A pair of siblings, Abbey and Betty, went to a book fair with six other friends. The eight of them stood in a queue in random order to get the autograph of a well-known writer. Find the probability that Abbey is first and Betty is sixth in the queue, either Abbey is first or Betty is second in the queue (or both). A man owns a wooden chest with three drawers. One drawer contains two gold coins, another contains one gold coin and one silver coin, and the last drawer contains two silver coins. A drawer is selected randomly and a coin is then selected randomly from that drawer. Given that a gold coin is selected, find the probability that the chosen drawer is the one that contains two gold coins. The man's two sons take turns to select a coin each from a drawer in the same manner, starting with the older son. What is the probability that the older son picks a gold coin, and the younger one picks a silver coin?

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 $1$.

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