Conditional probability of gates. > 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?

> 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.