Mathematics 9709/62 — May/June 2019

Walkthrough

This problem asks whether two events on a probability space are independent. The definition we use is that $S$ and $T$ are independent exactly when $P(S \cap T) = P(S)P(T)$; an equivalent statement is $P(S \mid T) = P(S)$.

Begin by fixing the sample space: two fair dice give $6 \times 6 = 36$ equally likely outcomes, so every probability is a count divided by 36.

Finding $P(S)$: a sum is even when both dice show the same parity (both even or both odd). For a fair die there are 3 even and 3 odd faces, and counting shows exactly half of the 36 outcomes are even sums, so $P(S) = 18/36 = 1/2$. This earns the first B1.

Finding $P(T)$: the wording "less than 6, or a multiple of 4, or both" tells us to take the union of the two sets. The sums less than 6 are 2, 3, 4, 5; the multiples of 4 are 4, 8, 12. The number 4 appears in both, but the union should contain it only once, so $T = \{2, 3, 4, 5, 8, 12\}$. Now count the ordered pairs giving each sum: $1 + 2 + 3 + 4 + 5 + 1 = 16$, hence $P(T) = 16/36 = 4/9$. That is the second B1.

Finding $P(S \cap T)$: intersect the even sums with $T$. The even sums in $T$ are 2, 4, 8 and 12, totalling $1 + 3 + 5 + 1 = 10$ outcomes, so $P(S \cap T) = 10/36 = 5/18$. This earns the M1. A crucial point from the mark scheme is that $P(S \cap T)$ must be obtained by direct counting — multiplying $P(S)$ by $P(T)$ scores M0 because doing so assumes independence, which is the very thing under test.

Testing independence: multiply $P(S)$ by $P(T)$ to get $\frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{8}{36}$, and compare with $P(S \cap T) = \frac{10}{36}$. Because $\frac{8}{36} \neq \frac{10}{36}$, the product rule fails and the events are not independent; stating this comparison and the conclusion earns the A1. The alternative accepted route is to observe $P(S \mid T) = \frac{10}{16} = \frac{5}{8}$, which differs from $P(S) = \frac{1}{2}$ (or $P(T \mid S) = \frac{10}{18} \neq \frac{4}{9}$), giving the same verdict.

Key Takeaways

The central idea is the product rule for independence: $A$ and $B$ are independent if and only if $P(A \cap B) = P(A)P(B)$, with the conditional form $P(A \mid B) = P(A)$ as an equivalent test. The problem also reinforces careful enumeration over the 36-element two-dice sample space, and correct handling of a union described with "or both" so that overlapping elements (the sum 4) are not double counted.

Common Mistakes

• Finding $P(S \cap T)$ by multiplying $P(S)$ and $P(T)$: this builds the independence assumption into the working (circular reasoning) and scores M0 under the mark scheme.
• Double counting the sum 4 when forming $T$, because it is simultaneously less than 6 and a multiple of 4; it must be counted once.
• Forgetting sum 12 as a multiple of 4, or misremembering the number of outcomes for a given sum (e.g. sum 8 has 5 outcomes).
• Concluding that the events are independent, or stating the conclusion without showing the comparison $\frac{8}{36} \neq \frac{10}{36}$.

Things to Be Careful About

• Count the ordered pairs for each sum carefully: sum 2 (1 pair), 3 (2), 4 (3), 5 (4), 8 (5), 12 (1). Missing any pair changes the counts.
• The phrase "or both" signals a union requiring the inclusion–exclusion idea: the overlap (sum 4) is counted only once.
• Independence demands the strict equality $P(S)P(T) = P(S \cap T)$; any inequality means dependence. Write both sides explicitly to secure the A1.
• The mark scheme requires the working to be entirely correct (www) for the final mark — an unsupported or wrong answer is not accepted.