1 - p = \frac{1}{2}$.Th

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$\frac{15}{2^{13}}$

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(B) Let $p$ be the probability of getting a head,so $p = \frac{1}{2}$.Let $q$ be the probability of not getting a head,so $q = 1 - p = \frac{1}{2}$.The probability of getting $x$ heads in $n$ tosses is given by the binomial distribution formula: $P(X=x) = {}^{n}C_{x} p^x q^{n-x} = {}^{n}C_{x} (\frac{1}{2})^n$.Given that $P(X=7) = P(X=9)$,we have:${}^{n}C_{7} (\frac{1}{2})^n = {}^{n}C_{9} (\frac{1}{2})^n$.This implies ${}^{n}C_{7} = {}^{n}C_{9}$.Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get $n = 7 + 9 = 16$.Now,we need to find the probability of getting $2$ heads,which is $P(X=2)$:$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{16} = \frac{16 	imes 15}{2} 	imes \frac{1}{2^{16}} = 120 	imes \frac{1}{2^{16}} = \frac{15 	imes 8}{2^{16}} = \frac{15}{2^{13}}$.

$A$ fair coin is tossed for a fixed number of times. If the probability of getting $7$ heads is equal to the probability of getting $9$ heads,then the probability of getting $2$ heads is:

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