The diagram below shows a circular target board of radius 40 cm, divided by two concentric circles of radii 12 cm and 24 cm, into three regions, Gold, Red and Blue. Arrows are shot at the target board and the scores obtained for hitting the Gold, Red and Blue regions are 10 points, 6 points and 3 points respectively. Arrows that land outside the target board score 0 points. Alex shoots one arrow at the target board. The probability that his shot lands on the target board is $p$, where $p > 0.5$. If his shot lands on the target board, the arrow is equally likely to hit any position on the board. Let $X$ represent the score obtained from a single shot. You may assume that the arrow does not land on the boundary of any region.

1. Calculate the total area of the target board and the area of the Red region.

Area of target board $= \pi(40)^2 = 1600\pi\text{ cm}^2$

Area of Red region $= \pi(24)^2 - \pi(12)^2 = 576\pi - 144\pi = 432\pi\text{ cm}^2$

2. Determine $\mathrm{P}(X=6)$ by multiplying the probability of landing on the board by the conditional probability of landing in the Red region.

$\mathrm{P}(\text{Red} \mid \text{lands on board}) = \frac{432\pi}{1600\pi} = \frac{27}{100} = 0.27$

$\mathrm{P}(X=6) = \mathrm{P}(\text{lands on board}) \times \mathrm{P}(\text{Red} \mid \text{lands on board})$
$\mathrm{P}(X=6) = p \times 0.27 = 0.27p$

1. Determine the full probability distribution of $X$.

$\mathrm{P}(X=10) = \frac{\pi(12)^2}{\pi(40)^2} p = \frac{144\pi}{1600\pi} p = 0.09p$
From part (a), $\mathrm{P}(X=6) = 0.27p$.
$\mathrm{P}(X=3) = \frac{\pi(40)^2 - \pi(24)^2}{\pi(40)^2} p = \frac{1600\pi - 576\pi}{1600\pi} p = 0.64p$
$\mathrm{P}(X=0) = 1 - p$

2. Calculate the expected value, $\operatorname{E}(X)$.

$\operatorname{E}(X) = 10(0.09p) + 6(0.27p) + 3(0.64p) + 0(1-p)$
$\operatorname{E}(X) = 0.9p + 1.62p + 1.92p = 4.44p$

3. Calculate the expected value of the squares, $\operatorname{E}(X^2)$.

$\operatorname{E}(X^2) = 10^2(0.09p) + 6^2(0.27p) + 3^2(0.64p) + 0^2(1-p)$
$\operatorname{E}(X^2) = 100(0.09p) + 36(0.27p) + 9(0.64p)$
$\operatorname{E}(X^2) = 9p + 9.72p + 5.76p = 24.48p$

4. Form the variance equation and solve for $p$.

$\operatorname{Var}(X) = \operatorname{E}(X^2) - [\operatorname{E}(X)]^2$
$5.464476 = 24.48p - (4.44p)^2$
$5.464476 = 24.48p - 19.7136p^2$
$19.7136p^2 - 24.48p + 5.464476 = 0$
Using the quadratic formula:
$p = \frac{-(-24.48) \pm \sqrt{(-24.48)^2 - 4(19.7136)(5.464476)}}{2(19.7136)}$
$p = \frac{24.48 \pm \sqrt{168.3724237}}{39.4272}$
$p = 0.95 \quad \text{or} \quad p \approx 0.2918$
Since it is given that $p > 0.5$, we reject $p \approx 0.2918$.
$p = 0.95$