Probability Of a 4 sided die A fair $4$-sided die is rolled twice and we assume that all sixteen
possible outcomes are equally likely. Let $X$ and $Y$ be the result of the $1^{\large\text{st}}$ and the
$2^{\large\text{nd}}$ roll, respectively. We wish to determine the conditional probability $P(A|B)$
where
$A = \max(X,Y)=m$
and
$B= \min(X,Y)=2,\quad m\in\{1,2,3,4\}$.
Can somebody first explain me this question and then explain its answer. I'm having trouble in approaching it.
 A: Initially there are 16 equally likely possibilities for the two dice rolls:
            First
       1    2    3    4

S  1   X    X    X    X
e
c  2   X    X    X    X
o
n  3   X    X    X    X
d
   4   X    X    X    X

If the minimum roll is $2$ then there are 5 equally likely possibilities for the two dice rolls:
            First
       1    2    3    4

S  1        
e
c  2        X    X    X
o
n  3        X        
d
   4        X        

So for the conditional probability of the maximum


*

*$P(A=1|B=2)=0$, 

*$P(A=2|B=2)=\frac{1}{5}$, 

*$P(A=3|B=2)=\frac{2}{5}$, 

*$P(A=4|B=2)=\frac{2}{5}$. 

A: Q: Initially you have 16 possible outcomes.A: (1,1) (1,2) (1,3) ... (4,3) (4,4)
Q: Assume that B is true. (How many outcomes are there left?)A: min(X,Y)=2. So 5 outcomes left: (2,2) (2,3) (2,4) (3,2) (4,2)
Q: What values can m take? A: m is the max of two numbers up to 4. So m can take values 1,2,3,4
Q: How many possible outcomes for each value of m?A: Taking the 3 outcomes left: max(2,2)=2. max(2,3)=max(3,2)=3. max(2,4)=max(4,2)=4
m=1: P(A|B)=0/5
m=2: P(A|B)=1/5
m=3: P(A|B)=2/5
m=4: P(A|B)=2/5

