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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am performing a product trial. The trial has three possible outcomes with the following probabilities:

Outcome A = .3

Outcome B = .5

Outcome C = .2

If I perform five trials, what are the odds of B and C occurring at least once throughout testing?

share|cite|improve this question
up vote 3 down vote accepted

The complement of this event is that only $A$ and $C$ occur or only $A$ and $B$ occur. The probabilities for those are $(.3+.2)^5=.5^5$ and $(.3+.5)^5=.8^5$, respectively. However, adding these two double-counts the possibility that only $A$ occurs, so we have to subtract $.3^5$, for a total of

$$ .5^5+.8^5-.3^5=0.3565\;. $$

Thus the probability of your event is $1-0.3565=0.6435$.

share|cite|improve this answer
Thank you Joriki!! – CodyBugstein Sep 27 '12 at 12:23
@Imray: You're welcome! – joriki Sep 27 '12 at 12:25
I'm a bit confused now... can you explain why we're overcounting A? Isn't it a separate possible outcome that it can be A every time? i.e., there are three possible complementary situation, only A's and B's, only A's & C's, and only A's. What am I missing? – CodyBugstein Sep 27 '12 at 15:59
To add to my previous comment, you say that the complementary outcomes are only A, C or only A,B. What about only A occurring? – CodyBugstein Sep 27 '12 at 16:06

This is the complement of the probability that all five are outcome A. Assuming independence this complement is 1 - .3^5 = .99757

Sorry this is for B or C occurring at least once. Please downvote more.

share|cite|improve this answer
There should be a delete button at the bottom of your answer. If not, click on flag and ask a moderator to remove this. – Ross Millikan Sep 27 '12 at 3:36
@RossMillikan Can you explain to me why Joriki omitted the possibility of all outcomes resulting in A? Or is it a typo? – CodyBugstein Sep 28 '12 at 4:18
@Imray: not a typo. joriki is pretty careful. The point is that only A occurs got counted twice: it is both in the only A and B and also in the only A and , so you need to add back one of them. You could see – Ross Millikan Sep 28 '12 at 4:22
Ok I'm pretty sure I got it... basically, within $(.3 + .2)^5$, there is the possibility of only the ".3" portion showing up in the outcome. Since we add the same thing in $(.3 + .5)^5$, we have to subtract a full $.3^5$ from the total – CodyBugstein Sep 28 '12 at 4:40
@Imray: Yes, that's right. Sorry I didn't get around to responding earlier myself. – joriki Sep 28 '12 at 15:31

Your Answer


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.