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.

Let $G_i$ be the event "gate $i$ has failed". As the $G_i$ are independent, knowing $G_1$ does not change the probability of an event that depends only on $G_2$ and $G_3$. It means that $$P(G_2\cap G_3|G_1) = P(G_2 \cap G_3)$$ By independence, this probability is equal to $P(G_2)P(G_3) = 0.1\times0.1 = 0.01$.