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A jar has 5 red and 10 blue marbles. I pick a handful of 4 marbles out. Let \(X\) be the number of red marbles in my hand. Find the pmf of \(X\), find its expected value and its variance.
in Data Science & Statistics by Diamond (40,719 points) | 30 views

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Best answer
The probability that you draw \(x\) red marbles is
\[
P(X=x)=\frac{\left({ }_{5} C_{x}\right)\left({ }_{10} C_{4-x}\right)}{{ }_{15} C_{4}}
\]
So the probability mass function is given by
\begin{equation}
\begin{array}{c|c}
x & f(x) \\
\hline 0 & (1)(210) / 1365=2 / 13 \\
1 & (5)(120) / 1365=40 / 91 \\
2 & (10)(45) / 1365=30 / 91 \\
3 & (10)(10) / 1365=20 / 273 \\
4 & (5)(1) / 1365=1 / 273
\end{array}
\end{equation}
The expected value is
\[
E(X)=0\left(\frac{2}{13}\right)+1\left(\frac{40}{91}\right)+2\left(\frac{30}{91}\right)+3\left(\frac{20}{273}\right)+4\left(\frac{1}{273}\right)=\frac{4}{3}
\]
Variance is calculated by
\[
\begin{aligned}
\operatorname{Var}(X) &=E\left(X^{2}\right)-E(X)^{2} \\
&=0^{2}\left(\frac{2}{13}\right)+1^{2}\left(\frac{40}{91}\right)+2^{2}\left(\frac{30}{91}\right)+3^{2}\left(\frac{20}{273}\right)+4^{2}\left(\frac{1}{273}\right)-\left(\frac{4}{3}\right)^{2} \\
& \approx .6984
\end{aligned}
\]
by Diamond (40,719 points)

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