# Probability problem

Discussion in 'Math' started by boks, Jan 7, 2009.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0

The numbers represent the probability that each component works.

a) What's the probability that the entire system works?
b) Given that the system works, what's the probability that the component A is not working?

a) P=.75112, easy
b) Don't know what to do here....

2. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Which one is component A? They're all labeled with numbers, not letters.
Also, could you outline how you got the answer you showed for part a?

3. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
I mean part 1...

a) Either both 1 and 2, or all of 3, 4 and 5 have to work for the entire system to work.

4. ### BillO Distinguished Member

Nov 24, 2008
985
136
I think we are missing some info here.

Oh, okay I see it...never mind.

Last edited: Jan 7, 2009
5. ### steveb Senior Member

Jul 3, 2008
2,433
469
You know the probablity of the system working is 0.75112, call this P0.

There are only two ways that the system is working but part 1 is not working. That is, 3 4 and 5 must be working, part 1 must not be working, and part 2 can either be working or not working. Call the probabilities of these two situations, without knowledge that the system is working or not, P1a and P1b.

The probablity that the 1 is not working, given the system is working is
(P1a+P1b)/P0

You should be able to calculate P1a and P1b easily.

6. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,699
909
I took a simpler, maybe wrong approach. The probability that A (1,2) is not working is (1- probability it is working) = 1 - 0.49 = 0.51

The statement that the system is "working" is essentially a red herring. You have already calculated (correctly) what the probability of that is.

John

7. ### BillO Distinguished Member

Nov 24, 2008
985
136
Not true. Every additional piece of information you know about the system changes the probabilities. If the system is operating we know either1 and 2 are working or 3,4 and 5 are working, or both. That is not the same situation as in the first question where we do not know the operational state of the system.

For instance, if I tell you the system is working and component 3 is not, then we know without a doubt that both 1 and 2 are working.

8. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,699
909
True, as I pointed out.

The question is not the operational state of the system. It is simply what the probability of failure in the top ("A") path is.

John