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I have four independent events, each with three possible outcomes.

Probabilities of each outcome:

A: $10\%$ B: $70\%$ C: $20\%$

How to calculate the probability of $2$ B's, $1$ A, and $1$ C? Or any other combination.

If one were to make a table of all outcome combinations, should the probabilities sum to $100\%$?

I had an answer, but they didn't add to $100\%$, so I am checking with you guys.

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2 Answers 2

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I'll address your main question first, then the sidenotes:

2B's 1A and 1C, you can have it in multiple ways, ABBC, BBAC, etc.

Of the 4 events you have to choose 2 to be B = $\binom{4}{2}$ ways Of the remaining 2, you choose 1 to be and 1 to be C = 2 ways

So there are $\binom{4}{2} * 2 = 12 $ ways for this to happen, and the probability of a given combination is $0.1 * 0.7^2 * 0.2$ = 0.0098, making the total probability 12*0.0098 = 0.1176

For the sidenotes:

There are $3^4 = 81$ possible outcomes (with unequal probability weights. e.g. AAAA is very unlikely compared to BBBB)

For each event you have 3 choices to make (A, B or C)

As there are 4 events, you have 3*3*3*3 = 81 possibilities and yes they will all sum to 100%

Note: I read your comment on an answer, and I'll try to explain why that listing is not correct:

Thanks for that. I understand how you got 12, perhaps I didnt word the question correctly, as I got 15 AAAA AAAB AAAC AABB AABC AACC ABBB ABBC ABCC ACCC BBBB BBBC BBCC BCCC CCCC

You are not paying attention to the order of events, which is not right because AACC can happen in 6 ways, which makes it's probability the sum of the probabilities of:

ACAC ACCA CCAA CAAC AACC CACA

The probability of each event is the same 0.1 * 0.2 * 0.1 * 0.2, so the total probability of your "AACC" event (ignoring order) will be 6*0.0004 = 0.0024

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  • $\begingroup$ Thank you, your last point was what I missing. $\endgroup$
    – eredd
    Apr 11, 2016 at 19:39
  • $\begingroup$ @eredd glad it helped you! $\endgroup$
    – sanketalekar
    Apr 11, 2016 at 19:59
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This can be broken down into two problems:

Given 2 Bs, 1 A, and 1 C, in how many ways can I arrange these events?

This is equivalent to asking how many unique ways I can permute $ABBC$, which is equal to $\frac{4!}{2!1!1!}$ = 12.

Now that I have every orientation of $ABBC$, what's the chance that it happened in the first place? That's equal to $Pr = (0.1){(0.7)^2}(0.2) = 0.0098$

Multiplying these together, we get: $12 * 0.0098 = 0.1176$

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  • $\begingroup$ Thanks for that. I understand how you got 12, perhaps I didnt word the question correctly, as I got 15 AAAA AAAB AAAC AABB AABC AACC ABBB ABBC ABCC ACCC BBBB BBBC BBCC BCCC CCCC $\endgroup$
    – eredd
    Apr 11, 2016 at 18:20

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