#"2 components in series work if both work, so the reliability"#
#"of the last two branches is 0.95*0.95 = 0.9025."#
#"The last two branches are in parallel. Two components in"#
#"parallel work if at least one works, so the reliability of the two"#
#"last branches together is : 0.9025 + 0.9025 - 0.9025²"#
#"= 0.9025 (2 - 0.9025) = 0.99049"#
#"(because P[A or B]=P[A]+P[B]-P[A and B])"#
#"The first two with reliability 0.9 are also in parallel, so the"#
#"overall reliability of them is : 0.9 + 0.9 - 0.9² = 0.9 (2 - 0.9)"#
#"= 0.99."#
#"So we have three reliabilities in series left so the overall"# #"reliability is : 0.99 * 0.95 * 0.99049 = 0.93156."#