Problem Statement
Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) Less than or equal to 12?
Verified Solution & Marking Scheme
Total Possible Outcomes
Total outcomes when throwing two dice $= 6 \times 6 = 36$.
Part (i): Sum is 8
Favorable outcomes: $(2,6), (3,5), (4,4), (5,3), (6,2)$ (5 outcomes).
$P(\text{Sum } 8) = \frac{5}{36}$
Part (ii): Sum is 13
Maximum possible sum is $6 + 6 = 12$. An outcome with sum 13 is impossible (0 outcomes).
$P(\text{Sum } 13) = 0$
Part (iii): Sum ≤ 12
All 36 outcomes have a sum less than or equal to 12. This is a sure event.
$P(\text{Sum } \le 12) = \frac{36}{36} = 1$