(i) The probability of taking a toffee from Susan's bag is \(\frac{5}{12}\). After placing it in Ahmad's bag, the probability of taking a boiled sweet from Ahmad's bag is \(\frac{2}{10}\). Therefore, the probability that the two sweets taken are a toffee from Susan’s bag and a boiled sweet from Ahmad’s bag is:

\(P(T, B) = \frac{5}{12} \times \frac{2}{10} = \frac{1}{12}\)

(ii) Let \(C_S\) be the event that a chocolate is taken from Susan's bag, and \(C_A\) be the event that a chocolate is taken from Ahmad's bag. We need to find \(P(C_S | C_A)\).

First, calculate \(P(C_S \cap C_A)\):

\(P(C_S \cap C_A) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120}\)

Next, calculate \(P(C_A)\):

\(P(C_A) = \frac{7}{12} \times \frac{4}{10} + \frac{5}{12} \times \frac{3}{10} = \frac{28}{120} + \frac{15}{120} = \frac{43}{120}\)

Finally, calculate \(P(C_S | C_A)\):

\(P(C_S | C_A) = \frac{P(C_S \cap C_A)}{P(C_A)} = \frac{28/120}{43/120} = \frac{28}{43}\)

(i) The total number of balls in box B after transferring one ball from box A is \(5 + x + 1 = x + 6\). Therefore, the probability of choosing a yellow ball from box B is \(\frac{x}{x+6}\).

(ii) The tree diagram is completed as follows:
From box A to box B:
- White to White: \(\frac{6}{x+6}\)
- White to Yellow: \(\frac{x}{x+6}\)
- Yellow to White: \(\frac{5}{x+6}\)
- Yellow to Yellow: \(\frac{x+1}{x+6}\).

(iii) Given that the probability of choosing a white ball from box B after a white ball from box A is \(\frac{1}{3}\), we have:
\(\frac{6}{x+6} = \frac{1}{3}\)
Solving for \(x\):
\(6 = \frac{x+6}{3}\)
\(18 = x + 6\)
\(x = 12\).

(iv) To find the conditional probability that the ball chosen from box A was yellow given that the ball chosen from box B is yellow, we use:
\(P(A_Y | B_Y) = \frac{P(A_Y \cap B_Y)}{P(B_Y)}\)
\(P(B_Y) = \frac{8}{10} \times \frac{12}{18} + \frac{2}{10} \times \frac{13}{18} = \frac{61}{90}\)
\(P(A_Y \cap B_Y) = \frac{2}{10} \times \frac{13}{18} = \frac{13}{90}\)
\(P(A_Y | B_Y) = \frac{13/90}{61/90} = \frac{13}{61}\).

(i) The probability that Fabio chooses Americano and leaves it to drink later is given by the product of the probability of choosing Americano and the probability of leaving it to drink later. This is calculated as:

\(P(A \text{ Later}) = 0.5 \times 0.2 = 0.1\)

(ii) To find the probability that Fabio chose Latte given that he drinks his coffee immediately, we use Bayes' theorem. Let \(L\) be the event that he chooses Latte, and \(I\) be the event that he drinks immediately. We need to find \(P(L \mid I)\).

First, calculate \(P(I)\), the total probability of drinking immediately:

\(P(I) = (0.5 \times 0.8) + (0.3 \times 0.6) + (0.2 \times 0.1) = 0.4 + 0.18 + 0.02 = 0.6\)

Now, calculate \(P(L \cap I)\), the probability that he chooses Latte and drinks immediately:

\(P(L \cap I) = 0.2 \times 0.1 = 0.02\)

Finally, apply Bayes' theorem:

\(P(L \mid I) = \frac{P(L \cap I)}{P(I)} = \frac{0.02}{0.6} = 0.0333\)

Let \(E\), \(L\), and \(OT\) represent the events that Ana is early, late, and on time, respectively. Let \(B\) represent the event that Ana eats a banana, and \(\overline{B}\) represent the event that she does not eat a banana.

We need to find \(P(OT | \overline{B})\).

Using the law of total probability, we have:

\(P(\overline{B}) = P(\overline{B} | E)P(E) + P(\overline{B} | L)P(L) + P(\overline{B} | OT)P(OT)\)

Given:

• \(P(E) = 0.05\)
• \(P(L) = 0.75\)
• \(P(OT) = 1 - P(E) - P(L) = 0.2\)
• \(P(\overline{B} | E) = 1 - 0.7 = 0.3\)
• \(P(\overline{B} | L) = 1\)
• \(P(\overline{B} | OT) = 1 - 0.4 = 0.6\)

Substitute these values into the equation:

\(P(\overline{B}) = 0.3 \times 0.05 + 1 \times 0.75 + 0.6 \times 0.2\)

\(P(\overline{B}) = 0.015 + 0.75 + 0.12 = 0.885\)

Now, use Bayes' theorem to find \(P(OT | \overline{B})\):

\(P(OT | \overline{B}) = \frac{P(\overline{B} | OT)P(OT)}{P(\overline{B})} = \frac{0.6 \times 0.2}{0.885}\)

\(P(OT | \overline{B}) = \frac{0.12}{0.885} \approx 0.136\)

Thus, the probability that Ana is on time given that she does not eat a banana is \(0.136 \left( \frac{8}{59} \right)\).

(i) The probability that the radio is set to station 3 and the presenter is male is given by:

\(0.25p = 0.075\)

Solving for \(p\):

\(p = \frac{0.075}{0.25} = 0.3\)

(ii) To find the probability that the radio was set to station 2 given that Maria hears a male presenter, we use conditional probability:

\(P(2|M) = \frac{P(2 \text{ and } M)}{P(M)}\)

Where:

\(P(2 \text{ and } M) = 0.45 \times 0.85 = 0.3825\)

\(P(M) = 0.3 \times 0.1 + 0.45 \times 0.85 + 0.25 \times 0.3 = 0.4875\)

Thus,

\(P(2|M) = \frac{0.3825}{0.4875} = 0.785\)