Question
Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the two dice is:
- Equal to 7
- Greater than 9
- Less than or equal to 3
Solution — Step by Step
The key to dice problems is the sample space. Two dice produce ordered pairs, not just sums — this is where students make mistakes.
Step 1: Write out the sample space.
When two dice are rolled, each die can show 1 to 6. The total number of outcomes = 6 × 6 = 36 equally likely outcomes.
These are ordered pairs (die 1, die 2). The pair (2, 5) is different from (5, 2) — the first die showed 2 and the second showed 5, and vice versa.
Step 2: Find favourable outcomes for sum = 7.
List all pairs (a, b) where a + b = 7:
- (1, 6): 1 + 6 = 7 ✓
- (2, 5): 2 + 5 = 7 ✓
- (3, 4): 3 + 4 = 7 ✓
- (4, 3): 4 + 3 = 7 ✓
- (5, 2): 5 + 2 = 7 ✓
- (6, 1): 6 + 1 = 7 ✓
Number of favourable outcomes = 6
P(sum = 7) = 6/36 = 1/6
Step 3: Find favourable outcomes for sum > 9.
Sums greater than 9: sum = 10, 11, or 12.
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 favourable = 3 + 2 + 1 = 6
P(sum > 9) = 6/36 = 1/6
Step 4: Find favourable outcomes for sum ≤ 3.
Sum = 2: (1,1) → 1 outcome Sum = 3: (1,2), (2,1) → 2 outcomes
Total favourable = 1 + 2 = 3
P(sum ≤ 3) = 3/36 = 1/12
Answers
P(sum = 7) = 1/6
P(sum > 9) = 1/6
P(sum ≤ 3) = 1/12
Why This Works
Each ordered pair (die 1, die 2) is an equally likely outcome because the dice are fair and independent. We use the classical definition:
P(event) = (number of favourable ordered pairs) / 36
The total is always 36 for two dice — not 11 (the possible sum values). Thinking of it as 11 outcomes (sums from 2 to 12) is wrong because these sums are not equally likely. Sum 7 has 6 ways to occur; sum 2 has only 1 way.
Common Mistake
⚠️ Common Mistake
Mistake: Thinking the sample space has 11 elements (sums 2 through 12).
The sums 2 to 12 are NOT equally likely. Sum 7 has 6 ways, but sum 2 has only 1 way. If you assign equal probability to each sum, you'd get P(sum = 7) = 1/11, which is wrong. Always use the 36 ordered pairs as your sample space for two dice. The probability P(sum = 7) = 6/36 = 1/6 is correct.