Stats Probability Question. Find P(A|X), P(B|X), and P(C|X).

Question

Suppose we have three mutually exclusive events A, B, and C such that P(A) = .4, P(B) = .4, and P(C) = .2.

Also, for some event X, P(X|A) = .75, P(X|B) = .5, and P(X|C) = .5.

Find P(A|X), P(B|X), and P(C|X).

Answer 1

Since events A, B, C are mutually exclusive and P(A) + P(B) + P(C) =1,

regardless of what type of event X is,  we must have

 

P(A|X) +  P(B|X) + P(C|X) = 1.      Thus sum condition  can be combined with Bayes's theorem to find the desired conditional probabilities. 

 

Bayes's theorem  is:

 

P(A|X) = P(X|A) P(A) / P(X)  ;

P(B|X) = P(X|B) P(B) / P(X)  ;

P(C|X) = P(X|C) P(C) / P(X)

 

Plugging in the given probabilities, the sum condition gives

 

1 = .75 x .4 /P(X) + .5 x .4 /P(X) + .5 x .2 /P(X)

This equation can be solved for P(X)  yielding   P(X) = .6

 

Plugging this into the Bayes formulas gives:

 

P(A|X) = .75 x .4 / .6 = 0.5   (= 1/2)

P(B|X)  = .5 x .4 / .6 =  .3333   ( = 1/3)

P(C|X)  = .5 x .2 / .6 = /16666   (= 1/6)

 

Answer 2

Hi Kristin,

 

You can use Bayes' Theorem to find these probabilities.

 

P(A|X) = P(X|A)*P(A) / P(X)

or B or C are the same, just replace the A's with B or C.

 

The only thing you're missing is P(X). However, you can find P(X) because you know that A, B and C are mutually exclusive, and they coverall the possibilities (P(A) + P(B) + P(C) = 1).

P(X) = P(X|A)*P(A) + P(X|B)*P(B) + P(X|C)*P(C)

 = 0.75*0.4 + 0.5*0.4 + 0.5*0.2 = 0.3 + 0.2 + 0.1 = 0.6

 

So, back to Bayes' Theorem,

 

P(A|X) = P(X|A)*P(A) / P(X)

 = 0.75*0.4/0.6 = 0.5

 

And so on, for B and C.