Question

If $X$ and $Y$ are independent Binomial variates $B(5,1/2)$ and $B(7,1/2)$, then $P(X+Y=3)$ is equal to:

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

Answer 1

The correct option is B

$\frac{55}{1024}$

Explanation for the correct option.

Step 1: Find the value of $P\left(X=0,Y=3\right)$

We know that, $P(X+Y=k)=\sum _{a,b}{}^{n}C_a{p}^n·{}^{m}C_b{q}^m$

$\begin{array}{rcl}P\left(X=0,Y=3\right)& =& {}^{5}C_0{\left(\frac{1}{2}\right)}^5\times {}^{7}C_3{\left(\frac{1}{2}\right)}^7\\ & =& 1\times \frac{1}{32}\times \frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}\times \frac{1}{128}\\ & =& \frac{35}{1096}.........\left(1\right)\end{array}$

Step 2: Find the value of $P\left(X=1,Y=2\right)$

$\begin{array}{rcl}P\left(X=1,Y=2\right)& =& {}^{5}C_1{\left(\frac{1}{2}\right)}^5\times {}^{7}C_2{\left(\frac{1}{2}\right)}^7\\ & =& 5\times \frac{1}{32}\times \frac{7\times 6\times 5!}{2\times 1\times 5!}\times \frac{1}{128}\\ & =& \frac{105}{1096}.........\left(2\right)\end{array}$

Step 3: Find the value of $P\left(X=2,Y=1\right)$

$\begin{array}{rcl}P\left(X=2,Y=1\right)& =& {}^{5}C_2{\left(\frac{1}{2}\right)}^5\times {}^{7}C_1{\left(\frac{1}{2}\right)}^7\\ & =& \frac{5\times 4\times 3!}{2\times 1\times 3!}\times \frac{1}{32}\times 7\times \frac{1}{128}\\ & =& \frac{70}{1096}.........\left(3\right)\end{array}$

Step 4: Find the value of $P\left(X=3,Y=0\right)$

$\begin{array}{rcl}P\left(X=3,Y=0\right)& =& {}^{5}C_3{\left(\frac{1}{2}\right)}^5\times {}^{7}C_0{\left(\frac{1}{2}\right)}^7\\ & =& \frac{5\times 4\times 3\times 2!}{3\times 2\times 1\times 2!}\times \frac{1}{32}\times 1\times \frac{1}{128}\\ & =& \frac{10}{1096}.........\left(4\right)\end{array}$

Step 5: Find the value of $P(X+Y=3)$

By adding $\left(1\right),\left(2\right),\left(3\right),\left(4\right)$ we get,

$\begin{array}{rcl}P(X+Y& =& 3)=\frac{35}{4096}+\frac{105}{4096}+\frac{70}{4096}+\frac{10}{4096}\\ & =& \frac{220}{4096}\\ & =& \frac{55}{1024}\end{array}$

Hence, option B is correct.