2022 H2 Mathematics Paper 2 Question 9

The Binomial Distribution

Answers

(a)

2p3q+12p4.{2p^3q + \frac{1}{2}p^4.}

(b)

4−8p+5p2(4−3p)2.{\frac{4 - 8 p + 5 p^2}{(4 - 3 p)^2}.}

(c)

p=0.710.{p=0.710.}

Full solutions

(a)

Let X{X} denote the number of left turns a counter takes (from either L{L} or R.{R.})
R∼B(4,p)R \sim \mathrm{B}(4, p)
To arrive at B,{B,} we can either take 3{3} left turns from L{L} or 4{4} left turns from R.{R.}
P(B)=12P(X=3)+12P(X=4)=12(43)p3q+12p4=2p3q+12p4  ■\begin{align*} \textrm{P}(B) &= \frac{1}{2} \textrm{P}(X=3) + \frac{1}{2} \textrm{P}(X=4) \\ &= \frac{1}{2} {4 \choose 3} p^3 q + \frac{1}{2}p^4 \\ &= 2p^3q + \frac{1}{2} p^4 \; \blacksquare \end{align*}

(b)

P(LLLLR)=12p3qP(LLLRL)=12p3qP(LLLRL)=12p3qP(LLRLL)=12p3qP(LRLLL)=12p3qP(RLLLL)=12p4\begin{align*} \textrm{P}(LLLLR) &= \frac{1}{2}p^3q \\ \textrm{P}(LLLRL) &= \frac{1}{2}p^3q \\ \textrm{P}(LLLRL) &= \frac{1}{2}p^3q \\ \textrm{P}(LLRLL) &= \frac{1}{2}p^3q \\ \textrm{P}(LRLLL) &= \frac{1}{2}p^3q \\ \textrm{P}(RLLLL) &= \frac{1}{2}p^4 \end{align*}
P(J,K same route∣both B)=P(J,K same route∩both B)P(both B)=4(12p3q)2+(12p4)2(2p3q+12p4)2=p6q2+14p814p6(4q+p)2=14p6(4q2+p2)14p6(4q+p)2=4(1−p)2+p2(4(1−p)+p)2=4−8p+5p2(4−3p)2  ■\begin{align*} & \textrm{P}\left( J,K \textrm{ same route} \mid \textrm{both } B \right) \\ &= \frac{\textrm{P}\left( J,K \textrm{ same route} \cap \textrm{both } B \right)}{\textrm{P}(\textrm{both } B)} \\ &= \frac{4 \left( \frac{1}{2}p^3q \right)^2 + \left( \frac{1}{2}p^4 \right)^2}{(2p^3q + \frac{1}{2}p^4)^2} \\ &= \frac{p^6 q^2 + \frac{1}{4}p^8}{\frac{1}{4}p^6 (4q + p)^2} \\ &= \frac{\frac{1}{4}p^6 (4q^2 + p^2)}{\frac{1}{4}p^6 (4q + p)^2} \\ &= \frac{4(1-p)^2 + p^2}{(4(1-p)+p)^2} \\ &= \frac{4 - 8 p + 5 p^2}{(4 - 3 p)^2} \; \blacksquare \end{align*}

(c)

P(C)=12P(X=2)+12P(X=3)=12(42)p2q2+12(43)p3q=3p2q2+2p3q  ■\begin{align*} \textrm{P}(C) &= \frac{1}{2} \textrm{P}(X=2) + \frac{1}{2} \textrm{P}(X=3) \\ &= \frac{1}{2} {4 \choose 2} p^2 q^2 + \frac{1}{2} {4 \choose 3} p^3 q \\ &= 3p^2q^2 + 2p^3q \; \blacksquare \end{align*}
Since probability that a counter arrives at B{B} is the same as the probability that it arrives at C,{C,}
2p3q+12p4=3p2q2+2p3qp2=6q2p2=6(1−p)2p2=6−12p+6p2\begin{align*} 2p^3q + \frac{1}{2}p^4 &= 3p^2q^2 + 2p^3q \\ p^2 &= 6q^2 \\ p^2 &= 6(1-p)^2 \\ p^2 &= 6 - 12 p + 6 p^2 \end{align*}
6−12p+5p2=0 6 - 12 p + 5 p^2 = 0
From GC, since 0<p<1,{0 < p < 1,}
p=0.710  ■p = 0.710 \; \blacksquare

Question Commentary

This question is remarkably similar to 2018 paper 2 question 6 so we hope students who have attempted that question will be able to apply the concepts here.

The main difficulty in parts (a) and (c) is in translating the question into a binomial distribution. Part (b) is potentially trickier. We hope students will be able to identify the conditional probability and apply the formula. Thereafter, the numerator term will require considering the different cases for both events to happen.