2.盒中有3个白球和2个黑球,从中随机抽取2个,X,Y分别是抽到的2个球中的白球数和黑球数,求X,Y之间的相关系数 p_(X,Y).

设 $X$ 为白球数，$Y$ 为黑球数，显然 $Y = 2 - X$。 计算 $X$ 的期望和方差： - $E(X) = 0 \times \frac{1}{10} + 1 \times \frac{3}{5} + 2 \times \frac{3}{10} = \frac{6}{5}$ - $E(X^2) = 0^2 \times \frac{1}{10} + 1^2 \times \frac{3}{5} + 2^2 \times \frac{3}{10} = \frac{9}{5}$ - $D(X) = E(X^2) - [E(X)]^2 = \frac{9}{5} - \left(\frac{6}{5}\right)^2 = \frac{9}{25}$ 由 $Y = 2 - X$，得： - $E(Y) = 2 - E(X) = \frac{4}{5}$ - $D(Y) = D(X) = \frac{9}{25}$ 计算协方差： - $E(XY) = E[2X - X^2] = 2E(X) - E(X^2) = \frac{3}{5}$ - $\text{Cov}(X, Y) = E(XY) - E(X)E(Y) = \frac{3}{5} - \frac{6}{5} \times \frac{4}{5} = -\frac{9}{25}$ 相关系数： - $p_{X,Y} = \frac{\text{Cov}(X, Y)}{\sqrt{D(X)} \sqrt{D(Y)}} = \frac{-\frac{9}{25}}{\frac{9}{25}} = -1$ 答案：$\boxed{-1}$