Question

If $X$ and $Y$ are independent binomial variate of binomial distribution $A(5,\frac{1}{2})$ and $B(7,\frac{1}{2})$ respectively, then $P(X+Y=3)$ is

• A. $\frac{55}{1024}$
• B. $\frac{55}{4098}$
• C. $\frac{55}{2048}$
• D. none of these

Answer 1

The correct option is A $\frac{55}{1024}$

$P(X+Y=3)=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)$

$=P(X=0)P(Y=3)+P(X=1)P(Y=2)+P(X=2)P(Y=1)+P(X=3)P(Y=0)$

$(X$ and $Y$ are independent$)$

${=}^5{C}_0.{\left(\frac{1}{2}\right)}^5$${}^7{C}_3.{\left(\frac{1}{2}\right)}^7$${+}^5{C}_1.{\left(\frac{1}{2}\right)}^5$${}^7{C}_2.{\left(\frac{1}{2}\right)}^7$

${+}^5{C}_2.{\left(\frac{1}{2}\right)}^5$${}^7{C}_1.{\left(\frac{1}{2}\right)}^7+$ ${}^5{C}_3.{\left(\frac{1}{2}\right)}^5$ ${}^7{C}_0.{\left(\frac{1}{2}\right)}^7$

$={\left(\frac{1}{2}\right)}^{12}\{\left(1\right)\left(35\right)+\left(5\right)\left(21\right)+\left(10\right)\left(7\right)+\left(10\right)\left(1\right)\}$

$=\frac{220}{2^{12}}=\frac{55}{1024}$