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. Calculate the probability of the specific arrangement

$\text{Probability} = \frac{\text{Number of ways to arrange the remaining 6 people}}{\text{Total number of ways to arrange 8 people}} = \frac{6!}{8!} \\ = \frac{1}{8 \times 7} = \frac{1}{56}$

1. Find the individual probabilities of Abbey being first and Betty being second, as well as the probability of both occurring together.

Let $A$ be the event that Abbey is first, and $B$ be the event that Betty is second.
$P(A) = \frac{1}{8}$
$P(B) = \frac{1}{8}$
The probability that both occur (Abbey is first and Betty is second) is:
$P(A \cap B) = P(A) \times P(B|A) = \frac{1}{8} \times \frac{1}{7} = \frac{1}{56}$

2. Apply the addition rule of probability to find the probability of either event occurring.

Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$P(A \cup B) = \frac{1}{8} + \frac{1}{8} - \frac{1}{56}$
$P(A \cup B) = \frac{7}{56} + \frac{7}{56} - \frac{1}{56}$
$P(A \cup B) = \frac{13}{56}$

1. Calculate the probability of picking a gold coin from the drawer with two gold coins

Let $D_1$ be the event of choosing the drawer with two gold coins.
$\text{P}(D_1 \cap \text{Gold}) = \text{P}(D_1) \times \text{P}(\text{Gold} \mid D_1)$
$\text{P}(D_1 \cap \text{Gold}) = \frac{1}{3} \times 1 = \frac{1}{3}$

2. Calculate the total probability of picking a gold coin from any drawer

Let $D_2$ and $D_3$ be the events of choosing the drawers with one gold coin and no gold coins, respectively.
$\text{P}(\text{Gold}) = \text{P}(D_1 \cap \text{Gold}) + \text{P}(D_2 \cap \text{Gold}) + \text{P}(D_3 \cap \text{Gold})$
$\text{P}(\text{Gold}) = \frac{1}{3} + \left(\frac{1}{3} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times 0\right)$
$\text{P}(\text{Gold}) = \frac{1}{3} + \frac{1}{6} + 0 = \frac{1}{2}$

3. Apply the conditional probability formula to find the required probability

$\text{P}(D_1 \mid \text{Gold}) = \frac{\text{P}(D_1 \cap \text{Gold})}{\text{P}(\text{Gold})}$
$\text{P}(D_1 \mid \text{Gold}) = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$

1. Find the probabilities of the older son picking a gold coin from the first and second drawers.

Let $D_1$ contain 2 Gold (GG), $D_2$ contain 1 Gold and 1 Silver (GS), and $D_3$ contain 2 Silver (SS).
The older son chooses a drawer at random (probability $\frac{1}{3}$) and then a coin at random.
$P(\text{Older picks Gold from } D_1) = P(D_1) \times P(\text{Gold} \mid D_1) = \frac{1}{3} \times 1 = \frac{1}{3}$
$P(\text{Older picks Gold from } D_2) = P(D_2) \times P(\text{Gold} \mid D_2) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$
(Note: The probability of picking Gold from $D_3$ is $0$).

2. Determine the probability of the younger son picking a silver coin if the older son took a gold coin from $D_1$.

If the older son took a Gold coin from $D_1$, the drawers now contain:

• $D_1$: 1 Gold
• $D_2$: 1 Gold, 1 Silver
• $D_3$: 2 Silver

The younger son selects a drawer at random (probability $\frac{1}{3}$ each) and a coin from it.
$P(\text{Younger picks Silver} \mid \text{Older picks Gold from } D_1)$
$= P(D_1)P(\text{Silver} \mid D_1) + P(D_2)P(\text{Silver} \mid D_2) + P(D_3)P(\text{Silver} \mid D_3)$
$= \frac{1}{3}(0) + \frac{1}{3}\left(\frac{1}{2}\right) + \frac{1}{3}(1)$
$= 0 + \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$

3. Determine the probability of the younger son picking a silver coin if the older son took a gold coin from $D_2$.

If the older son took a Gold coin from $D_2$, the drawers now contain:

• $D_1$: 2 Gold
• $D_2$: 1 Silver
• $D_3$: 2 Silver

$P(\text{Younger picks Silver} \mid \text{Older picks Gold from } D_2)$
$= P(D_1)P(\text{Silver} \mid D_1) + P(D_2)P(\text{Silver} \mid D_2) + P(D_3)P(\text{Silver} \mid D_3)$
$= \frac{1}{3}(0) + \frac{1}{3}(1) + \frac{1}{3}(1)$
$= 0 + \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

4. Calculate the total probability of the older son picking a gold coin and the younger son picking a silver coin.

Using the law of total probability to combine the mutually exclusive cases:
$P(\text{Older Gold and Younger Silver})$
$= P(\text{Older picks Gold from } D_1) \times P(\text{Younger picks Silver} \mid \text{Older picks Gold from } D_1) \\ \quad + P(\text{Older picks Gold from } D_2) \times P(\text{Younger picks Silver} \mid \text{Older picks Gold from } D_2)$
$= \left(\frac{1}{3} \times \frac{1}{2}\right) + \left(\frac{1}{6} \times \frac{2}{3}\right)$
$= \frac{1}{6} + \frac{2}{18}$
$= \frac{3}{18} + \frac{2}{18} = \frac{5}{18}$