Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes.
(i) What is the probability that the sum of the two numbers appearing on the top of the dice is:
(a) $8$?
(b) $13$?
(c) less than or equal to $12$?
(ii) Find the probability of getting a doublet of even numbers.
Step-by-Step Solution
Key Concept: Total outcomes $= 6 \times 6 = 36$.
Stepwise Solution:
Total outcomes $N = 36$. [1.0 Mark]
(i)(a) Sum $= 8$: outcomes are $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$ (5 outcomes).
$P(\text{sum } 8) = \dfrac{5}{36}$. [1.0 Mark]
(i)(b) Sum $= 13$: maximum possible sum is $6+6=12$, so 0 outcomes.
$P(\text{sum } 13) = \dfrac{0}{36} = 0$. [1.0 Mark]
(i)(c) Sum $\leq 12$: all $36$ outcomes have sum $\leq 12$.
$P(\text{sum } \leq 12) = \dfrac{36}{36} = 1$. [1.0 Mark]
(ii) Doublet of even numbers: $\{(2,2), (4,4), (6,6)\}$ (3 outcomes).
$P(\text{even doublet}) = \dfrac{3}{36} = \dfrac{1}{12}$. [1.0 Mark]
Marking Scheme:
• Stating sample space size 36: 1.0 Mark
• Part (i)(a) $P(\text{sum } 8) = 5/36$: 1.0 Mark
• Part (i)(b) $P(\text{sum } 13) = 0$: 1.0 Mark
• Part (i)(c) $P(\text{sum } \leq 12) = 1$: 1.0 Mark
• Part (ii) $P(\text{even doublet}) = 1/12$: 1.0 Mark
Correct Answer: