Question

If $X$ and $Y$ are the independent random variables $B(5,\frac{1}{2})$ and $B(7,\frac{1}{2})$, then $P(X+Y\ge 1)=$

• A. $\frac{4095}{4096}$
• B. $\frac{309}{4096}$
• C. $\frac{4032}{4096}$
• D. None of these

Answer 1

Answer: (A)

$\frac{4095}{4096}$

For random variable $X,n=5,p=\frac{1}{2}$ and for random variable $Y,n=7,p=\frac{1}{2}$
Now $P(X+Y\ge 1)=1-P(X+Y<1)=1-P(X+Y=0)$
$=1-P(X=0,Y=0)$
$=1-P(X=0)P(Y=0)$
$[\therefore X$ and $Y$ are independent$]$
$=1{-}^5{C}_0{\left(\frac{1}{2}\right)}^5{.}^7{C}_0{\left(\frac{1}{2}\right)}^7=1-{\left(\frac{1}{2}\right)}^{12}=\frac{4095}{4096}.$