# Probability question **URL:** \ **Category:** Factual Questions\ **Created:** [October 16, 2003, 11:43am UTC](https://boards.straightdope.com/t/probability-question/207762 "2003-10-16T11:43:18Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Grey](https://avatars.discourse-cdn.com/v4/letter/g/b782af/32.png) [@Grey](https://boards.straightdope.com/u/Grey)\ **Post date:** [October 16, 2003, 11:43am UTC](https://boards.straightdope.com/t/probability-question/207762/1 "2003-10-16T11:43:18Z")
If I have 4 possible states (A,B,C,D) that can lead to a system state F by the following relation (A OR B) AND (C OR D) = F how would I find the probability of the final failure state happening? I remember doing this in school, but I’ll be damned if I can remember how to do it now. ---
**Author:** ![Jabba](https://avatars.discourse-cdn.com/v4/letter/j/b5e925/32.png) [@Jabba](https://boards.straightdope.com/u/Jabba)\ **Post date:** [October 16, 2003, 11:54am UTC](https://boards.straightdope.com/t/probability-question/207762/2 "2003-10-16T11:54:54Z")
You need to know the probabilities of A, B, C and D, including whether or not they are independent. Assuming they are independent, the probability of (A or B) is P(A) + P(B) - P(A)P(B) and the probability of (C or D) is P© + P(D) - P©P(D) and the overall probability is the product of these two. ---
**Author:** ![croakdale](https://avatars.discourse-cdn.com/v4/letter/c/b77776/32.png) [@croakdale](https://boards.straightdope.com/u/croakdale)\ **Post date:** [October 16, 2003, 12:50pm UTC](https://boards.straightdope.com/t/probability-question/207762/3 "2003-10-16T12:50:29Z")
he didn’t say exlusive OR I think it should be ( P(A) + P(B) ).( P© + P(D) ) ---
**Author:** ![MartinL](https://avatars.discourse-cdn.com/v4/letter/m/ebca7d/32.png) [@MartinL](https://boards.straightdope.com/u/MartinL)\ **Post date:** [October 16, 2003, 12:52pm UTC](https://boards.straightdope.com/t/probability-question/207762/4 "2003-10-16T12:52:51Z")
In addition to what **Jabba** posted (I totally agree), the probability of (A or B) generally is P(A) + P(B) - P(A and B), where P(A and B) = P(A|B) \* P(B) = P(B|A) \* P(A) If they are independent, P(A|B) = P(A) and P(B|A) = P(B), which gives you the formulas as seen in the previous post. ---
**Author:** ![MartinL](https://avatars.discourse-cdn.com/v4/letter/m/ebca7d/32.png) [@MartinL](https://boards.straightdope.com/u/MartinL)\ **Post date:** [October 16, 2003, 12:55pm UTC](https://boards.straightdope.com/t/probability-question/207762/5 "2003-10-16T12:55:00Z")
> [@](#): > > \*Originally posted by croakdale \* > \*\*he didn’t say exlusive OR > I think it should be > ( P(A) + P(B) ).( P© + P(D) ) \*\* Neither **Jabba** nor myself assumed an exclusive OR. Subtracting P(A and B) merely makes sure that this probability isn’t counted twice (the intersection is contained both in P(A) and in P(B)). ---
**Author:** ![croakdale](https://avatars.discourse-cdn.com/v4/letter/c/b77776/32.png) [@croakdale](https://boards.straightdope.com/u/croakdale)\ **Post date:** [October 16, 2003, 1:06pm UTC](https://boards.straightdope.com/t/probability-question/207762/6 "2003-10-16T13:06:53Z")
whoops, you are right MartinL & Jabba, please disregard my post. Serves me right for jumping in… sorry! ---
**Author:** ![Grey](https://avatars.discourse-cdn.com/v4/letter/g/b782af/32.png) [@Grey](https://boards.straightdope.com/u/Grey)\ **Post date:** [October 16, 2003, 1:11pm UTC](https://boards.straightdope.com/t/probability-question/207762/7 "2003-10-16T13:11:16Z")
Thanks folks, helps a lot. ---
**Author:** ![Jurph](https://avatars.discourse-cdn.com/v4/letter/j/50afbb/32.png) [@Jurph](https://boards.straightdope.com/u/Jurph)\ **Post date:** [October 16, 2003, 1:19pm UTC](https://boards.straightdope.com/t/probability-question/207762/8 "2003-10-16T13:19:10Z")
If either (A or B) simultaneous with (C or D) can lead to the state F, then A and B are counted as parallel, as are C and D. The probability (A or B) is (A + (1-A)(B)).\* So if A can happen 90% of the time, and B can also happen 90% of the time, you have .90 (chance that A can happen) + .10 (chance that A doesn’t happen) \* .90 (chance that B will happen when A fails) = .99 The probability of the system achieving F on any one trial is (A + (1-A)(B)) \* (C + (1-C)(D)). It makes it easier to think of A, B, C, and D as “black boxes” that are chained together. A and B share an input and output cable, as do C and D. C-D’s input cable is directly connected to A-B’s output. Each box has a known probability of successfully passing the signal. Now you can simply think of this as a system reliability problem. If you send a signal into the ABCD box, what are the odds that you’ll get the same signal back out? If the A and B are both .90 reliable and C and D are both .80 reliable, then R(ab) = .99 and R(cd) = .96, and R(abcd) = .9506 . You’ve combined four components, all with reliability at or below 90%, and built a machine that is (only just) better than 95%. (by the way, I didn’t realize it, but **Jabba** ’s post contains the same formula as mine in a slightly different form). Enjoy, Jurph