Question

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

• A. $\frac{45}{1024}$
• B. $\frac{55}{1024}$
• C. $\frac{65}{1024}$
• D. $\frac{60}{1024}$

Answer 1

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

Solution - $n_A=5P_A=q_A=\frac{1}{2}{n}_B=r_B=\frac{1}{2}$

$P(x+y=3)=p(x=0)P(y=3)+p(x=1)P(y=2)$

$+p(x=2)p(y=1)+p(x=3)p(y=0)$

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

$=\frac{35+105+70+10}{2^{12}}$

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

$=\frac{55}{1.24}$

B is correct