Do not understand this basic conditional probability This example came from Tsitsiklis' Intro to Probability 2nd edition.
I don't understand why is $P(A|B)=2/5$ when $m=3$ or $m=4$. Surely it should be $4/5$?
I get that $P(A|B)=1/5$ when $m=2$, because when we restrict our domain to the highlighted area (i.e. $B$), only 1 out of a total of 5 combinations matches the description.
So if we apply the same logic, out of the 5 combinations, $(4,2),(3,2),(2,3),(2,4)$ fulfill the descriptions - so shouldn't it be $4/5$?
In fact in the MITx course that goes with this textbook, Tsitsiklis mentioned that $P(M=3|B)=2/5$ which makes sense; so how can the probability stays constant, when $m$ is now larger(since we allow $m=3$ or $m=4$), which surely should allow more possibilities?


 A: I think the confusion stems from the interpretation of the phrase $m = 3$ or $m = 4$.
The probability $P(A \mid B)$ when $m = 3$ is indeed $2/5$ because the only pairs that fall under event $A = \left\{\max(X, Y) = 3\right\}$ are $(2, 3)$ and $(3, 2)$. Similarly, the probability $P(A \mid B)$ when $m = 4$ is also $2/5$ because the only pairs that fall under $A = \left\{\max(X, Y) = 4\right\}$ are $(2, 4)$ and $(4, 2)$.
Hence, it is true that $P(\left\{\max(X, Y) = m\right\} \mid B)$ is $2/5$ when $m = 3$ or $m = 4$. Here, we note that the event $A$ is not defined as having $\max(X, Y) = 3 \text{ or } 4$.
But if we take $A = \left\{\max(X, Y) = 3 \text{ or } 4\right\}$, then in this case, the pairs that fall under event $A$ are $(2, 3)$, $(2, 4)$, $(3, 2)$ and $(4, 2)$. In this case, $P(A \mid B)$ would be $4/5$. I think you misunderstood the book to imply this case.
We have to be careful with how we interpret the phrase $m = 3$ or $m = 4$ in the book. Indeed, if we take $m = 3$ or $m = 4$, we get $2/5$ as the answer in both cases if we follow the definition of $A$ in the book. However, if we want to find the probability $P(A \mid B)$ with $A$ defined differently as the event where $\max(X, Y)$ is $3$ or $4$, then in this case, $4/5$ is the answer.
Also, at the end of the page, $P(\left\{\max(X, Y) = m\right\} \mid B)$ is presented as a piece-wise function that varies each answer for different values of $m$. In this case, the book is indeed correct in saying that $P(\left\{\max(X, Y) = 3\right\} \mid B) = 2/5$ when $m = 3$ or $m = 4$.
