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(D) Let $p$ be the probability of getting an odd number,$p = \frac{1}{2}$.Given $P(	ext{odd } 7 	ext{ times}) = P(	ext{odd } 9 	ext{ times})$.Usin

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$60$

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(D) Let $p$ be the probability of getting an odd number,$p = \frac{1}{2}$.Given $P(	ext{odd } 7 	ext{ times}) = P(	ext{odd } 9 	ext{ times})$.Using the binomial distribution formula $P(X=r) = {}^{n}C_{r} p^r q^{n-r}$:${}^{n}C_{7} (\frac{1}{2})^7 (\frac{1}{2})^{n-7} = {}^{n}C_{9} (\frac{1}{2})^9 (\frac{1}{2})^{n-9}$${}^{n}C_{7} = {}^{n}C_{9}$Since ${}^{n}C_{a} = {}^{n}C_{b}$ implies $a+b=n$ or $a=b$,we have $n = 7+9 = 16$.Now,we need the probability of getting even numbers twice in $16$ rolls. The probability of an even number is $q = \frac{1}{2}$.$P(	ext{even } 2 	ext{ times}) = {}^{16}C_{2} (\frac{1}{2})^2 (\frac{1}{2})^{16-2} = {}^{16}C_{2} (\frac{1}{2})^{16}$$= \frac{16 	imes 15}{2} 	imes \frac{1}{2^{16}} = 120 	imes \frac{1}{2^{16}} = \frac{120}{2 	imes 2^{15}} = \frac{60}{2^{15}}$.Comparing with $\frac{k}{2^{15}}$,we get $k = 60$.

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2^{15}}$,then $k$ is equal to:

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