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Binomial Expansion: Combined trials; conditional — JPJC 2025 H2 Math Prelim Paper 2

Jurong Pioneer Junior College2025 PrelimPaper 2●●○ Standard8 marks

What this question tests

Combined trials; conditional.

Question

A company makes portable speakers. Some speakers turn out to be faulty. Each carton has \(24\) portable speakers.

  • (i) State, in context, two assumptions needed for the number of faulty portable speakers in a randomly chosen carton to be well modelled by a binomial distribution.

It is given that \(2\%\) of the portable speakers produced are faulty.

  • (ii) Find the probability that a randomly chosen carton contains at most one faulty portable speaker.
  • (iii) A carton with more than one faulty portable speaker is considered as `substandard'. Find the probability that, out of \(50\) cartons, there are no `substandard' cartons.

The company also makes flash drives packed in batches of one hundred. The probability that a flash drive is faulty is \(p\), where \(0<p<1\). For quality control purposes, a random sample of \(3\) flash drives is tested from each batch.

  • If there are \(2\) or \(3\) faulty flash drives, the batch is rejected.
  • If there is no faulty flash drive, the batch is accepted.
  • If there is one faulty flash drive in the sample, another random sample of \(3\) flash drives is tested and the batch is accepted if there are no faulty flash drives in the second sample. Otherwise the batch is rejected.

  • (iv) Find the range of \(p\) such that we accept at least \(95\%\) of the batches based on this quality control method.
Show full worked solution▾

(i) Two assumptions for the binomial model:

  1. Whether one portable speaker is faulty is independent of whether any other portable speaker is faulty.
  2. The probability that a randomly chosen portable speaker is faulty is constant.

(ii) Let \(X\) be the number of faulty speakers out of \(24\). Then \(X \sim \mathrm{B}(24, 0.02)\). \[ \mathrm{P}(X \le 1) = 0.91738 \approx 0.917. \]

(iii) A carton is “substandard” if \(X > 1\), with probability \(1 - 0.91738 = 0.082613\)

Method 1 Let \(Y\) be the number of substandard cartons out of \(50\). Then \(Y \sim \mathrm{B}(50,\ 0.082613)\): \[ \mathrm{P}(Y=0) = 0.013416 \approx 0.0134. \]

Method 2 By independence, \[ [\mathrm{P}(X\le 1)]^{50} = [0.917387]^{50} \approx 0.0134. \]

(iv) Let \(F\) be the number of faulty flash drives in a sample of 3. Then \(F \sim \mathrm{B}(3, p)\).

A batch is accepted in two disjoint ways:

  • \(F = 0\) on the first sample, OR
  • \(F = 1\) on the first sample AND \(F = 0\) on the second (independent) sample.
\[\begin{aligned} \mathrm{P}(\text{accept}) &= \mathrm{P}(F=0) + \mathrm{P}(F=1)\cdot\mathrm{P}(F=0) \\ &= (1-p)^{3} + \binom{3}{1}p\,(1-p)^{2}\cdot(1-p)^{3} \\ &= (1-p)^{3}\left[1 + 3p(1-p)^{2}\right]. \end{aligned}\] At least \(95\%\) accepted: \[ (1-p)^{3}\left[1 + 3p(1-p)^{2}\right] \ge 0.95. \] Solving with a GC (with a graph), \(0 < p \le 0.0703\) (3 s.f.).

Answer: (ii) \(0.917\). (iii) \(0.0134\). (iv) \(0 < p \le 0.0703\).
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