求解答这道概统题。。。。。。。。。。。。。
答案:1 悬赏:50 手机版
解决时间 2021-01-11 13:29
- 提问者网友：姑娘长的好罪过
- 2021-01-11 08:33
求解答这道概统题。。。。。。。。。。。。。
最佳答案
- 五星知识达人网友：胯下狙击手
- 2021-01-11 09:25
X1, X2各只有两种取值.
只需计算P(X1 = 0, X2 = 0), P(X1 = 0, X2 = 1), P(X1 = 1, X2 = 0), P(X1 = 1, X2 = 1).
首先若X2 = 1, 则Y > 2 > 1, 故X1 = 1. 因此P(X1 = 0, X2 = 1) = 0.
设Y的分布函数为F(x), 由Y服从指数分布e(1), F(x) = 1-e^(-x)对x > 0.
P(X1 = 0, X2 = 0) = P(Y ≤ 1) = F(1) = 1-1/e.
P(X1 = 1, X2 = 0) = P(1 < Y ≤ 2) = P(Y ≤ 2)-P(Y ≤ 1) = F(2)-F(1) = 1/e-1/e².
P(X1 = 1, X2 = 1) = P(Y > 2) = 1-P(Y ≤ 2) = 1-F(2) = 1/e².
只需计算P(X1 = 0, X2 = 0), P(X1 = 0, X2 = 1), P(X1 = 1, X2 = 0), P(X1 = 1, X2 = 1).
首先若X2 = 1, 则Y > 2 > 1, 故X1 = 1. 因此P(X1 = 0, X2 = 1) = 0.
设Y的分布函数为F(x), 由Y服从指数分布e(1), F(x) = 1-e^(-x)对x > 0.
P(X1 = 0, X2 = 0) = P(Y ≤ 1) = F(1) = 1-1/e.
P(X1 = 1, X2 = 0) = P(1 < Y ≤ 2) = P(Y ≤ 2)-P(Y ≤ 1) = F(2)-F(1) = 1/e-1/e².
P(X1 = 1, X2 = 1) = P(Y > 2) = 1-P(Y ≤ 2) = 1-F(2) = 1/e².
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