一对同胞 Abbey 和 Betty 还有其他六个朋友一起去参加书展。他们八个人随机排成一队等候一位著名作家的签名。求满足以下条件的概率： Abbey 排在队列第一位且 Betty 排在队列第六位， Abbey 排在队列第一位或 Betty 排在队列第二位（或两者兼有）。 一个人有一个带有三个抽屉的木柜。一个抽屉里装有两枚金币，另一个抽屉里装有一枚金币和一枚银币，最后一个抽屉里装有两枚银币。随机选择一个抽屉，然后从该抽屉中随机选择一枚硬币。 已知选中了一枚金币，求所选抽屉是装有两枚金币的那个抽屉的概率。 这个人的两个儿子轮流以同样的方式从抽屉中各取出一枚硬币，由大儿子先开始。大儿子拿到金币且小儿子拿到银币的概率是多少？

1. Determine the total number of possible arrangements and the number of favourable arrangements.

Total number of possible arrangements for the $8$ people $= 8!$

If Abbey is exactly 1st and Betty is exactly 6th, their positions are fixed. The remaining $6$ friends can be arranged in the other $6$ positions in $6!$ ways.
Number of favourable arrangements $= 6!$

2. Calculate the required probability.

$\text{Probability} = \frac{6!}{8!} = \frac{6!}{8 \times 7 \times 6!} = \frac{1}{56}$

1. Determine the individual probabilities of Abbey being first, Betty being second, and both occurring together.

Let $A$ be the event that Abbey is first, and $B$ be the event that Betty is second. The total number of ways to arrange the 8 people is $8!$.

If Abbey is first, the remaining 7 people can be arranged in $7!$ ways:
$P(A) = \frac{7!}{8!} = \frac{1}{8}$

If Betty is second, the remaining 7 people can be arranged in $7!$ ways:
$P(B) = \frac{7!}{8!} = \frac{1}{8}$

If Abbey is first and Betty is second, the remaining 6 people can be arranged in $6!$ ways:
$P(A \cap B) = \frac{6!}{8!} = \frac{1}{56}$

2. Apply the addition rule to find the probability of the union of the two events.

Using the formula for the probability of either or both events occurring:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{1}{8} + \frac{1}{8} - \frac{1}{56}$
$P(A \cup B) = \frac{7}{56} + \frac{7}{56} - \frac{1}{56} = \frac{13}{56}$

1. Calculate the total probability of selecting a gold coin.

Let $D_1$ be the drawer with two gold coins, $D_2$ be the drawer with one gold and one silver coin, and $D_3$ be the drawer with two silver coins. Let $G$ be the event of selecting a gold coin.
$\begin{aligned} \text{P}(G) &= \text{P}(D_1 \cap G) + \text{P}(D_2 \cap G) + \text{P}(D_3 \cap G) \\ &= \text{P}(D_1)\text{P}(G \mid D_1) + \text{P}(D_2)\text{P}(G \mid D_2) + \text{P}(D_3)\text{P}(G \mid D_3) \\ &= \left(\frac{1}{3}\right)(1) + \left(\frac{1}{3}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{3}\right)(0) \\ &= \frac{1}{3} + \frac{1}{6} \\ &= \frac{1}{2} \end{aligned}$

2. State the probability of selecting the drawer with two gold coins and successfully drawing a gold coin.

$\text{P}(D_1 \cap G) = \frac{1}{3} \times 1 = \frac{1}{3}$

3. Calculate the conditional probability that the chosen drawer is the one with two gold coins given a gold coin is selected.

$\begin{aligned} \text{P}(D_1 \mid G) &= \frac{\text{P}(D_1 \cap G)}{\text{P}(G)} \\ &= \frac{1/3}{1/2} \\ &= \frac{2}{3} \end{aligned}$

1. Determine which drawer must be chosen to satisfy the outcome

For the sons to draw one gold and one silver coin from the same drawer, the selected drawer must be the one containing exactly one gold and one silver coin (let's call this Drawer 2).

2. Calculate the joint probability of selecting the drawer and drawing the required sequence of coins

The probability of randomly selecting Drawer 2 is $\frac{1}{3}$.
Given Drawer 2 is selected, the probability that the older son picks the gold coin is $\frac{1}{2}$.
After the older son removes the gold coin, only the silver coin is left in the drawer. The probability that the younger son then picks the silver coin is $1$.

$\text{P(Older picks Gold and Younger picks Silver)} = \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6}$