题目
题型:不详难度:来源:
答案
则P(A1)=0.1,P(A2)=0.2,P(A3)=0.3.
由题意ξ有四个可能值0,1,2,3.
由于A1,A2,A3相互独立,
∴P(ξ=0)=P(
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P(ξ=1)=P(A1
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=0.1×0.8×0.7+0.9×0.2×0.7+0.9×0.8×0.3=0.398;
P(ξ=2)=P(A1A2
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=0.1×0.2×0.7+0.1×0.8×0.3+0.9×0.2×0.3=0.092;
P(ξ=3)=P(A1A2A3)=0.1×0.2×0.3=0.006.
∴Eξ=1×0.398+2×0.092+3×0.006=0.6,
Dξ=Eξ2-(Eξ)2=1×0.398+4×0.092+9×0.006-0.62=0.82-0.36=0.46.
核心考点
试题【一台设备由三大部件组成,在设备运转中,各部件需要调整的概率相应为0.10,0.20和0.30.假设各部件的状态相互独立,以ξ表示同时需要调整的部件数,试求ξ的数】;主要考察你对离散型随机变量均值与方差等知识点的理解。[详细]
举一反三
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