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A parallel system functions whenever at least one of its components works. Consider a parallel system of three components, and suppose that each component works independently with probability $0.5$.

Find the conditional probability that component 1 works given that the system is functioning.

lets say $A_i =$ the event that the $i$th component works. If they're not mutually exclusive, what would $A_1 \cap A_2$ be??

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    Of course, if two events are independent *and* mutually exclusive, then one of them is impossible...2012-09-20

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If $A_1$, $A_2$ are independent, then $\Pr(A_1\cap A_2) = \Pr(A_a)\cdot\Pr(A_2) = (0.5)(0.5) = 0.25 \ne 0.$ If $A_1$ and $A_2$ were mutually exclusive, then that probability would be $0$.

The question in your title asks why they are not mutually exclusive. The above should answer that.

In the body of your question you ask what $A_1\cap A_2$ is. It's just the event that the first two components both work.

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    this is a good answer Thanks!!!2012-09-20
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For the conditional probability, you can see that there are 8 equally likely outcomes possible for the functionality of the components, only one of which results in the system not working.

Of the seven working systems, three have component #1 not working and four have component #1 working.

So, the condition probability that component #1 is working, given that the whole system is working is $4/7$.

(This whole setup is just like a three coin flip experiment, where you are asking "If a friend flipped a coin three times and told you at least one head appeared, what is the probability a heads appeared on your friend's first flip?"