Binomial Expansion: Binomial; advanced — NYJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
A small ceramic workshop produces 100 plates each working day. Some of the plates turn out to be faulty.
- State, in context, two assumptions needed for the number of faulty plates made in a day to be well modelled by a binomial distribution.
Assume now that the number of faulty plates produced each working day has the distribution \(\mathrm{B}(100,p)\).
- Show that the probability that exactly 3 faulty plates are produced on a randomly chosen working day is \(161700p^3(1-p)^{97}\).
- Given that the most likely number of faulty plates produced on a working day is 3, find the possible range of values of \(p\), leaving your answer in exact form.
The workshop also produces bowls on each working day. The number of faulty bowls also follows a binomial distribution. The probability that a bowl is faulty is \(q\). Faults on plates are independent of faults on bowls. The plates and bowls are sold in sets of 2 randomly chosen bowls and 2 randomly chosen plates. In the case where \(p=0.01\), the probability that a set contains at most 1 faulty item is 0.88.
- Write down an equation satisfied by \(q\). Hence find the value of \(q\).
Show full worked solution▾
(a)
(b)
Let \(X\) be the number of faulty plates. \(X\sim\mathrm{B}(100,p)\).
\[ \mathrm{P}(X=3)=\binom{100}{3}p^3(1-p)^{97}=161700p^3(1-p)^{97}\quad\text{(shown)} \](c)
Since \(X=3\) is the mode, \(\mathrm{P}(X=3)>\mathrm{P}(X=2)\) and \(\mathrm{P}(X=3)>\mathrm{P}(X=4)\).
From \(\mathrm{P}(X=3)>\mathrm{P}(X=2)\): \[\begin{aligned} 161700p^3(1-p)^{97} &> 4950p^2(1-p)^{98}\\ \frac{98}{3}\,p &> 1-p\\ \frac{101}{3}\,p &> 1\\ p &> \frac{3}{101} \end{aligned}\]
From \(\mathrm{P}(X=3)>\mathrm{P}(X=4)\): \[\begin{aligned} 161700p^3(1-p)^{97} &> 3921225p^4(1-p)^{96}\\ \frac{161700}{3921225}(1-p) &> p\\ \frac{97}{4}(1-p) &> p \cdot\frac{97}{4}\\ \frac{97}{4} &> \frac{101}{4}\,p\\ p &< \frac{97}{101\cdot 4}\cdot 4 = \frac{4}{101} \end{aligned}\]
(d)
With \(p=0.01\): each plate is faulty with probability 0.01, each bowl with probability \(q\). A set has 2 bowls and 2 plates.
\(\mathrm{P}(\text{at most 1 faulty})=\mathrm{P}(\text{0 faulty})+\mathrm{P}(\text{exactly 1 faulty})=0.88\).
\[\begin{aligned} &(0.99)^2(1-q)^2+2(0.99)(0.01)(1-q)^2+(0.99)^2\cdot 2q(1-q)=0.88\\ &0.9999(1-q)^2+(1.9602)q(1-q)=0.88 \end{aligned}\]Using GC with \(0<q<1\):