2023 H2 Mathematics Paper 2 Question 6

Permutations and Combinations (P&C)
Discrete Random Variables (DRVs)

Answers

(a)

Show.

(b)

2r+5=b. 2r + 5 = b.
0.0516.0.0516.

Full solutions

(a)

Let RR denote the number of red counters Mei draws.

P(R=4)=P(R=3)(r4)(b8)(r+b12)=(r3)(b9)(r+b12)r!4!(r−4)!b!8!(b−8)!=r!3!(r−3)!b!9!(b−9)!3!(r−3)! 9!(b−9)!=4!(r−4)! 8!(b−8)!3!(r−3)(r−4)! 9⋅8!(b−9)!=4⋅3!(r−4)! 8!(b−8)(b−9)!9(r−3)=4(b−8)9r−27=4b−329r+5=4b  ■\begin{align*} \mathrm{P} \left( R=4 \right) &= \mathrm{P} \left( R=3 \right) \\ \frac{ {r \choose 4} {b \choose 8} }{r+b \choose 12 } &= \frac{ {r \choose 3} {b \choose 9} }{r+b \choose 12 } \\ \frac{r!}{4! \left( r-4 \right)!} \frac{b!}{8! \left( b-8 \right)!} &= \frac{r!}{3! \left( r-3 \right)!} \frac{b!}{9! \left( b-9 \right)!} \\ 3! \left( r-3 \right)! \, 9! \left( b-9 \right)! &= 4! \left( r-4 \right)! \, 8! \left( b-8 \right)! \\ 3! \left( r-3 \right) (r-4)! \, 9 \cdot 8! \left( b-9 \right)! &= 4 \cdot 3! \left( r-4 \right)! \, 8! \left( b-8 \right) (b-9)! \\ 9 \left( r-3 \right) &= 4 \left( b-8 \right) \\ 9r - 27 &= 4b - 32 \\ 9r + 5 &= 4b \; \blacksquare \end{align*}

(b)

Let RR denote the number of red counters Mei draws.

P(R=3)=53P(R=2)(r3)(b9)(r+b12)=53(r2)(b10)(r+b12)r!3!(r−3)!b!9!(b−9)!=53r!2!(r−2)!b!10!(b−10)!3⋅2!(r−2)! 10!(b−10)!=5⋅3!(r−3)! 9!(b−9)!30(r−2)=15(b−9)2r−4=b−92r+5=b  ■\begin{align*} \mathrm{P} \left( R=3 \right) &= \frac{5}{3} \mathrm{P} \left( R=2 \right) \\ \frac{ {r \choose 3} {b \choose 9} }{r+b \choose 12 } &= \frac{5}{3} \frac{ {r \choose 2} {b \choose 10} }{r+b \choose 12 } \\ \frac{r!}{3! \left( r-3 \right)!} \frac{b!}{9! \left( b-9 \right)!} &= \frac{5}{3} \frac{r!}{2! \left( r-2 \right)!} \frac{b!}{10! \left( b-10 \right)!} \\ 3 \cdot 2! \left( r-2 \right)! \, 10! \left( b-10 \right)! &= 5 \cdot 3! \left( r-3 \right)! \, 9! \left( b-9 \right)! \\ 30(r-2) &= 15(b-9) \\ 2r - 4 &= b - 9 \\ 2r + 5 &= b \; \blacksquare \end{align*}

Solving the two equations simultaneously,

9r+5=8r+20r=15b=35\begin{align*} 9 r + 5 &= 8 r + 20 \\ r &= 15 \\ b &= 35 \end{align*}
P(R=1)=(151)(3510)(5012)=0.0516  (3 sf)  ■\begin{align*} & \mathrm{P} \left( R=1 \right) \\ &= \frac{ {15 \choose 1} {35 \choose 10} }{50 \choose 12 } \\ &= 0.0516 \; \textrm{(3 sf)} \; \blacksquare \end{align*}

Question Commentary

This question is really similar to 2020 P2Q8 where we can use combinations to find the required probabilities.