Explanation

(a)
Let $X$ denotes the number of red balls. Here probability of getting red balls, $p = \frac{3}{7}$ and probability of getting red bills, $q = \frac{4}{7}$
$P_{1}(X = 0) = {\ ^{3}C}_{0}\left( \frac{3}{7} \right)^{0}\left( \frac{4}{7} \right)^{3} = \frac{64}{(7)^{3}}$
$P_{2}(X = 1) = {\ ^{3}C}_{1}\left( \frac{3}{7} \right)^{1}\left( \frac{4}{7} \right)^{2} = \frac{144}{(7)^{3}}$
$P_{3}\ (X = 2) = {\ ^{3}C}_{2}\left( \frac{3}{7} \right)^{2}\left( \frac{4}{7} \right)^{1}\frac{108}{(7)^{3}}$
$P_{4}\ (X = 3) = {\ ^{3}C}_{3}\left( \frac{3}{7} \right)^{3} = \frac{27}{(7)^{2}}$
$\therefore$ Variance$= \sum_{i = 0}^{3}{P_{i}x_{i}^{2} - \left( \sum_{i = 0}^{3}{P_{i}x_{i}} \right)^{2}}$
$= \left\lbrack \frac{64}{(7)^{3}} \times 0 + \frac{144}{(7)^{3}} \times (1)^{2} + \frac{108}{(7)^{3}} \times (2)^{2} + \frac{27}{(7)^{3}} \times (3)^{2} \right\rbrack$
$- \left\lbrack \frac{64}{(7)^{3}} \times 0 + \frac{144}{(7)^{3}} \times 1 + \frac{108}{(7)^{3}} \times 2 + \frac{27}{(7)^{3}} \times 3 \right\rbrack^{2}$
$= \left\lbrack 0 + \frac{144}{343} + \frac{432}{343} + \frac{243}{343} \right\rbrack - \left\lbrack 0 + \frac{144}{343} + \frac{216}{343} + \frac{81}{343} \right\rbrack^{2}$
$= \frac{819}{343} - \left( \frac{441}{343} \right)^{2}$
$= \frac{280917 - 194481}{(343)^{2}} = \frac{36}{49}$
Now, standard deviation $= \sqrt{variance}$
$= \sqrt{\frac{36}{49}} = \frac{6}{7}$