When can you not use inclusion-exclusion in probability?

Question

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?

Answer 1

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}}$

Answer 2

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 $2/3$.)