Statistics

Let $X$ be the random variable that denotes the number of blue marbles chosen. The probability of choosing $x$ blue marbles is

$P (X = x) = \frac{\binom{7}{4 - x} \times \binom{3}{x}}{\binom{10}{4}}$

Since there can be $0, 1, 2,$ or $3$ blue marbles in our selection, we can substitute $x = 0, 1, 2, 3$ in the above formula to find the distribution of $X$ .

$\begin{array} {|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline P(X = x) & \frac{\binom{7}{4} \times \binom{3}{0}}{\binom{10}{4}} = \frac{1}{6} & \frac{\binom{7}{3} \times \binom{3}{1}}{\binom{10}{4}} = \frac{1}{2} & \frac{\binom{7}{2} \times \binom{3}{2}}{\binom{10}{4}} = \frac{3}{10} & \frac{\binom{7}{3} \times \binom{3}{3}}{\binom{10}{4}} = \frac{1}{30} \\ \hline \end{array}$

The standard deviation of a distribution is given by $\sigma = \sqrt{ Var(X) }$

The variance of the distribution is $Var(X) = E[X^{2}] - (E[X])^{2}$

$\begin{aligned}E[X^{2}] & = 0^{2} \times \frac{1}{6} + 1^{2} \times \frac{1}{2} + 2^{2} \times \frac{3} {10} + 3^{2} \times \frac{1}{30} \\ & = 0 + \frac{1}{2} + \frac{6}{5} + \frac{3 } {10} \\ & = 2 \\ \\ E[X] & = 0 \times \frac{1}{6} + 1 \times \frac{1}{2} + 2 \times \frac{3} {10} + 3 \times \frac{1}{30} \\ & = \frac{1}{2} + \frac{3}{5} + \frac{1}{10} \\ & = 1.2 \\ \implies (E[X])^{2} & = 1.2^{2} = 1.44 \end{aligned}$

We get variance $= 2 - 1.44 = 0.56$ .
Standard deviation is the square root of variance $\sigma = \sqrt{0.56} = \sqrt{\frac{14}{25} } = \frac{\sqrt{14}}{5}$ . Thus, $14 + 5 = 19$ $_\square$