An automobile engine has three independent mounts, their probabilities of breaking in a 10-year period are 0.2, 0.4, and 0.3, respectively. If the driver of the vehicle hears the characteristic noise of two broken mounts, determine the probability that mounts one and two are broken.
I have done problems like this one but never with this kind of question where you have to find the probability of two things breaking at the same time. Can someone help please?
Okay, I'm going to give it a try
We know that for sure two of the mounts are broken. We also know that the mounts breaking are independent of each other(at least that's what I gather from the problem's wording). Let events $A, B, C$ denote the respective breaking of the mounts:
$$P(A) = .2, P(B) = .4, P(C) = .3$$
Let's first find the probability that two mounts(and only two mounts) break:
$$P(\text{Two mounts break}) = P(A\cap B) + P(A\cap C) + P(B\cap C) - 3P(A\cap B\cap C)$$ Where the last term eliminates the probability that all three of them break because we have only heard two mounts break.
The question asks us for $$P(A \cap B | \text{Two mounts break}) = \frac{P(\text{Two mounts break}|A \cap B)\times P(A \cap B)}{P(\text{Two mounts break})}$$ Where we can likely assume that $P(A \cap B) = .2 \times .3 = .06$
Can you take it from here?