Discrete Random Variables: Distribution; game payoff — TMJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
Wardrobe \(A\) contains \(4\) distinct blouses, labelled \(T_1, T_2, T_3\) and \(T_4\). Wardrobe \(B\) contains \(3\) distinct pairs of pants, labelled \(P_1, P_2\) and \(P_3\). The following table shows the formality (formal or casual) of all the clothing.
| Wardrobe | Clothing | Formal | Casual |
|---|---|---|---|
| \(A\) | Blouse | \(T_1, T_2\) | \(T_3, T_4\) |
| \(B\) | Pants | \(P_1, P_2\) | \(P_3\) |
Nancy selects an outfit at random, by first choosing one blouse from Wardrobe \(A\), followed by choosing one pair of pants from Wardrobe \(B\).
\(X\) denotes the score of a randomly chosen outfit.
- If the blouse and pants have the same formality, then the score of the outfit is the sum of their respective indices. For example, the score of an outfit comprising \(T_4\) and \(P_3\) is \(7\).
- If the blouse and pants have different formality, then the score of the outfit is the product of their respective indices. For example, the score of an outfit comprising \(T_2\) and \(P_3\) is \(6\).
(a) Show that \(\mathrm{P}(X = 4) = \tfrac{1}{6}\).
(b) Find the probability distribution of \(X\).
(c) Show that \(\mathrm{E}(X) = \tfrac{55}{12}\) and find \(\mathrm{Var}(X)\).
(d) Nancy selects three outfits in succession without replacement. Find the probability that there are exactly two outfits that comprise both a formal blouse and a pair of formal pants.
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(a) A score of \(4\) is achieved only by (i) blouse \(T_4\) (C) with pants \(P_1\) (F) [different formality, product \(= 4\)], or (ii) blouse \(T_2\) (F) with pants \(P_2\) (F) [same formality, sum \(= 4\)]: \[ \mathrm{P}(X = 4) = \mathrm{P}(T_4,P_1) + \mathrm{P}(T_2,P_2) = \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{4}\cdot\tfrac{1}{3} = \tfrac{1}{6} \quad\text{(shown).} \]
(b) Tabulating the score for each (blouse, pants) outcome (each occurring with probability \(\tfrac{1}{12}\)):
| Blouse \(\backslash\) Pants | \(P_1\) (F) | \(P_2\) (F) | \(P_3\) (C) |
|---|---|---|---|
| \(T_1\) (F) | 2 | 3 | 3 |
| \(T_2\) (F) | 3 | 4 | 6 |
| \(T_3\) (C) | 3 | 6 | 6 |
| \(T_4\) (C) | 4 | 8 | 7 |
Probability distribution of \(X\):
| \(x\) | 2 | 3 | 4 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(\dfrac{1}{12}\) | \(\dfrac{4}{12} = \dfrac{1}{3}\) | \(\dfrac{2}{12} = \dfrac{1}{6}\) | \(\dfrac{3}{12} = \dfrac{1}{4}\) | \(\dfrac{1}{12}\) | \(\dfrac{1}{12}\) |
(c) \[\begin{aligned} \mathrm{E}(X) &= 2\!\left(\tfrac{1}{12}\right) + 3\!\left(\tfrac{1}{3}\right) + 4\!\left(\tfrac{1}{6}\right) + 6\!\left(\tfrac{1}{4}\right) + 7\!\left(\tfrac{1}{12}\right) + 8\!\left(\tfrac{1}{12}\right)\\ &= \tfrac{55}{12} \quad\text{(shown).}\\[4pt] \mathrm{E}(X^2) &= 2^2\!\left(\tfrac{1}{12}\right) + 3^2\!\left(\tfrac{1}{3}\right) + 4^2\!\left(\tfrac{1}{6}\right) + 6^2\!\left(\tfrac{1}{4}\right) + 7^2\!\left(\tfrac{1}{12}\right) + 8^2\!\left(\tfrac{1}{12}\right)\\ &= \tfrac{293}{12}.\\[4pt] \mathrm{Var}(X) &= \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = \tfrac{293}{12} - \left(\tfrac{55}{12}\right)^{\!2} = \tfrac{491}{144}\\ &\approx 3.41 \quad\text{(3 s.f.)}. \end{aligned}\]
(d) A “formal-formal” outfit consists of a blouse in \(\{T_1,T_2\}\) and pants in \(\{P_1,P_2\}\). There are \(2\cdot 2 = 4\) such outfits out of \(12\).
The school presents the probability by tracking the formal blouses and formal pants used as each outfit is selected: \[\begin{aligned} \text{Required prob.} &= \Bigl(\underbrace{\tfrac{2}{4}\cdot\tfrac{2}{3}}_{\substack{\text{first outfit}\\\text{formal-formal}}}\Bigr) \cdot\Bigl(\underbrace{\tfrac{1}{3}\cdot\tfrac{1}{2}}_{\substack{\text{second outfit}\\\text{formal-formal}}}\Bigr) \cdot\Bigl(\underbrace{\tfrac{2}{2}\cdot\tfrac{1}{1}}_{\substack{\text{third outfit}\\\text{not formal-formal}}}\Bigr) \cdot\underbrace{\frac{3!}{2!\,1!}}_{\text{ordering}}\\ &= \tfrac{1}{6}. \end{aligned}\]