Question

you roll two fair dice.
a) if you were to list the sample space for this problem, how many equally likely outcomes would there be?
b) find the probability of getting a sum of 8. leave your answer as a fraction.
c) find the probability of getting a 4 on the first die. leave your answer as a fraction.
d) find the probability of getting a sum of 8 and a 4 on the first die. leave your answer as a fraction.
e) find the probability of getting a sum of 8 or a 4 on the first die. leave your answer as a fraction.
f) find the probability of getting doubles. leave your answer as a fraction.
g) find the probability of getting both dice to be even numbers. leave your answer as a fraction.
h) find the probability of getting doubles and both dice to be even numbers. leave your answer as a fraction.
i) find the probability of getting doubles or both dice to be even numbers. leave your answer as a fraction.

Explanation:

Step1: Calculate sample - space size

Each die has 6 possible outcomes. When rolling two dice, by the multiplication principle, the number of equally - likely outcomes in the sample space $n(S)=6\times6 = 36$.

Step2: Find outcomes with sum of 8

The pairs that sum to 8 are $(2,6),(3,5),(4,4),(5,3),(6,2)$. So $n(\text{sum}=8)=5$. The probability $P(\text{sum}=8)=\frac{n(\text{sum}=8)}{n(S)}=\frac{5}{36}$.

Step3: Find outcomes with 4 on first die

The pairs with 4 on the first die are $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$. So $n(\text{first die}=4)=6$. The probability $P(\text{first die}=4)=\frac{n(\text{first die}=4)}{n(S)}=\frac{6}{36}=\frac{1}{6}$.

Step4: Find outcomes with sum of 8 and 4 on first die

The pair that has a sum of 8 and 4 on the first die is $(4,4)$. So $n(\text{sum}=8\cap\text{first die}=4)=1$. The probability $P(\text{sum}=8\cap\text{first die}=4)=\frac{n(\text{sum}=8\cap\text{first die}=4)}{n(S)}=\frac{1}{36}$.

Step5: Use the addition rule for sum of 8 or 4 on first die

By the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here $A$ is the event of getting a sum of 8 and $B$ is the event of getting 4 on the first die. So $P(\text{sum}=8\cup\text{first die}=4)=\frac{5}{36}+\frac{6}{36}-\frac{1}{36}=\frac{5 + 6-1}{36}=\frac{10}{36}=\frac{5}{18}$.

Step6: Find outcomes of doubles

The doubles are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$. So $n(\text{doubles})=6$. The probability $P(\text{doubles})=\frac{n(\text{doubles})}{n(S)}=\frac{6}{36}=\frac{1}{6}$.

Step7: Find outcomes of both dice even

The first die has 3 even numbers (2, 4, 6) and the second die has 3 even numbers. So $n(\text{both even})=3\times3 = 9$. The probability $P(\text{both even})=\frac{n(\text{both even})}{n(S)}=\frac{9}{36}=\frac{1}{4}$.

Step8: Find outcomes of doubles and both dice even

The pairs that are doubles and both even are $(2,2),(4,4),(6,6)$. So $n(\text{doubles}\cap\text{both even})=3$. The probability $P(\text{doubles}\cap\text{both even})=\frac{n(\text{doubles}\cap\text{both even})}{n(S)}=\frac{3}{36}=\frac{1}{12}$.

Step9: Use the addition rule for doubles or both dice even

By the addition rule $P(\text{doubles}\cup\text{both even})=P(\text{doubles})+P(\text{both even})-P(\text{doubles}\cap\text{both even})=\frac{6}{36}+\frac{9}{36}-\frac{3}{36}=\frac{6 + 9-3}{36}=\frac{12}{36}=\frac{1}{3}$.

Answer:

a) 36
b) $\frac{5}{36}$
c) $\frac{1}{6}$
d) $\frac{1}{36}$
e) $\frac{5}{18}$
f) $\frac{1}{6}$
g) $\frac{1}{4}$
h) $\frac{1}{12}$
i) $\frac{1}{3}$