Probability of B and C happening at least once, in five independent trials.

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?

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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 $1-0.3565=0.6435$.

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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.

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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 en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle – 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