# 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?

• You're fairly new here, and hence don't know. You should not post unsearchable pictures of text because the site's search functions cannot find the problem. Perhaps this exact question has already been asked and answered. Or will be asked and answered. Please typeset using MathJax. Dec 16, 2020 at 5:02
• oops sorry! Ill type it out Dec 16, 2020 at 5:13

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

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

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 $$2/3$$.)