Binomial Expansion: B\((n,p)\); mean and variance — RVHS 2025 H2 Math Prelim Paper 2
What this question tests
Question
A machine produces a large number of camera lenses daily. During the process, lenses could get scratched and become defective.
- State, in the context of the question, one assumption needed for the number of defective lenses made to be modelled by a binomial distribution and explain why the assumption may not hold in this context.
It is assumed that the number of defective lenses produced daily has a binomial distribution and that on average, 1 in 20 lenses produced are defective.
- 30 independent samples of 100 lenses each are taken. Find the probability that there are more than 10 samples in which there are exactly 3 defective lenses in each sample.
- 2 independent samples of 100 lenses are taken. It is given that exactly one of the samples was found to have exactly 3 defective lenses. Find the probability that there are a total of 10 defective lenses in the 2 samples.
- Every day, lenses are randomly selected for quality checks on the production. An initial sample of 100 lenses is obtained.
- The entire production of lenses for that day would be deemed to have passed the quality check when there are at most 2 defective lenses in the initial sample.
- If there are more than 3 defective lenses found in the initial sample, then the entire production of lenses for that day will be rejected.
- If the initial sample contains exactly 3 defective lenses, another sample of 100 lenses will be taken. If the 2 samples contain an average of at most 2 defective lenses, then the entire production of lenses for that day would be deemed to have passed the quality check.
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(a) Assume that the probability in which a lens is defective is constant throughout the entire population of lenses produced.
This may not be valid, since machine parts degrade gradually which affects the manufacturing process and thus probability of a lens being defective changes over time.
OR Assume that the event in which a lens is defective is independent to all other lenses.
This may not be appropriate, since it is possible that a defective lens could cause other lenses to be defective downstream (e.g. a defective lens occurring during the process of production causes misalignment in subsequent lenses and thus causes more defective lenses).
(b) Let \(X\) denote the number of defective lenses, out of 100. \[ X \sim \mathrm{B}\!\left(100,\,\frac{1}{20}\right) \] \[ \mathrm{P}(X = 3) = 0.13958 \]
Let \(Y\) be the random variable for the number of samples that has exactly 3 defective lenses, out of 30. \[ Y \sim \mathrm{B}(30,\,0.13958) \] \[\begin{aligned} \mathrm{P}(Y > 10) &= 1 - \mathrm{P}(Y \leq 10)\\ &= 0.00163393\\ &\approx 0.00163 \end{aligned}\]
(c) \[\begin{aligned} &\mathrm{P}(X_1+X_2=10 \mid X_1=3 \text{ or } X_2=3 \text{ but not both})\\[6pt] &= \frac{\mathrm{P}\bigl((X_1+X_2=10)\cap(X_1=3 \text{ or } X_2=3)\bigr)} {\mathrm{P}(X_1=3 \text{ or } X_2=3 \text{ but not both})}\\[6pt] &= \frac{\mathrm{P}(X_1=3\cap X_2=7) + \mathrm{P}(X_2=3\cap X_1=7)} {\mathrm{P}(X_1=3)+\mathrm{P}(X_2=3)-2\mathrm{P}(X_1=3)\mathrm{P}(X_2=3)}\\[6pt] &= \frac{(0.13958)(0.10603)\times 2}{0.13958+0.13958-2(0.13958)^2}\\[6pt] &= 0.12323\\[4pt] &\approx 0.123 \end{aligned}\]
Denominator can also be computed via \(\mathrm{P}(Y=1) = 0.2401883\), where \(Y\) is the number of samples out of 2 to have exactly 3 defective lenses, \(Y\sim\mathrm{B}(2,\,0.13958)\).
(d) \[ \frac{X_1+X_2}{2} \leq 2 \implies \frac{3+X_2}{2} \leq 2 \implies X_2 \leq 1 \]
Required Probability \[\begin{aligned} &= \mathrm{P}(X_1 \leq 2) + \mathrm{P}(X_1=3)\mathrm{P}(X_2\leq 1)\\ &= 0.11826 + (0.13958)(0.0370815)\\ &= 0.123 \end{aligned}\]