In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{5}{18}$
1. Total Possible Outcomes
When two dice are thrown, each die has 6 possible outcomes. Therefore, the total number of combinations is:
$6 \times 6 = 36$
2. Identify Favorable Outcomes
A "total more than 8" means the sum must be 9, 10, 11, or 12. Let's list the pairs $(d1, d2)$ for each sum:
- 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 number of favorable outcomes = $4 + 3 + 2 + 1 = \mathbf{10}$
3. Calculate the Probability
The probability ($P$) is the ratio of favorable outcomes to total outcomes:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P = \frac{10}{36}$
Now, simplify the fraction by dividing both the numerator and the denominator by 2:
$P = \frac{5}{18}$