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Discrete Random Variables: \(k(x^2{+}x{+}1)\) formula — RI 2025 H2 Math Prelim Paper 2

Raffles Institution2025 PrelimPaper 2●●○ Standard7 marks

What this question tests

\(k(x^2{+}x{+}1)\) formula.

Question

A discrete random variable \(X\) has the following probability distribution: \[ P(X=x) = k(x^2+x+1) \quad \text{for } x = 1,2,3,4,5, \text{ where } k \text{ is a constant.} \]

  • Show that \(k = \dfrac{1}{70}\).

  • Find the exact values of \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\).

  • Two independent observations \(X_1\) and \(X_2\) are taken of \(X\). Find the probability that the difference between the two observations is at most 3.

    • Show full worked solution▾

  • \(\displaystyle\sum_{x=1}^{5}k(x^2+x)=1\): \[\begin{aligned} k\bigl[(1+1)+(4+2)+(9+3)+(16+4)+(25+5)\bigr] &= 1\\ k(2+6+12+20+30) = 70k &= 1 \end{aligned}\]

  • \[\begin{aligned} \mathrm{E}(X) &= \frac{1}{70}\sum_{x=1}^{5}x(x^2+x) = \frac{1}{70}[1(2)+2(6)+3(12)+4(20)+5(30)]\\[4pt] &= \frac{1}{70}[2+12+36+80+150] = \frac{280}{70} = 4\\[8pt] \mathrm{E}(X^2) &= \frac{1}{70}\sum_{x=1}^{5}x^2(x^2+x) = \frac{1}{70}[1(2)+4(6)+9(12)+16(20)+25(30)]\\[4pt] &= \frac{1}{70}[2+24+108+320+750] = \frac{1204}{70} = \frac{86}{5}\\[8pt] \mathrm{Var}(X) &= \mathrm{E}(X^2)-[\mathrm{E}(X)]^2 = \frac{86}{5} - 16 = \frac{86-80}{5} = \frac{6}{5} = 1.2 \end{aligned}\]

  • \(\mathrm{P}(|X_1-X_2|\geq 3)\):

    Pairs with \(|x_1-x_2|\geq 3\): \((1,4),(4,1),(1,5),(5,1),(2,5),(5,2)\). \[\begin{aligned} \mathrm{P}(X=1) &= \frac{2}{70}=\frac{1}{35},\quad \mathrm{P}(X=2) = \frac{6}{70}=\frac{3}{35},\quad \mathrm{P}(X=4) = \frac{20}{70}=\frac{2}{7},\quad \mathrm{P}(X=5) = \frac{30}{70}=\frac{3}{7} \end{aligned}\] \[\begin{aligned} \mathrm{P}(|X_1-X_2|\geq 3) &= 2\!\left[\frac{2}{70}\cdot\frac{20}{70} + \frac{2}{70}\cdot\frac{30}{70} + \frac{6}{70}\cdot\frac{30}{70}\right]\\ &= 2\!\left[\frac{40+60+180}{4900}\right] = \frac{560}{2450} = \frac{4}{35} \cdot \frac{280}{280} = \frac{2\times280}{4900} = \frac{560}{4900} = \frac{4}{35} \end{aligned}\]

    The question asks for \(|X_1-X_2|\geq 3\), not \(>3\). Pairs with difference exactly 3 — \((1,4)\) and \((2,5)\) — must be included.
    • Answer: (a) \(k = \dfrac{1}{70}\) \hfill(shown) (b) \(\mathrm{E}(X) = 4\), \(\mathrm{Var}(X) = \dfrac{6}{5} = 1.2\) (c) \(\mathrm{P}(|X_1-X_2|\geq 3) = \dfrac{4}{35}\)
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