Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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??

share|improve this question
1  
Your question title doesn't seem to match your question. $A_1 \cap A_2$ is the event that the $1$st an $2$nd components both work. –  Clive Newstead Sep 20 '12 at 16:08
    
for two events to be mutually exclusive $A_1$ $\cap $ $A_2$ must be nothing. However, I've been told in this question, those two events are not mutually exclusive. I just need some evidence to show that they're not mutually exclusive... –  user133466 Sep 20 '12 at 16:10
2  
Mutual exclusivity would mean that if $A_1$ works then $A_2$ cannot; is there anything in the question that implies that this is so? [In fact, there's something in the statement of the question that implies that it isn't so!] –  Clive Newstead Sep 20 '12 at 16:12
    
what statement implies that it isn't so? –  user133466 Sep 20 '12 at 16:26
    
The events are independent (and have nonzero probabilities). –  Clive Newstead Sep 20 '12 at 16:27
show 7 more comments

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
    
this is a good answer Thanks!!! –  user133466 Sep 20 '12 at 16:34
add comment

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?"

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.