Question

an ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. this sum is recorded as the outcome of a single trial of a random experiment. compute the probability of each of the following events. event a: the sum is greater than 8. event b: the sum is not divisible by 6. write your answers as fractions. p(a) = \square p(b) = \square

Explanation:

Step1: Determine total possible outcomes

When rolling a die twice, each roll has 6 outcomes. So total outcomes = \(6\times6 = 36\).

Step2: Find outcomes for Event A (sum > 8)

Possible sums when rolling two dice:

• Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
• Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes
• Sum = 11: (5,6), (6,5) → 2 outcomes
• Sum = 12: (6,6) → 1 outcome

Total favorable for A: \(4 + 3 + 2 + 1 = 10\)? Wait, no: Wait, sum >8 means sum ≥9. Wait, let's list all pairs:
Sum 9: (3,6),(4,5),(5,4),(6,3) → 4
Sum 10: (4,6),(5,5),(6,4) → 3
Sum 11: (5,6),(6,5) → 2
Sum 12: (6,6) → 1
Wait, 4+3+2+1=10? Wait no, wait sum >8: 9,10,11,12. Wait, actually, let's recheck:
Possible pairs for sum 9: (3,6), (4,5), (5,4), (6,3) → 4
Sum 10: (4,6), (5,5), (6,4) → 3
Sum 11: (5,6), (6,5) → 2
Sum 12: (6,6) → 1
Total: 4+3+2+1=10? Wait, no, wait sum >8: 9,10,11,12. Wait, but wait, sum 9: 4, sum 10:3, sum11:2, sum12:1. Total 4+3+2+1=10? Wait, no, wait, let's list all possible two-dice sums:
The minimum sum is 2 (1+1), maximum 12 (6+6). The number of ways to get each sum:
Sum 2:1, sum3:2, sum4:3, sum5:4, sum6:5, sum7:6, sum8:5, sum9:4, sum10:3, sum11:2, sum12:1.
So sum >8: sum 9,10,11,12. Number of ways: 4+3+2+1=10. Wait, but sum 8 is 5 ways, sum >8: 4+3+2+1=10. Total outcomes 36. So P(A) = 10/36? Wait no, wait sum >8: 9,10,11,12. Wait, sum 9:4, sum10:3, sum11:2, sum12:1. 4+3=7, 7+2=9, 9+1=10. Yes. So 10/36? Wait, no, wait sum 9:4, sum10:3, sum11:2, sum12:1. 4+3+2+1=10. So P(A) = 10/36 = 5/18? Wait, no, wait 4+3+2+1=10? Wait, 4 (sum9) +3 (sum10)=7, +2 (sum11)=9, +1 (sum12)=10. Yes. So 10/36 simplifies to 5/18? Wait, no, 10 and 36 have a common factor of 2: 10÷2=5, 36÷2=18. So 5/18. Wait, but let's check again. Wait sum >8: sum 9,10,11,12. The number of combinations:
Sum 9: (3,6), (4,5), (5,4), (6,3) → 4
Sum 10: (4,6), (5,5), (6,4) → 3
Sum 11: (5,6), (6,5) → 2
Sum 12: (6,6) → 1
Total: 4+3+2+1=10. So 10/36 = 5/18.

Step3: Find outcomes for Event B (sum not divisible by 6)

First, find sums divisible by 6: possible sums when rolling two dice: 6, 12.
Sum 6: number of ways: (1,5),(2,4),(3,3),(4,2),(5,1) → 5 ways
Sum 12: (6,6) → 1 way
Total favorable for "divisible by 6": 5 + 1 = 6 ways
Thus, outcomes not divisible by 6: total outcomes - 6 = 36 - 6 = 30
So P(B) = 30/36 = 5/6.

Wait, let's verify sum divisible by 6: possible sums are 6, 12 (since 6×1=6, 6×2=12; next would be 18, which is impossible with two dice). So sum 6: 5 ways, sum12:1 way. Total 6. So not divisible by 6: 36-6=30. 30/36 = 5/6.

Answer:

For \( P(A) \):
The total number of possible outcomes when rolling a die twice is \( 6 \times 6 = 36 \). The favorable outcomes for the sum being greater than 8 (i.e., sums 9, 10, 11, 12) are:

• Sum 9: 4 outcomes (\((3,6), (4,5), (5,4), (6,3)\))
• Sum 10: 3 outcomes (\((4,6), (5,5), (6,4)\))
• Sum 11: 2 outcomes (\((5,6), (6,5)\))
• Sum 12: 1 outcome (\((6,6)\))

Total favorable outcomes: \( 4 + 3 + 2 + 1 = 10 \). Thus, \( P(A) = \frac{10}{36} = \frac{5}{18} \).

For \( P(B) \):
The sums divisible by 6 when rolling two dice are 6 and 12.

• Sum 6: 5 outcomes (\((1,5), (2,4), (3,3), (4,2), (5,1)\))
• Sum 12: 1 outcome (\((6,6)\))

Total outcomes where the sum is divisible by 6: \( 5 + 1 = 6 \). The number of outcomes where the sum is not divisible by 6 is \( 36 - 6 = 30 \). Thus, \( P(B) = \frac{30}{36} = \frac{5}{6} \).

\( P(A) = \boxed{\dfrac{5}{18}} \)
\( P(B) = \boxed{\dfrac{5}{6}} \)