When can you not use inclusion-exclusion in probability? I came across this question:
Alice and Bob both try to climb a rope. Alice and Bob will get to the top with probability 1/3 and 1/4 respectively. given that exactly one person got to the top, what is the probability that the person is Alice.
The way the textbook solved it, they did $\frac13\times\frac34+\frac23\times\frac14$.
And I got that answer, but I was thinking, why can't we do inclusion-exclusion on the probability that exactly on person makes it to the top. ie, why cant we do $P[\text{Alice} ]+P[\text{Bob}]-P[\text{both}] = 1/3+1/4-1/12$ The calculation yields 2/3, not 5/12
So I'm confused -- when can we not use inclusion-exclusion?
 A: Actually, both answers are wrong.
Breaking up into mutually exclusive cases,
P(only A reaches top) $= \frac 1 3\cdot \frac 3 4 = \frac 3 {12}$
P(only B reaches top) $= \frac 1 4\cdot \frac 2 3 = \frac 2{12}$
P(both reach top) $= \frac 1 3\cdot \frac 1 4 = \frac 1 {12}$
P(neither reach top) $= \frac 2 3 \cdot \frac 3 4 = \frac6 {12}$
The book has computed P(exactly one has reached top), you have computed P(at least one has reached  top),
whereas  you are asked  to find out P(A has reached top |exactly one has reached top)
You should be able to proceed now to get the correct answer
The correct answer is

  P(A reaches top | exactly one reaches top) = $ \dfrac 3 {12}\over {\dfrac3 {12} + \dfrac 2 {12}}$

A: You can use inclusion-exclusion to find $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
But you need the probability that exactly one event happens, which is $P(A\cup B)-P(A\cap B)$. So calculating this with inclusion-exclusion gives $P(A)+P(B)-2P(A\cap B)=5/12$, the same as the other method.
(You also have an error in calculation: $1/3+1/4-1/12=1/2$, not $2/3$.)
