2023 H2 Mathematics Paper 2 Question 10

The Binomial Distribution

Answers

(ai)

The probability of an ornament is faulty is the same for each ornament.
Whether an ornament is faulty is independent of any other ornament.

(aii)
np−npq=0.08. np - npq = 0.08.
(aiii)
0.677.0.677.
(aiv)
0.805.0.805.
(av)
0.583.0.583.

(b)

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Full solutions

(ai)

The probability of an ornament is faulty is the same for each ornament   ■\; \blacksquare
Whether an ornament is faulty is independent of any other ornament   ■\; \blacksquare

(aii)

Let XX denote the random variable of the number of faulty ornaments, out of 50.

X∼B(50,0.04){X \sim \textrm{B}\left(50, 0.04\right)}
Mean−Variance=np−npq=50×0.04−50×0.04×0.96=0.08  ■\begin{align*} & \text{Mean} - \text{Variance} \\ &= np - npq \\ &= 50 \times 0.04 - 50 \times 0.04 \times 0.96 \\ &= 0.08 \; \blacksquare \end{align*}

(aiii)
P(X≤2)=0.67671=0.677 (3 sf)  ■\begin{align*} \mathrm{P}(X \leq 2) &= 0.67671 \\ &= 0.677 \text{ (3 sf)} \; \blacksquare \end{align*}

(aiv)

Let YY denote the random variable of the number of days, out of 5, that no more than 2 faulty ornaments are produced

Y∼B(5,0.67671){Y \sim \textrm{B}\left(5, 0.67671\right)}
P(Y≥3)=1−P(Y≤2)=0.80477=0.805 (3 sf)  ■\begin{align*} \mathrm{P}(Y \geq 3) &= 1 - \mathrm{P}(Y\leq 2) \\ &= 0.80477 \\ &= 0.805 \text{ (3 sf)} \; \blacksquare \end{align*}

(av)

Let WW denote the random variable of the number of faulty items produced in 5 days

W∼B(250,0.04){W \sim \textrm{B}\left(250, 0.04\right)}
P(W≤10)=0.58306=0.583 (3 sf)  ■\begin{align*} \mathrm{P}(W \leq 10) &= 0.58306 \\ &= 0.583 \text{ (3 sf)} \; \blacksquare \end{align*}

(b)

Let LL denote the random variable of the number of faulty pens in a sample of 66

L∼B(6,1−p){ L \sim \text{B}(6, 1-p)}
Proportion of boxes Mr Lu accepts=P(L≤1)=P(L=0)+P(L=1)=p6+(61)(1−p)p6−1=p6+6(1−p)p6−1\begin{align*} & \text{Proportion of boxes Mr Lu accepts} \\ &= \mathrm{P}\left( L \leq 1 \right) \\ &= \mathrm{P}\left( L = 0 \right) + \mathrm{P}\left( L = 1 \right) \\ &= p^6 + {6 \choose 1}(1-p)p^{6-1} \\ &= p^6 + 6(1-p)p^{6-1} \end{align*}

Let MM denote the random variable of the number of faulty pens in a sample of 33

M∼B(3,1−p){ M \sim \text{B}(3, 1-p)}
Proportion of boxes Mrs Ming accepts=P(M=0)+P(M=1)⋅P(M=0)=p3+(31)(1−p)p3−1⋅p3=p3+3(1−p)p5\begin{align*} & \text{Proportion of boxes Mrs Ming accepts} \\ &= \mathrm{P}\left( M = 0 \right) + \mathrm{P}\left( M = 1 \right) \cdot \mathrm{P}\left( M = 0 \right) \\ &= p^3 + {3 \choose 1}(1-p)p^{3-1} \cdot p^{3} \\ &= p^3 + 3(1-p)p^{5} \end{align*}
Proportion of boxes Mr Ming accepts−Proportion of boxes Mr Lu accepts=p3+3(1−p)p5−p6−6(1−p)p5=p3−p6−3(1−p)p5=p3(1−p3)−3(1−p)p5=p3(1−p)(1+p+p2)−3(1−p)p5=p3(1−p)(1+p+p2−3p2)=p3(1−p)(1+p−2p2)=p3(1−p)(1+2p)(1−p)>0since 0<p<1\begin{align*} & \text{Proportion of boxes Mr Ming accepts} \\ & \quad - \text{Proportion of boxes Mr Lu accepts} \\ &= p^3 + 3(1-p)p^5 - p^6 - 6(1-p)p^5 \\ &= p^3 - p^6 - 3(1-p)p^5 \\ &= p^3 (1-p^3) - 3(1-p)p^5 \\ &= p^3 (1-p)(1+p+p^2) - 3(1-p)p^5 \\ &= p^3 (1-p)(1+p+p^2 - 3p^2 ) \\ &= p^3 (1-p) (1 + p - 2p^2) \\ &= p^3 (1-p) (1 + 2p)(1 - p) \\ & > 0 \quad \text{since } 0 < p < 1 \end{align*}

Hence Mrs Ming accepts a greater proportion of boxes than Mr Lu   ■\; \blacksquare

Question Commentary

Part (a) and all its subparts are a pretty standard question about the binomial distribution.

However, part (b) was novel and proving algebraically can be a challenge. Hopefully most students were able to use the binomial pdf formula to get the proportion of boxes Mr Lu and Mrs Ming accepts separately.

Proving which proportion is greater will require a strategy that, while common at higher levels, is not often seen at this level. In general, one approach to show x>yx > y is to show x−y>0.x - y > 0.

This is the approach we take above, and coupled with algebraic manipulation, factorization and the cubic factorization formula

x3−y3=(x−y)(x2+xy+y2),{x^3 - y^3 = (x-y)(x^2 + xy + y^2),}

we are able to show that the difference is positive using the condition 0<p<1.0 < p < 1.