Specimen 2017 H2 Mathematics Paper 2 Question 8

Discrete Random Variables (DRVs)

Answers

{y=2.}

{\mathrm{P}(S=1)=\frac{3}{20}.}

{\mathrm{P}(S=2)=\frac{7}{20}.}

{\mathrm{P}(S=3)=\frac{7}{20}.}

{\mathrm{P}(S=4)=\frac{3}{20}.}

To get

{S=3,}

we either have 3 blue counters or 1 of each colour

\begin{align*} & \mathrm{P}(S=3) \\ & = \frac{3}{y + 4}\cdot\frac{2}{y + 3}\cdot\frac{1}{y + 2} + \frac{3}{y + 4}\cdot\frac{1}{y + 3}\cdot\frac{y}{y + 2} \times 3! \\ & = \frac{6+18y}{(y + 4)(y + 3)(y + 2)} \\ & = \frac{6(3 y + 1)}{(y + 4)(y + 3)(y + 2)} \; \blacksquare \end{align*}

\begin{gather} \quad \quad \frac{6(3 y + 1)}{(y + 4)(y + 3)(y + 2)} = \frac{7}{20} \end{gather}

Solving

{(1)}

with a GC,

y=2 \; \blacksquare

To get

{S=1,}

we have 1 blue and 2 yellow counters

\begin{align*} & \mathrm{P}(S=1) \\ & = \frac{3}{6}\cdot\frac{2}{5}\cdot\frac{1}{4} \times \frac{3!}{2!} \\ & = \frac{3}{20} \; \blacksquare \end{align*}

To get

{S=2,}

we either have 2 blue, 1 yellow or 1 red, 2 yellow

\begin{align*} & \mathrm{P}(S=2) \\ & = \frac{3}{6}\cdot\frac{2}{5}\cdot\frac{2}{4} \times \frac{3!}{2!} + \frac{1}{6}\cdot\frac{2}{5}\cdot\frac{1}{4} \times \frac{3!}{2!} \\ & = \frac{7}{20} \; \blacksquare \end{align*}

To get

{S=4,}

we have 2 blue and 1 red counters

\begin{align*} & \mathrm{P}(S=4) \\ & = \frac{3}{6}\cdot\frac{2}{5}\cdot\frac{1}{4} \times \frac{3!}{2!} \\ & = \frac{3}{20} \; \blacksquare \end{align*}