NJC 2025 H2 Math Prelim Paper 2 — Discrete Random Variables (worked solution)

\begin{center} \begin{tabular}{|c|c|c|c|c|} \hline $t$ & 1 & 2 & 3 & 4 or more \\ \hline $\mathrm{P}(T=t)$ & $\dfrac{1}{6}$ & $\dfrac{1}{9}$ & $\dfrac{5}{36}$ & $\dfrac{7}{12}$ \\[4pt] \hline \end{tabular} \end{center} \enspace $\mathrm{E}(

Discrete Random Variables: Distribution; expected score — NJC 2025 H2 Math Prelim Paper 2

National Junior College2025 PrelimPaper 2●●○ Standard10 marks

What this question tests

Distribution; expected score.

Question

A circular board has 6 sectors labelled $A$, $B$, $C$, $D$, $E$ and $F$. A game is played by placing a pawn on sector $A$. It is then moved clockwise around the board according to the number shown on the top face of a fair die when it is tossed. The die has faces labelled 2, 2, 3, 4, 4, 5. For example, an initial toss of 4 would move the pawn from sector $A$ to sector $E$. The player continues tossing the die and moving the pawn accordingly. When the pawn lands on sector $F$, the game ends.

Let $T$ be the number of tosses required to move the pawn until the game ends.

Show that $\mathrm{P}(T=3) = \dfrac{5}{36}$. Hence, copy and complete the probability distribution table for $T$:
$t$ 1 2 3 4 or more
$\mathrm{P}(T=t)$
The score $S$ for each game is given by $S = 5T$, for $t = 1, 2, 3$ and $s = 0$ for $t \geq 4$. Find $\mathrm{E}(S)$ and $\mathrm{Var}(S)$.

\(t\)	1	2	3	4 or more
\(\mathrm{P}(T=t)\)

Show full worked solution▾

(a)

Label sectors $A=0,\, B=1,\, C=2,\, D=3,\, E=4,\, F=5$ in clockwise order. The pawn starts at $A$; the game ends when the cumulative move $\equiv 5\pmod{6}$ (i.e. lands on $F$).

$T=1$: a single move of 5: $\mathrm{P}(T=1) = \tfrac{1}{6}$.

$T=2$: two tosses summing to $5\pmod 6$. The only pair is $\{2,3\}$: \[\begin{aligned} \mathrm{P}(T=2) &= \mathrm{P}(\text{a 2 and a 3 in any order})\\ &= \!\left(\frac{2}{6}\right)\!\left(\frac{1}{6}\right)\times 2! = \frac{1}{9} \end{aligned}\]

$T=3$: three tosses summing to $5\pmod 6$ without early win — valid combinations $\{2,4,5\}$, $\{3,3,5\}$, $\{3,4,4\}$: \[\begin{aligned} \mathrm{P}(T=3) &= \!\left(\frac{2}{6}\right)^{\!2}\!\!\left(\frac{1}{6}\right)\times(3!-2) + \!\left(\frac{1}{6}\right)^{\!3}\!\times\!\left(\frac{3!}{2!}-1\right) + \!\left(\frac{1}{6}\right)\!\!\left(\frac{2}{6}\right)^{\!2}\!\times\frac{3!}{2!}\\ &= \frac{5}{36}\quad<strong>(Shown)</strong> \end{aligned}\]

$\mathrm{P}(T\geq4) = 1 - \tfrac{1}{6} - \tfrac{1}{9} - \tfrac{5}{36} = \dfrac{7}{12}$.

(b)

$S = 5T$ for $t=1,2,3$ and $S=0$ for $t\geq 4$.

$s$	5	10	15	0
$\mathrm{P}(S=s)$	$\dfrac{1}{6}$	$\dfrac{1}{9}$	$\dfrac{5}{36}$	$\dfrac{7}{12}$

\[\begin{aligned} \mathrm{E}(S) &= 5\!\left(\frac{1}{6}\right) + 10\!\left(\frac{1}{9}\right) + 15\!\left(\frac{5}{36}\right) = \frac{30}{36}+\frac{40}{36}+\frac{75}{36} = \frac{145}{36}\\[4pt] \mathrm{E}(S^2) &= 25\!\left(\frac{1}{6}\right) + 100\!\left(\frac{1}{9}\right) + 225\!\left(\frac{5}{36}\right) = \frac{150}{36}+\frac{400}{36}+\frac{1125}{36} = \frac{1675}{36}\\[4pt] \mathrm{Var}(S) &= \mathrm{E}(S^2)-[\mathrm{E}(S)]^2 = \frac{1675}{36} - \!\left(\frac{145}{36}\right)^{\!2} = \frac{39\,275}{1296} \end{aligned}\]

Answer:

$t$	1	2	3	4 or more
$\mathrm{P}(T=t)$	$\dfrac{1}{6}$	$\dfrac{1}{9}$	$\dfrac{5}{36}$	$\dfrac{7}{12}$ [4pt]

\enspace $\mathrm{E}(S) = \dfrac{145}{36} \approx 4.03$ $\mathrm{Var}(S) = \dfrac{39\,275}{1296} \approx 30.3$ (3 s.f.)

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