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(D) Given $X \sim B(10, 1/2)$ and $Y \sim B(8, 1/2)$. Since $X$ and $Y$ are independent,$X+Y$ follows a binomial distribution $B(n_1+n_2, p)$ if $p$ i

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$\frac{153}{4^{9}}$

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(D) Given $X \sim B(10, 1/2)$ and $Y \sim B(8, 1/2)$. Since $X$ and $Y$ are independent,$X+Y$ follows a binomial distribution $B(n_1+n_2, p)$ if $p$ is the same. Here $p=1/2$ for both,so $X+Y \sim B(18, 1/2)$.Alternatively,we calculate $P(X+Y=2)$ by considering all possible cases:$P(X+Y=2) = P(X=0, Y=2) + P(X=1, Y=1) + P(X=2, Y=0)$$= \left( \binom{10}{0} (1/2)^{10} 	imes \binom{8}{2} (1/2)^8 ight) + \left( \binom{10}{1} (1/2)^{10} 	imes \binom{8}{1} (1/2)^8 ight) + \left( \binom{10}{2} (1/2)^{10} 	imes \binom{8}{0} (1/2)^8 ight)$$= \frac{1}{2^{18}} \left( 1 	imes 28 + 10 	imes 8 + 45 	imes 1 ight)$$= \frac{28 + 80 + 45}{2^{18}} = \frac{153}{2^{18}} = \frac{153}{4^9}$

If $X$ and $Y$ are two independent binomial variates,satisfying $B(10, 1/2)$ and $B(8, 1/2)$ respectively,then the probability $P(X + Y = 2)$ is

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