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；发生2次故障的利润0元；发生3次或3次以上故障要亏损1万元,问这台机器一周内可获利的均值是多少

5个工作日获利 X 万元,X = 5,2.5,0,-1P(X=5) = (1-0.1)^5 = 0.59049 P(X=2.5) = C(5,1) * 0.1^1 * (1-0.1)^4 = 0.32805P(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...