X and Y are independent binomial variatesB5,12 and B7,12then P(X+Y=3) is

x :3210Y :0123 Req. =P(X=3)⋅P(Y=0)+P(X=2)P(X=1)+P(X=1)⋅P(X=2)+P(X=0)P(X=3)=5C31257C0127+5C2125 7C1127+5C11257C2127+5C01257C3127=12125C37C0+5C27C1+5C17C2+5C07C3=12C31212=551024

X and Y are independent binomial variatesB5,12 and B7,12then P(X+Y=3) is

x :3210Y :0123 Req. =P(X=3)⋅P(Y=0)+P(X=2)P(X=1)+P(X=1)⋅P(X=2)+P(X=0)P(X=3)=5C31257C0127+5C2125 7C1127+5C11257C2127+5C01257C3127=12125C37C0+5C27C1+5C17C2+5C07C3=12C31212=551024

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X and Y are independent binomial variates $B (5, \frac{1}{2}) and B (7, \frac{1}{2})$ then P(X+Y=3) is

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$\frac{45}{1024}$

$\frac{55}{1024}$

$\frac{65}{1024}$

$\frac{60}{1024}$

answer is B.

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Detailed Solution

x :	3	2	1	0
Y :	0	1	2	3

$\begin{array}{l} Req. = P (X = 3) \cdot P (Y = 0) + P (X = 2) P (X = 1) \\ + P (X = 1) \cdot P (X = 2) + P (X = 0) P (X = 3) \\ = 5 C_{3} {(\frac{1}{2})}^{5} 7 C_{0} {(\frac{1}{2})}^{7} +^{5} C_{2} {(\frac{1}{2})}^{5}^{7} C_{1} {(\frac{1}{2})}^{7} + 5 C_{1} {(\frac{1}{2})}^{5} 7 C_{2} {(\frac{1}{2})}^{7} \\ + 5 C_{0} {(\frac{1}{2})}^{5} 7 C_{3} {(\frac{1}{2})}^{7} \\ = {(\frac{1}{2})}^{12} [5 C_{3} 7 C_{0} + 5 C_{2} 7 C_{1} + 5 C_{1} 7 C_{2} + 5 C_{0} 7 C_{3}] \\ =^{12} C_{3} {(\frac{1}{2})}^{12} = \frac{55}{1024} \end{array}$