A hydraulic structure has four gates which operate independently. The probability of failure of each gate is 0.2. given that gate 1 has failed. Find the probability that both gates 2 and 3 will fail.

A hydraulic structure has four gates which operate independently. The probability of failure of each gate is 0.2. given that gate 1 has failed, the probability that both gates 2 and 3 will fail is 

(A) 0.240 

(B) 0.200 

(C) 0.040 

(D) 0.008 

(GATE 2004) 

Answer: (C) 0.040 

Explanation: 

Let $G_1$, $G_2$, $G_3$ and $G_4$ be the events of failure of the four gate respectively. 

Here, $G_1$, $G_2$, $G_3$ and $G_4$ are independent and $P(G_1) = 0.2$, $P(G_2) = 0.2$, $P(G_3) = 0.2$ and $P(G_4) = 0.2$. 

Therefore, the probability that both gates 2 and 3 will fail is 

$= P(G_1 \cap G_3)$ 

$= P(G_1) \times P(G_3)$, since $G_2$ and $G_3$ are independent events 

$= 0.2 \times 0.2$ 

$= 0.04$