Discussion
There are two identical dice with a single letter on each of the faces. The following six letters: Q, R, S, T, U, and V, one on each of the faces. Any of the six outcomes are equally likely.
The two dice are thrown once independently at random.
What is the probability that the outcomes on the dice were composed only of any combination of the following possible outcomes: Q, U and V?
Answer: A
The probability is 1/4.
Each die has six equally likely faces, three of which (Q, U, V) are allowed. The dice are independent, so both must show an allowed letter:
\[ \begin{aligned} P &= \frac{3}{6} \times \frac{3}{6} \\ &= \frac{9}{36} = \frac{1}{4} \end{aligned} \]
Counting directly gives the same: 3 × 3 = 9 favourable ordered pairs out of 36. 3/4 is the chance that at least one die shows an allowed letter, which is a weaker event. 1/6 and 5/36 come from counting only particular pairs rather than every combination of Q, U and V.
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