1.4

盒子里装有 3 只黑球、 2 只红球、 2 只白球,在其中任选 4 只球,以 $X$ 表示取到黑球的只数,以 $Y$ 表示取到红球的只数,求 $X$ 和 $Y$ 的联合分布律.

解

$(X, Y)$ 的所有可能取值为 $(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1)$ , $(2, 2), (3, 0), (3, 1), (3, 2)$ .

按古典概型,显有

P {X = 0, Y = 2} = \frac{C _{3}^{0} \times C _{2}^{2} \times C _{2}^{2}}{C _{7}^{4}} = \frac{1}{35}

P {X = 1, Y = 1} = \frac{C _{3}^{1} \times C _{2}^{1} \times C _{2}^{2}}{C _{7}^{4}} = \frac{6}{35}

P {X = 1, Y = 2} = \frac{C _{3}^{1} \times C _{2}^{2} \times C _{2}^{1}}{C _{7}^{4}} = \frac{6}{35}

P {X = 2, Y = 1} = \frac{C _{3}^{2} \times C _{2}^{1} \times C _{2}^{1}}{C _{7}^{4}} = \frac{12}{35}

P {X = 2, Y = 0} = \frac{C _{3}^{2} \times C _{2}^{0} \times C _{2}^{2}}{C _{7}^{4}} = \frac{3}{35}

P {X = 2, Y = 2} = \frac{C _{3}^{2} \times C _{2}^{2} \times C _{2}^{0}}{C _{7}^{4}} = \frac{3}{35}

P {X = 3, Y = 0} = \frac{C _{3}^{3} \times C _{2}^{0} \times C _{2}^{1}}{C _{7}^{4}} = \frac{2}{35}

P {X = 3, Y = 1} = \frac{C _{3}^{3} \times C _{2}^{1} \times C _{2}^{0}}{C _{7}^{4}} = \frac{2}{35}

则 $X$ 和 $Y$ 的联合分布律为:

$X Y$	0	1	2	3
0	0	0	$\frac{3}{35}$	$\frac{2}{35}$
1	0	$\frac{6}{35}$	$\frac{12}{35}$	$\frac{2}{35}$
2	$\frac{1}{35}$	$\frac{6}{35}$	$\frac{3}{35}$	0