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Probability: Tree; conditional cases — RI 2025 H2 Math Prelim Paper 2

Raffles Institution2025 PrelimPaper 2●●● Challenging12 marks

What this question tests

Tree; conditional cases.

Question

A bag contains 4 identical red and 6 identical blue balls.

  • In the first game, the balls are randomly picked by a player from the bag, one at a time, without replacement, until there are no balls left in the bag.

  • Find the probability that the last ball picked is red.

  • Find the probability that in the first 5 picks, exactly 2 blue balls are picked given that at least 3 red balls are picked.

  • In the second game, each ball picked by a player will be placed back into the bag by the colour of the ball and each pick is recorded. The player makes a total of 20 picks from the bag.

  • Find the probability that the colour red is recorded exactly 4 times.

  • Find the probability that the colour red is recorded 4 times but not more than 8 times.

  • Find the probability that the 8th pick is the 6th time the colour blue is recorded.

  • The second game is played by 50 randomly chosen players. Estimate the probability that the average number of times the colour red is recorded is more than 8.5.

    • Show full worked solution▾

    Bag: 4 red, 6 blue balls (identical within colour).

  • By symmetry, each ball is equally likely to be last. There are 10 balls, 4 red. \[\begin{aligned} \mathrm{P}(\text{last ball is red}) &= \frac{4}{10} = \frac{2}{5} \end{aligned}\]

  • In the first 5 picks, given at least 3 red balls are picked:

    Events: \(A\) = “exactly 2 blue (and 3 red) in first 5 picks”; \(B\) = “at least 3 red in first 5 picks”.

    Since there are only 4 red balls, “at least 3 red in 5 picks” means exactly 3 or exactly 4 red.

    \[\begin{aligned} \mathrm{P}(\text{3 red, 2 blue in 5 picks}) &= \frac{\binom{4}{3}\binom{6}{2}}{\binom{10}{5}} = \frac{4\cdot 15}{252} = \frac{60}{252}\\[6pt] \mathrm{P}(\text{4 red, 1 blue in 5 picks}) &= \frac{\binom{4}{4}\binom{6}{1}}{\binom{10}{5}} = \frac{1\cdot 6}{252} = \frac{6}{252}\\[6pt] \mathrm{P}(B) &= \frac{60+6}{252} = \frac{66}{252} \end{aligned}\] \[\begin{aligned} \mathrm{P}(A|B) &= \frac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)} = \frac{60/252}{66/252} = \frac{60}{66} = \frac{10}{11} \end{aligned}\]

  • Replacement game: each pick is independent, \(\mathrm{P}(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5}\).

    Let \(X\) = number of red picks in 20 picks. \(X\sim B\!\left(20,\dfrac{2}{5}\right)\).

  • \(\mathrm{P}(X=4) = \dbinom{20}{4}\!\left(\dfrac{2}{5}\right)^4\!\left(\dfrac{3}{5}\right)^{16} = 0.0350\) (3 s.f.)

  • \(\mathrm{P}(4\leq X\leq 8) = \mathrm{P}(X\leq 8)-\mathrm{P}(X\leq 3)\)

    From GC: \(\approx 0.596 - 0.0160 = 0.580\) (3 s.f.)

  • The 8th pick is the 6th blue. So in the first 7 picks: exactly 5 blue and 2 red; and the 8th pick is blue.

    \[\begin{aligned} \mathrm{P} &= \binom{7}{5}\!\left(\frac{3}{5}\right)^5\!\left(\frac{2}{5}\right)^2\times\frac{3}{5}\\[4pt] &= 21\times\frac{243}{3125}\times\frac{4}{25}\times\frac{3}{5}\\[4pt] &= 21\times\frac{2916}{390625} = \frac{61236}{390625} \approx 0.157 \end{aligned}\]

  • \(R \sim \mathrm{B}\!\left(20, \dfrac{2}{5}\right)\), so \(\mathrm{E}(R) = 20\!\left(\dfrac{2}{5}\right) = 8\)

    \[ \mathrm{Var}(R) = 20\!\left(\frac{2}{5}\right)\!\left(\frac{3}{5}\right) = \frac{24}{5} \quad \text{or} \quad 4.8 \]

    Since \(n = 50\) is large, by Central Limit Theorem, \[ \bar{R} \sim \mathrm{N}\!\left(8,\ \frac{4.8}{50}\right) \text{ approximately, i.e.} \] \[ \bar{R} \sim \mathrm{N}(8,\ 0.096) \text{ approximately} \] \[ \mathrm{P}(\bar{R} > 8.5) = 0.0533 \text{ (3 s.f.)} \]

    Alternatively,

    Since \(n = 50\) is large, by Central Limit Theorem, \[ R_1 + \ldots + R_{50} \sim \mathrm{N}(50 \times 8,\ 50 \times 4.8) \text{ approximately, i.e.} \] \[ R_1 + \ldots + R_{50} \sim \mathrm{N}(400,\ 240) \text{ approximately.} \] \[ \mathrm{P}(R_1 + \ldots + R_{50} > 8.5 \times 50) = 0.0533 \text{ (3 s.f.)} \]

    • Answer: (a)(i) \(\dfrac{2}{5}\) (a)(ii) \(\dfrac{10}{11}\) (b)(i) \(\mathrm{P}(X=4) \approx 0.0350\) (b)(ii) \(\mathrm{P}(4\leq X\leq 8) \approx 0.596\) (b)(iii) \(\approx 0.157\) (b)(iv) \(\mathrm{P}(\bar{T}>8.5)\approx 0.0533\)
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