随机变量x～B（2、0.5）则P（x≥1）怎么求 设随机变量X服从二项分布B(2, p)，已知P{X≥1}=5...
\u8bbe\u968f\u673a\u53d8\u91cfX~B(2\uff0cP)\uff0c\u4e14P{X\u22651}=5/9\u8bbe\u968f\u673a\u53d8\u91cfX\uff5eB(2,p),\u82e5P(X>=1)=5/9,\u5219p=?
\u8bbe\u968f\u673a\u53d8\u91cfX\uff5eB(2,p),\u82e5P(X>=1)=5/9,\u5219p=?
\u89e3:x 0 1 2
p (1-p)^2 p(1-p) p^2
P(X>=1)=5/9
P(x=0)=(1-p)^2=1-P(X>=1)=1-5/9
\uff081-p\uff09=2/3
p=1/3
P{X\u22651}=5/9\u2192P{X=0}=1-P{X\u22651}=4/9
P{X=0}=1*p^0*(1-p)^2=4/9\u2192p=1/3
即x服从二项分布
那么x的取值只能是0，1，2
所以P(x≥1)= 1 -P(x=0)
而P(x=0)=0.5*0.5=0.25
所以
P(x≥1)
=1 -P(x=0)
=1 -0.25=0.75
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