4.1

设

今从甲袋中任意取一只球放入乙袋中, 再从乙袋中任意取一只球, 问取到白球的概率是多少?

先从第一盒中任取 2 只球放入第二个盒子中, 然后从第二个盒子中任取一只球, 求取到白球的概率.

解

设

则 $A = B A + \overset{ˉ}{B} A$

P (A) = P (B) P (A ∣ B) + P (\overset{ˉ}{B}) P (A ∣ \overset{ˉ}{B}) = \frac{n}{m + n} \cdot \frac{N + 1}{N + M + 1} + \frac{m}{m + n} \cdot \frac{N}{M + N + 1} .

设

则

P (A) = P (B_{0}) P (A ∣ B_{0}) + P (B_{1}) P (A ∣ B_{1}) + P (B_{2}) P (A ∣ B_{2}) = \frac{( 2 5 )}{( 2 9 )} \cdot \frac{5}{11} + \frac{( 1 5 ) ( 1 4 )}{( 2 9 )} \cdot \frac{6}{11} + \frac{( 2 4 )}{( 2 9 )} \cdot \frac{7}{11} = \frac{53}{99} .

从数 1,2,3,4 中任取一个数, 记为 $X$ ,再从 $1, \dots, X$ 中任取一个数, 记为 $Y$ ,则 $P {Y = 2} =$ _____.

令 $A_{i} = {X = i}, i = 1, 2, 3, 4$ , 则 $A_{1}, A_{2}, A_{3}, A_{4}$ 构成一个完备事件组, 且

P (A_{i}) = \frac{1}{4}, i = 1, 2, 3, 4

而

P {Y = 2 ∣ A_{1}} = 0, P {Y = 2 ∣ A_{i}} = \frac{1}{i}, i = 2, 3, 4,

那么由全概率公式得

P {Y = 2} = i = 1 \sum 4 P (A_{i}) P {Y = 2 ∣ A_{i}} = \frac{1}{4} (0 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = \frac{13}{48} .