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.
$\endgroup$2 Answers
$\begingroup$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 XIf 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
- $P(A=1|B=2)=0$,
- $P(A=2|B=2)=\frac{1}{5}$,
- $P(A=3|B=2)=\frac{2}{5}$,
- $P(A=4|B=2)=\frac{2}{5}$.
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
