设随机变量x~b(10,0.3),概率最大值点

\u8bbe\u968f\u673a\u53d8\u91cfX~B(3,0.4),\u6c42Y=|X-1|\u7684\u6982\u7387\u5206\u5e03

X= 0 1 2 3
Y= 1 0 1 2
\u628aX=0\uff0c2\u5408\u5e76\u5373\u53ef

x~b(3,0.3)
\u6240\u4ee5n=3\uff0cp=0.3
P(X=2)=3C2*0.3^2*(1-0.3)^(3-2)=3*0.09*0.7=0.189

你好!二项分布的概率最大值点是[(n+1)p],这里n=10,p=0.3,(n+1)p=3.3,所以答案是3。经济数学团队帮你解答,请及时采纳。谢谢!

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