N trials, 3 possible outcomes

#1
I have n independent trials, where the result of each trial is either A, B or C, with probabilities a, b and c respectively, so that a + b + c = 1.

What is the probability that all results are either A or B (in any combination/permutation, but no C), including at least one A and at least one B?

Is it a*b*(a+b)^(n-2)?

Or n*(n-1)*a*b*(a+b)^(n-2)?

Or something else?

Getting confused, please help!
 
Last edited:

BGM

TS Contributor
#2
Note that \( A, B, C \sim \text{Multinomial}(n; a, b, c) \)

The required probability is

\( \Pr\{A \geq 1, B \geq 1, C = 0\} \)

\( = \Pr\{C = 0\} - \Pr\{A = 0 \cup B = 0, C = 0\} \)

\( = \Pr\{C = 0\} - \Pr\{A = 0, C = 0\} - \Pr\{B = 0, C = 0\}
+ \Pr\{A = 0, B = 0, C = 0\} \)

\( = (a + b)^n - b^n - a^n \)
 
#3
Thanks for answering, but now I'm *really* confused...

Why not n*(n-1)*a*b*(a+b)^(n-2), based on the probability that one result is A, one result is B, and the rest (n-2) are either A or B?

Isn't that the same outcome as described in my original post? So why the discrepancy???

I'm just trying to understand why we don't get the same answer, if there's a flaw in my (or your) logic, which answer is correct and why?

Please help! :-(
 
Last edited:
#4
Perhaps the problem is that I try to calculate the multinomial probability as Pr(A=1, B=1, A or B = n-2), but is that not right?

Is it because A, B and (A or B) aren't mutually exclusive outcomes as they should be? I'm just guessing...

Please explain what I'm doing wrong.

HELP!
 
Last edited:

rogojel

TS Contributor
#5
hi,
sorry to increase the confusion but I would look at this in a simple way as a binomial trial with an event C of probability c and an event NotC with a probability (1-c) a.k.a. a+b. The probability of not having any Cs in N trials is simply (1-c)^N - which is (a+b)^N. Do I miss something?
 
#6
"Do I miss something?" - Yes, not only should all results be A or B (no C), but at least one result should be A (not B), and at least one result should be B (not A).

In other words, all-A and all-B are excluded, so that makes it a bit more complicated.
 

rogojel

TS Contributor
#7
Oh, I see.

Then it is clear :) The event of No Cs can happen in three ways that are mutually exclusive: NC= PA or PB or MAB where PA is a series of pure As, PB the same for Bs and MAB is a mixed series with As and Bs. Because the 3 are mutually exclusive P(NC)=P(PA)+P(PB)+P(MAB).

P(PA)=a^N, P(PB)=b^N and P(NC)=(a+b)^N. So, all you need is to express P(MAB) :)

Regards
 
#8
Thanks, that makes a lot of sense, so now I can see (and more importantly, understand) that the correct answer is indeed (a+b)^n - a^n - b^n ...

... which means that the original probability I calculated is wrong, and I guess (still not sure) this is because the multinomial formula can only be used when the elements are mutually exclusive?
 

rogojel

TS Contributor
#9
Why not n*(n-1)*a*b*(a+b)^(n-2), based on the probability that one result is A, one result is B, and the rest (n-2) are either A or B?
I think the problem is the way you count the cases - I just learned to not trust my intuition there.