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Solution

There are 2 red balls in total, so when drawing 3 balls, possible red counts are 0 (no red), 1 (one red), or 2 (both reds). $X$ can take values $\{0, 1, 2\}$.

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Solution

$\binom{100}{50}p^{50}(1-p)^{50} = \binom{100}{51}p^{51}(1-p)^{49}$ $\Rightarrow \dfrac{\binom{100}{50}}{\binom{100}{51}} \cdot \dfrac{1-p}{p} = 1$ $\dfrac{\binom{100}{50}}{\binom{100}{51}} = \dfrac{51}{50}$ $\Rightarrow \dfrac{1-p}{p} = \dfrac{50}{51}$ $\Rightarrow p = \dfrac{51}{101}$

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Solution

Numbers are $2,4,\dots,100=2\cdot{1,\dots,50}$. $\operatorname{Var}(1,\dots,50)=\dfrac{50^2-1}{12}=\dfrac{2499}{12}=\dfrac{833}{4}$. Scaling by $2$: $\operatorname{Var}=4\cdot\dfrac{833}{4}=833$.

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Solution

Let $n_1 = n_2 = 5, ; \sigma_1^2 = 4, ; \sigma_2^2 = 5, ; \bar{x}_1 = 2, ; \bar{x}_2 = 4$ Then $\sigma^2 = \dfrac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1+n_2} + \dfrac{n_1(\bar{x}_1 - \bar{x})^2 + n_2(\bar{x}_2 - \bar{x})^2}{n_1+n_2}$ where $\bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2} = 3$ $\Rightarrow \sigma^2 = \dfrac{5(4) + 5(5)}{10} + \dfrac{5(1)^2 + 5(1)^2}{10}$ $= 4.5 + 1 = 5.5 = \boxed{11/2}$

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Solution

29/72

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Solution

E(X)=5×0.3

E(X)=1.5