Looks like a simple probability calculation but am I getting it right?

[EliteLegend]

All of a sudden, I seem to be confused about the union of probabilities. I wanted to see how this would work out but ended up confusing myself even more. Here's a scenario I'm considering:

I have three friends. I know that if a friend posts me a letter upon being asked, it reaches me with a probability of p. I can contact my first friend A, directly but I would have to contact C through B. I wanted to calculate the probability of getting a letter from one of A or C. So, in the end, I have to get one letter atmost. To do this, I ended up saying the following:

P(getting at most one letter)
= P(I receive a letter from A or I receive a letter from C)
= P(I receive a letter from A) + P(I receive a letter from C)
= P(choosing A)*P(letter from A reaching me) + P(choosing C)*P(letter from C reaching me)
= \frac{1}{2}p + P(choosing C)*P(B receives the letter from C)*P(I receive the letter from B)
= \frac{1}{2}p + \frac{1}{2}p*p
= \frac{1}{2}p + \frac{1}{2}p²

I felt something was wrong so I ended up subtracting the P(getting a letter from both A and C) to make the solution as \frac{1}{2}p + \frac{1}{2}p² - \frac{1}{4}p³ from the axiom of probability of unions.

I've complicated the problem so much that I'm now confused... I am making a fundamental mistake in understanding the problem and hope someone can help me out in understanding it right...

PS: I made up this problem myself to try out something interesting so I might have been wrong in framing the question itself... If that's the case, please advice...

 

[Elucidus]

You are trying to find the probability that you receive a letter from exactly one person. I am assuming here that you are contacting both A and B (with the message to be delivered to C). If this is correct, then this is not a conditional probability question since you are not selecting who might reply (they are). Your assumption about the 50-50 split between A and B is off.

The probability that you will receive a reply from A is p; the probability that you'll receive a reply from C (as I understand your question) is p². Since receiving replies from A or C are independent (your question does not give any hint they shouldn't be) then we have:

P(AC)=P(A) \cdot P(C)

(Here AC = A \cap C.)

As always P(XY') = P(X) - P(XY)

You need

P(AC' \cup A'C)= \left[ P(A)-P(AC) \right] + \left[ P(C) - P(AC) \right]

So

P(AC' \cup A'C) = (p - p \cdot p^2) + (p^2 - p \cdot p^2)

= p + p^2 - 2p^3.



--Elucidus

 

[EliteLegend]

Thanks so much... Because the formulation involves union of probabilities, am I right in assuming that if the number of persons increase, then the calculation become more and more complex?

To make it clear, if I add a fourth person with a probability p again, I would end up calculating:

P(AB'C' \cup A'BC' \cup A'B'C) = 3*P(ABC) + P(A) + P(B) + P(C) - 2(P(AB) + P(BC) + P(CA))

in which case, the calculation turns to be crazy. Do you think I'm on the right track? (If I am, is there a general expansion that can be used to do this?)

 

[Elucidus]

Thanks so much... Because the formulation involves union of probabilities, am I right in assuming that if the number of persons increase, then the calculation become more and more complex?

To make it clear, if I add a fourth person with a probability p again, I would end up calculating:

P(AB'C' \cup A'BC' \cup A'B'C) = 3*P(ABC) + P(A) + P(B) + P(C) - 2(P(AB) + P(BC) + P(CA))

in which case, the calculation turns to be crazy. Do you think I'm on the right track? (If I am, is there a general expansion that can be used to do this?)

In the case where the probabilities of each A, B, C, etc are the same, then maybe. Otherwise, no. There are always n combinations that include exactly one success, but calculating the probabilities becomes an exercise in tedium using the inclusion/exclusion principle.

--Elucidus

EDIT: Sorry, forgot to mention your calculation is correct.