select the correct answer from each drop - down menu. two six - sided fair dice are rolled simultaneously…

Explanation:

Step1: Find total outcomes

When two six - sided dice are rolled, the total number of outcomes is $n(S)=6\times6 = 36$.

Step2: Define events for first part

Let $A$ be the event that the sum of the two numbers is greater than 3 and $B$ be the event that the sum of the two numbers is not greater than 5. The possible sums not greater than 5 are 2, 3, 4, 5. The pairs for sum = 2: $(1,1)$; sum = 3: $(1,2),(2,1)$; sum = 4: $(1,3),(3,1),(2,2)$; sum = 5: $(1,4),(4,1),(2,3),(3,2)$. So $n(B)=1 + 2+3 + 4=10$. The pairs in $B$ that also satisfy $A$ (sum > 3) are the pairs with sum 4 and 5, so $n(A\cap B)=3 + 4 = 7$. By the formula for conditional probability $P(A|B)=\frac{n(A\cap B)}{n(B)}$, we have $P(A|B)=\frac{7}{10}$.

Step3: Define events for second part

Let $C$ be the event that the sum of the two numbers is an odd number and $D$ be the event that one of the numbers is a 6. The pairs where one of the numbers is 6 are: $(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$. So $n(D)=11$. The pairs in $D$ where the sum is odd are: $(1,6),(3,6),(5,6),(6,1),(6,3),(6,5)$. So $n(C\cap D)=6$. By the formula for conditional probability $P(C|D)=\frac{n(C\cap D)}{n(D)}$, we have $P(C|D)=\frac{6}{11}$.

Answer:

The probability that the sum of the two numbers rolled is greater than 3, given that the sum of the numbers is not greater than 5, is $\frac{7}{10}$. The probability that the sum of the two numbers rolled is an odd number, given that one of the numbers is a 6, is $\frac{6}{11}$.