<p>A bag contains $r$ red counters and $b$ blue counters, where $r > 12$ and $b > 12$. Mei randomly removes $12$ counters from the bag. The probability that there are $4$ red counters among Mei's $12$ counters is the same as the probability that there are $3$ red counters.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $9r+5=4b$.</div><p style="text-align: right;">[2]</p></li></ol>
<p>The probability that there are $3$ red counters among Mei's $12$ counters is $\frac{5}{3}$ times the probability that there are $2$ red counters.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">Derive an equation similar to the equation in part <strong>(a)</strong> and hence find the probability that just one of the $12$ counters removed is red.</div><p style="text-align: right;">[6]</p></li></ol>

A bag contains $r$ red counters and $b$ blue counters, where $r > 12$ and $b > 12$. Mei randomly removes $12$ counters from the bag. The probability that there are $4$ red counters among Mei's $12$ counters is the same as the probability that there are $3$ red counters.

The probability that there are $3$ red counters among Mei's $12$ counters is $\frac{5}{3}$ times the probability that there are $2$ red counters.

Derive an equation similar to the equation in part <strong>(a)</strong> and hence find the probability that just one of the $12$ counters removed is red.

Probability

H2 Math

H1 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2023 P2 Q6

$P(4 \text{ Red}, 8 \text{ Blue}) = P(3 \text{ Red}, 9 \text{ Blue})$

\[495 r(r-1)(r-2)(r-3) b(b-1)\cdots(b-7) = 220 r(r-1)(r-2) b(b-1)\cdots(b-7)(b-8)\]

$n(4\text{Red}, 8\text{Blue}) = n(3\text{Red}, 9\text{Blue})$

\[{}^rC_4 \times {}^bC_8 = {}^rC_3 \times {}^bC_9\]

\[\dfrac{r!}{4!(r-4)!} \cdot \dfrac{b!}{8!(b-8)!} = \dfrac{r!}{3!(r-3)!} \cdot \dfrac{b!}{9!(b-9)!}\]

\[4(r-4)!(b-8)(b-9)! = 9(r-3)(r-4)!(b-9)!\]

\[{}^rC_3 \times {}^bC_9 = \dfrac{5}{3} \cdot {}^rC_2 \times {}^bC_{10}\]

\[\dfrac{1}{3(b-9)} = \dfrac{5}{3(r-2)(10)}\]

Solving (1) & (2): $b = 35$, $r = 15$

Required Probability $= P(1R, 11B) = \dfrac{{}^{15}C_1 \times {}^{35}C_{11}}{{}^{50}C_{12}}$

N2023 P2 Q6

N2023 P2 Q6

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