A jar contains 7 white marbles and 3 blue mar | vedclass

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Question

(B) Let $X$ be the random variable representing the number of blue marbles chosen. The total number of marbles is $10$. The number of ways to choose $

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(B) Let $X$ be the random variable representing the number of blue marbles chosen. The total number of marbles is $10$. The number of ways to choose $4$ marbles is $\binom{10}{4} = 210$.The probability distribution of $X$ is given by $P(X=k) = \frac{\binom{3}{k} \binom{7}{4-k}}{\binom{10}{4}}$:$X$$P(X=k)$$0$$\frac{\binom{3}{0} \binom{7}{4}}{210} = \frac{35}{210} = \frac{1}{6}$$1$$\frac{\binom{3}{1} \binom{7}{3}}{210} = \frac{3 	imes 35}{210} = \frac{1}{2}$$2$$\frac{\binom{3}{2} \binom{7}{2}}{210} = \frac{3 	imes 21}{210} = \frac{3}{10}$$3$$\frac{\binom{3}{3} \binom{7}{1}}{210} = \frac{1 	imes 7}{210} = \frac{1}{30}$$E[X] = 0(\frac{1}{6}) + 1(\frac{1}{2}) + 2(\frac{3}{10}) + 3(\frac{1}{30}) = 0 + 0.5 + 0.6 + 0.1 = 1.2 = \frac{6}{5}$.$E[X^2] = 0^2(\frac{1}{6}) + 1^2(\frac{1}{2}) + 2^2(\frac{3}{10}) + 3^2(\frac{1}{30}) = 0 + 0.5 + 1.2 + 0.3 = 2.0$.$Var(X) = E[X^2] - (E[X])^2 = 2.0 - (1.2)^2 = 2.0 - 1.44 = 0.56 = \frac{56}{100} = \frac{14}{25}$.Standard Deviation $\sigma = \sqrt{Var(X)} = \sqrt{\frac{14}{25}} = \frac{\sqrt{14}}{5}$.Here $a = 14$ and $b = 5$. Since $14$ and $5$ are co-prime and $14$ is square-free,$a + b = 14 + 5 = 19$.

$A$ jar contains $7$ white marbles and $3$ blue marbles. Given that $4$ marbles are chosen from the jar at the same time,the standard deviation of the number of blue marbles chosen is $\frac{\sqrt{a}}{b}$,where $a$ and $b$ are co-prime numbers and $a$ is square-free. Then $a + b$ is:

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