3
$\begingroup$

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??

  • 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.2012-09-20
  • 0
    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...2012-09-20
  • 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!]2012-09-20
  • 0
    what statement implies that it isn't so?2012-09-20
  • 0
    The events are independent (and have nonzero probabilities).2012-09-20
  • 0
    @user133466 : Independence implies it isn't so.2012-09-20
  • 0
    @CliveN. independence does not guarantee mutual exclusivity!2012-09-20
  • 0
    @user133466: Yes it does. If $A$ and $B$ are independent, and $\mathbb{P}(A) > 0$ and $\mathbb{P}(B) > 0$, then $\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B) > 0$. However, $\mathbb{P}(\varnothing) = 0$; so in particular, $A \cap B \ne \varnothing$. [I should really have put all this in an answer.]2012-09-20
  • 0
    @CliveN. user133466 just exclaimed "$I$ does not imply $M$!" but you just correctly proved that "$I$ implies not $M$". So you both appear to be on the same page :)2012-09-20
  • 0
    @rschwieb: Hah, oh dear, I need to get my eyes checked.2012-09-20
  • 0
    in fact this question is an example that components are independent, but not mutually exclusive =)2012-09-20
  • 0
    Of course, if two events are independent *and* mutually exclusive, then one of them is impossible...2012-09-20

2 Answers 2

2

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.

  • 0
    this is a good answer Thanks!!!2012-09-20
0

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?"