一个圆形靶盘由3个同心圆组成，如下图所示。 这些圆具有相同的圆心，其半径分别为 10 cm、20 cm 和 30 cm。一名选手使用这个靶盘进行飞镖游戏。选手完全脱靶（未击中靶盘）的概率为 $\dfrac{1}{5}$，如果他脱靶，则得 0 分。 可以假设，如果他击中靶盘，飞镖落在靶盘上任何位置的可能性是相等的。 如果飞镖落在最内侧的圆内，选手得 5 分；如果飞镖落在最内侧圆和中间圆之间，得 3 分；如果飞镖落在中间圆和最外侧圆之间，得 1 分。可以假设飞镖不会落在圆周上。

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

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