Question
[Maximum mark: 6]
The following diagram shows an archery target which is divided into three regions A, B and C.
A contest consists of an archer shooting one arrow at the target. The probability of hitting each region is given in the following table.
$
\begin{array}{|l|c|c|c|}
\hline \text { Region } & \text { A } & \text { B } & \text { C } \\
\hline \text { Probability } & \frac{1}{24} & \frac{4}{24} & \frac{7}{24} \\
\hline
\end{array}
$
(a) Find the probability that the arrow does not hit the target. [2]
An archer scores points on the contest as shown in the following table.
$
\begin{array}{|c|c|c|c|c|}
\hline \text { Region } & \text { A } & \text { B } & \text { C } & \text { Missed Target } \\
\hline \text { Points } & 10 & 6 & k & -4 \\
\hline
\end{array}
$
(b) Given that the contest is fair, find the value of $k$. [4]
Answer/Explanation
(a) We have
$
\begin{aligned}
\operatorname{Pr}(\mathrm{A})+\operatorname{Pr}(\mathrm{B})+\operatorname{Pr}(\mathrm{C})+\operatorname{Pr}(\text { Missed Target }) & =1 \\
\frac{1}{24}+\frac{4}{24}+\frac{7}{24}+\operatorname{Pr}(\text { Missed Target }) & =1 \\
\frac{12}{24}+\operatorname{Pr}(\text { Missed Target }) & =1 \\
\operatorname{Pr}(\text { Missed Target }) & =\frac{1}{2}
\end{aligned}
$
(b) Let $X$ be the number of points scored by an archer on the contest.
Hence, using the formula for the expected value of $X$, we get
$
\begin{aligned}
\mathrm{E}(X) & =0 \quad \text { [fair contest] } \\
10 \times \frac{1}{24}+6 \times \frac{4}{24}+k \times \frac{7}{24}+(-4) \times \frac{12}{24} & =0 \\
\frac{10+24+7 k-48}{24} & =0 \\
7 k-14 & =0 \\
k & =2
\end{aligned}
$
Question
[Maximum mark: 6]
A national park is in the form of a triangle, $\mathrm{ABC}$, as shown in the following diagram. Side length $\mathrm{AB}$ is $49 \mathrm{~km}$ and side length $\mathrm{BC}$ is $66 \mathrm{~km}$. Angle $\mathrm{A\hat{B}}\mathrm{C}$ is $58^{\circ}$.
(a) Calculate the side length $\mathrm{AC}$ in $\mathrm{km}$. [3]
(b) Calculate the area of the park. [3]
Answer/Explanation
(a) Using the cosine rule, we get
$
\begin{aligned}
\mathrm{AC}^2 & =(\mathrm{AB})^2+(\mathrm{BC})^2-2(\mathrm{AB})(\mathrm{BC}) \cos (\mathrm{A\hat{B}}\mathrm{C}) \\
& =49^2+66^2-2(49)(66) \cos \left(58^{\circ}\right) \\
\mathrm{AC} & =\sqrt{49^2+66^2-2(49)(66) \cos \left(58^{\circ}\right)} \\
& \approx 57.7 \mathrm{~km}
\end{aligned}
$
(b) Using the area of a triangle formula, we get
$
\begin{aligned}
A & =\frac{1}{2}(\mathrm{AB})(\mathrm{BC}) \sin (\mathrm{A\hat{B}}\mathrm{C}) \\
& =\frac{1}{2}(49)(66) \sin \left(58^{\circ}\right) \\
& \approx 1370 \mathrm{~km}^2
\end{aligned}
$