91Ó°ÊÓ

Here X is the random variable that denotes the proportion of the red balls in the sample. Let Y denote the number of red balls among seven balls, then 0≤ Y ≤5. Y is following hypergeometric distribution that is \(Y \sim Hyp\left( {N = 15,r = 5,n = 7} \right)\)

Therefore, the pmf of the following distribution is:

\(p\left( y \right) = \frac{{{}^5{C_y}{}^{10}{C_{7 - y}}}}{{{}^{15}{C_7}}},y = 0,1,2, \ldots \)

We know that the expectation of a hypergeometric distribution is:

\(\)\(\begin{array}{c}E\left( Y \right) = n\frac{r}{N}\\ = \frac{{7 \times 5}}{{15}}\\ \approx 2.33\end{array}\)

We know that the variance of a hypergeometric distribution is:

\(\begin{array}{c}V\left( Y \right) = n\frac{r}{N}\frac{{N - r}}{N}\frac{{N - n}}{{N - 1}}\\ = \frac{{7 \times 5}}{{15}} \cdot \frac{{15 - 5}}{{15}} \cdot \frac{{15 - 7}}{{15 - 1}}\\ \approx 0.88\end{array}\)

\(\)\(\begin{array}{c}Let,\;X = \frac{n}{7}\\E\left( X \right) = \frac{{E\left( n \right)}}{7}\\ = \frac{1}{3}\end{array}\)

\(\begin{array}{c}V\left( X \right) = \frac{{V\left( n \right)}}{{49}}\\ = \frac{8}{{441}}\end{array}\)

Therefore, the expectation and variance of X that denotes the proportion of red balls in the sample is

\(E(x) = \frac{1}{3}{\rm{ and }}V(x) = \frac{8}{{441}}\)

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