TITLE: Conditional probability of gates.
QUESTION [6 upvotes]: A hydraulic stucture has four gates which operate independently . The probability of failure of each gate is 0.1.
Given that gate 1 has failed, what is the probability that gate 2 and gate 3 will fail ?


I think the answer is $P(G_2 \cap G_3) / P(G_1) = (0.1*0.1)/0.1 = 0.1$
Am I correct?

REPLY [8 votes]: I think the answer is $P(G_2 \cap G_3) / P(G_1) = (0.1*0.1)/0.1 = 0.1$
Am I correct?

No; not quite.
The definition of conditional probability sais: $\mathsf P(B\mid A)=\mathsf P(A\cap B)\big/\mathsf P(A)$, so therefore:
$$\mathsf P(G_2\cap G_3\mid G_1) = \dfrac{\mathsf P(G_1\cap G_2\cap G_3)}{\mathsf P(G_1)}$$
Otherwise, you correctly used the product rule for the intersection of independent events.  So then:
$$\require{cancel}\mathsf P(G_2\cap G_3\mid G_1) ~{= \dfrac{\cancel{~\mathsf P(G_1)~}\mathsf P(G_2)~\mathsf P(G_3)}{\cancel{~\mathsf P(G_1)~}}\\=~0.01}$$
That is all.