Discrete Random Variables: Distribution; mean, variance — JPJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
In a game, a player selects one ball at random from a bag which contains four balls numbered `\(1\)', `\(1\)', `\(2\)' and `\(3\)'. The player notes the number on the selected ball, and then tosses that number of fair coins. The number of heads obtained is denoted by \(X\) and has a probability distribution as follows:
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(\mathrm{P}(X=x)\) | \(a\) | \(\tfrac{15}{32}\) | \(b\) | \(\tfrac{1}{32}\) |
- (i) Find the exact values of \(a\) and \(b\).
- (ii) Find \(\mathrm{E}(X)\) and show that \(\mathrm{Var}(X)=\dfrac{39}{64}\).
- (iii) The player plays \(50\) games. Find the probability that the average number of heads obtained per game exceeds \(1\).
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(i) The ball drawn determines how many fair coins are tossed.
\(\mathrm{P}(\text{ball}=1) = \tfrac{2}{4} = \tfrac{1}{2}\),
\(\mathrm{P}(\text{ball}=2) = \tfrac{1}{4}\),
\(\mathrm{P}(\text{ball}=3) = \tfrac{1}{4}\).
\[\begin{aligned} a = \mathrm{P}(X=0) &= \left(\tfrac{1}{2}\cdot\tfrac{1}{2}\right) + \left(\tfrac{1}{4}\cdot\left(\tfrac{1}{2}\right)^{2}\right) + \left(\tfrac{1}{4}\cdot\left(\tfrac{1}{2}\right)^{3}\right) \\ &= \tfrac{1}{4} + \tfrac{1}{16} + \tfrac{1}{32} = \tfrac{11}{32}.\\[2pt] b = \mathrm{P}(X=2) &= \left(\tfrac{1}{4}\cdot\left(\tfrac{1}{2}\right)^{2}\right) + \left(\tfrac{1}{4}\cdot\binom{3}{2}\!\left(\tfrac{1}{2}\right)^{3}\right) \\ &= \tfrac{1}{16} + \tfrac{3}{32} = \tfrac{5}{32}. \end{aligned}\] Check: \(\tfrac{11}{32}+\tfrac{15}{32}+\tfrac{5}{32}+\tfrac{1}{32}=1.\)
(ii) Using the completed table: \[\begin{aligned} \mathrm{E}(X) &= 0\cdot\tfrac{11}{32} + 1\cdot\tfrac{15}{32} + 2\cdot\tfrac{5}{32} + 3\cdot\tfrac{1}{32} = \tfrac{28}{32} = \tfrac{7}{8}.\\ \mathrm{E}(X^{2}) &= 0 + 1\cdot\tfrac{15}{32} + 4\cdot\tfrac{5}{32} + 9\cdot\tfrac{1}{32} = \tfrac{44}{32} = \tfrac{11}{8}.\\ \mathrm{Var}(X) &= \mathrm{E}(X^{2}) - [\mathrm{E}(X)]^{2} = \tfrac{11}{8} - \left(\tfrac{7}{8}\right)^{2} = \tfrac{88-49}{64} = \tfrac{39}{64}. \qquad\text{(Shown)} \end{aligned}\]
(iii) For \(n=50\) games, by the Central Limit Theorem: \[ \overline{X} \sim \mathrm{N}\!\left(\tfrac{7}{8},\ \tfrac{39/64}{50}\right) = \mathrm{N}\!\left(\tfrac{7}{8},\ \tfrac{39}{3200}\right) \text{ approximately.} \] \[ \mathrm{P}(\overline{X} > 1) = 0.12876\ldots \approx 0.129. \]