A circular target board is made up of 3 concentric circles as shown below. The circles have the same centre and their radii are 10 cm, 20 cm and 30 cm respectively. A player plays a game of darts using this target board. The probability of the player missing the target board completely is $\dfrac{1}{5}$ and the player scores no point if he misses the target board. It can be assumed that he is equally likely to land his dart anywhere on the target board if he does not miss the target board. The player scores 5 points if the dart lands on the inner most circle, 3 points if the dart lands between the inner most and the middle circles, and 1 point if the dart lands between the middle and the outer most circles. It can be assumed that the dart will not land on the circumference of the circles.

1. Find the probability of landing a dart on the target board.

\text{P}(\text{hit target board}) = 1 - \text{P}(\text{miss}) = 1 - \frac{1}{5} = \frac{4}{5}

2. Calculate the total area and the area of each scoring region.

\text{Total area} = \pi(30)^2 = 900\pi \text{ cm}^2

\text{Area for 5 points} = \pi(10)^2 = 100\pi \text{ cm}^2

\text{Area for 3 points} = \pi(20)^2 - \pi(10)^2 = 400\pi - 100\pi = 300\pi \text{ cm}^2

\text{Area for 1 point} = \pi(30)^2 - \pi(20)^2 = 900\pi - 400\pi = 500\pi \text{ cm}^2

3. Show the required probabilities by multiplying the probability of a hit by the respective area ratios.

\text{P}(5 \text{ points}) = \frac{4}{5} \times \frac{100\pi}{900\pi} = \frac{4}{45} \text{ (shown)}

\text{P}(3 \text{ points}) = \frac{4}{5} \times \frac{300\pi}{900\pi} = \frac{4}{15} \text{ (shown)}

\text{P}(1 \text{ point}) = \frac{4}{5} \times \frac{500\pi}{900\pi} = \frac{4}{9} \text{ (shown)}

1. Identify all possible scores ($X$) and their corresponding probabilities to construct the probability distribution table.

The problem defines the score of a player in a throw as $X$. The player scores $0$ points if they miss the target board, which has a given probability of $\frac{1}{5}$.

Using the probabilities derived in part (a) for the other scores, we have:
$P(X=0) = \frac{1}{5}$
$P(X=1) = \frac{4}{9}$
$P(X=3) = \frac{4}{15}$
$P(X=5) = \frac{4}{45}$

Tabulating these values gives the probability distribution of $X$:

1. Apply the expectation formula using the probability distribution of $X$

$\text{E}(X) = \sum x \text{P}(X=x)$
$\text{E}(X) = (0)\left(\frac{1}{5}\right) + (1)\left(\frac{4}{9}\right) + (3)\left(\frac{4}{15}\right) + (5)\left(\frac{4}{45}\right)$

2. Evaluate the exact numerical value of the expectation

$\text{E}(X) = 0 + \frac{4}{9} + \frac{12}{15} + \frac{20}{45}$
$\text{E}(X) = \frac{20}{45} + \frac{36}{45} + \frac{20}{45}$
$\text{E}(X) = \frac{76}{45}$