Explanation

(b)
For $X$ binomial variate B$\left( 5,\frac{1}{2} \right)$
$\Rightarrow p = \frac{1}{2},n = 5,q = \frac{1}{2}$
For $Y$ binomial variate B $\left( 7,\frac{1}{2} \right)$
$\Rightarrow p = \frac{1}{2},n = 7,q = \frac{1}{2}$
Now, $X + Y = 3$
(i) When $X = 0,Y = 3$, possible cases
$= {\ ^{5}C}_{0}\left( \frac{1}{2} \right)^{5}.{\ ^{7}C}_{3}\left( \frac{1}{2} \right)^{7}$
$= 35\left( \frac{1}{2} \right)^{12}$
(ii) When $X = 1,Y = 2,\ possible\ cases$
$=^{5}C_{1}\left( \frac{1}{2} \right)^{5}.{\ ^{7}C}_{2}\left( \frac{1}{2} \right)^{7}$
$= 105\left( \frac{1}{2} \right)^{12}$
(iii) When $X = 2,Y = 1$, possible cases
$= {\ ^{5}C}_{2}\left( \frac{1}{2} \right)^{5}.{\ ^{7}C}_{1}\left( \frac{1}{2} \right)^{7}$
$= 70\left( \frac{1}{2} \right)^{12}$
(iv) When $X = 3,\ Y = 0$, possible cases
$= {\ ^{5}C}_{3}\left( \frac{1}{2} \right)^{3}.{\ ^{7}C}_{0}\left( \frac{1}{2} \right)^{7}$
$= 10\left( \frac{1}{2} \right)^{12}$
$\therefore$ Total cases$= \left( \frac{1}{2} \right)^{12}\lbrack 35 + 105 + 70 + 10\rbrack$
$= \frac{220}{2^{12}} = \frac{55}{1024}$