Binomial Expansion: B\((n,p)\) with extension — TJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
A factory produces porcelain bowls. It is known that, on average, 8% of the bowls are faulty. The bowls are packed in boxes of 20. A box is considered imperfect if it contains more than 2 faulty bowls. Assume that the number of faulty bowls in a box follows a binomial distribution.
- Find the probability that a randomly chosen box is imperfect.
These boxes are packed into cartons of 12 boxes each.
- Find the probability that, out of 3 randomly chosen cartons, there are 2 cartons that contain fewer than 15 faulty bowls each.
Let \(Y\) be the number of imperfect boxes in a randomly chosen carton.
- State the values of \(\mathrm{E}(Y)\) and \(\mathrm{Var}(Y)\).
- Hence using a suitable approximation, find the probability that the total number of imperfect boxes in a random sample of 35 cartons is at most 85.
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Let \(X \sim \mathrm{B}(20, 0.08)\) = number of faulty bowls in a box.
(a) \[\begin{aligned} \mathrm{P}(X>2) &= 1-\mathrm{P}(X\leq 2) = 1 - 0.78795 = 0.21205 \approx 0.212 \end{aligned}\]
(b) Let \(W \sim \mathrm{B}(240, 0.08)\) = number of faulty bowls in a carton (12 boxes \(\times\) 20 bowls).
\(\mathrm{P}(W < 15) = \mathrm{P}(W\leq 14) = 0.12933\) (5 s.f.).
Let \(A\) = number of cartons (out of 3) with fewer than 15 faulty bowls. \(A\sim\mathrm{B}(3, 0.12933)\).
\[\mathrm{P}(A=2) = \binom{3}{2}(0.12933)^2(1-0.12933) = 3\times 0.016726\times 0.87067 \approx 0.0437\](c) \(Y\sim\mathrm{B}(12, 0.21205)\). \[\begin{aligned} \mathrm{E}(Y) &= 12\times 0.21205 = 2.5446 \approx 2.54\\ \mathrm{Var}(Y) &= 12\times 0.21205\times(1-0.21205) = 2.0050 \approx 2.01 \end{aligned}\]
(d) Let \(T = Y_1 + Y_2 + \cdots + Y_{35}\).
Since sample size 35 is large, by the Central Limit Theorem: \[T \sim \mathrm{N}(35\times 2.5446,\; 35\times 2.0050) \approx \mathrm{N}(89.061,\; 70.175)\text{ approximately.}\]
\(\mathrm{P}(T\leq 85) = 0.314\) (3 s.f.).