Question

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

• A. $\frac{35}{47}$
• B. $\frac{55}{1024}$
• C. $\frac{220}{512}$
• D. $\frac{11}{204}$

Answer 1

Answer: (B)

$\frac{55}{1024}$

$B(5,\frac{1}{2})\Rightarrow n=5,p=\frac{1}{2},q=\frac{1}{2}$
$B(7,\frac{1}{2})\Rightarrow n=7,p=\frac{1}{2},q=\frac{1}{2}$
since $x$ and $y$ are independent events
$x+y=3$ means
$x=0,y=3,x=1,y=2;x=2,y=1;x=0,y=3$
$\therefore P(x+y=3)={{}^{5}C}_0{\left(\frac{1}{2}\right)}^5.{{}^{7}C}_3{\left(\frac{1}{2}\right)}^7+{{}^{5}C}_1{\left(\frac{1}{2}\right)}^5{{}^{7}C}_2{\left(\frac{1}{2}\right)}^7+{{}^{5}C}_2{\left(\frac{1}{2}\right)}^5{{}^{7}C}_1{\left(\frac{1}{2}\right)}^7+{{}^{5}C}_3{\left(\frac{1}{2}\right)}^5{{}^{7}C}_0{\left(\frac{1}{2}\right)}^7$

$=\frac{55}{1024}$ (on simplification)