Question

the first outcome is one of X, Y, and Z, and the second outcome in each case is one of A and B.

Pr[X] = [ 4/7]
Pr[Y] = [ 1/7]
Pr[Z] = [ 2/7]
Pr[A | X] = [ 2/5]
Pr[B | X] = [ 3/5]
Pr[A | Y] = [ 1/3]
Pr[B | Y] = [ 2/3]
Pr[A | Z] = [ 5/8]
Pr[B | Z] = [ 3/8]

Find the missing probability: Pr[(X or Y) | B]

Answer 1

First, recognize from the rule for unions that: \(P(X~or~Y|B)=P(X \cup Y|B)=P(X|B)+P(Y|B)-P(X \cap Y|B)\). But, since the first event can only X or Y or Z , you can't have two of these events occurring simultaneously meaning that \(P(X \cap Y|B)=0\)

Since X, Y and Z form a partition over the sample space, you can use Bayes' Theorem, and find that
\[P(X|B)=\frac{P(B|X)P(X)}{P(B|X)P(X)+P(B|Y)P(Y)+P(B|Z)P(Z)} \\
= \frac{(3/5)(4/7)}{(3/5)(4/7)+(2/3)(1/7)+(3/8)(2/7)}\]
Similarly,
\[P(Y|B)=\frac{P(B|Y)P(Y)}{P(B|X)P(X)+P(B|Y)P(Y)+P(B|Z)P(Z)} \\
= \frac{(2/3)(1/7)}{(3/5)(4/7)+(2/3)(1/7)+(3/8)(2/7)} \]

Answer 2

then just add those 2 probabilities together

Answer 3

This was correct, great explanation - thank you so much!

Answer 4

awesome :)