Question

fill in the blank 20 points
a) a pair of dice are tossed. find the probability that the sum on the two dice is 4 given that the sum is even. write the answer as a fraction in simplest form.
type your answer...
b) a pair of dice are tossed. what is the probability that the sum on the two dice is 8 given that the sum is more than 7? write the answer as a fraction in simplest form.
type your answer...
(chart: outcome of first die (1 - 6) and outcome of second die (1 - 6) with sums in the grid)

Explanation:

Part A

Step1: Identify total even sums

When two dice are tossed, the total number of outcomes is \(6\times6 = 36\). The even sums are 2, 4, 6, 8, 10, 12. Count the number of ways to get each:

• Sum = 2: (1,1) → 1 way
• Sum = 4: (1,3), (2,2), (3,1) → 3 ways
• Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 ways
• Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways
• Sum = 10: (4,6), (5,5), (6,4) → 3 ways
• Sum = 12: (6,6) → 1 way

Total even sum outcomes: \(1 + 3 + 5 + 5 + 3 + 1 = 18\)

Step2: Find conditional probability

Let \(A\) be the event that sum is 4, \(B\) be the event that sum is even. We need \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Since \(A\subseteq B\), \(P(A\cap B)=P(A)\). Number of ways for \(A\) is 3, for \(B\) is 18. So \(P(A|B)=\frac{3}{18}=\frac{1}{6}\)

Step1: Identify sums more than 7

Sums more than 7 are 8, 9, 10, 11, 12. Count the number of ways:

• Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways
• Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 ways
• Sum = 10: (4,6), (5,5), (6,4) → 3 ways
• Sum = 11: (5,6), (6,5) → 2 ways
• Sum = 12: (6,6) → 1 way

Total outcomes for sum >7: \(5 + 4 + 3 + 2 + 1 = 15\)

Step2: Find conditional probability

Let \(C\) be the event that sum is 8, \(D\) be the event that sum >7. \(P(C|D)=\frac{P(C\cap D)}{P(D)}\). Since \(C\subseteq D\), \(P(C\cap D)=P(C)\). Number of ways for \(C\) is 5, for \(D\) is 15. So \(P(C|D)=\frac{5}{15}=\frac{1}{3}\)

Answer:

\(\frac{1}{6}\)

Part B