Question

一台机器一天内发生故障的概率为0.1,若这台机器一周5个工作日不发生故障,可获利5万元；发生1次故障仍可获利2.5万元；

Answer 1

5个工作日获利 X 万元,X = 5,2.5,0,-1

P(X=5) = (1-0.1)^5 = 0.59049

P(X=2.5) = C(5,1) * 0.1^1 * (1-0.1)^4 = 0.32805

P(X=0) = C(5,2) * 0.1^2 * (1-0.1)^3 = 0.0729

P(X=-1) = C(5,3) * 0.1^3 * (1-0.1)^2 + C(5,4) * 0.1^4 * (1-0.1) + C(5,5) * 0.1^5

= 0.00856

这台机器一周内可获利的均值 E(X) = 5*P(X=5)+2.5*P(X=2.5)+0+(-1)*P(X=-1)

= 5*0.59049 + 2.5*0.32805 - 0.00856

= 3.764015

这台机器一周内可获利的均值是 37640.15元