(i) Let \(P(BH)\) be the probability that Khalid hurts himself while riding his bicycle, and \(P(SH)\) be the probability that he hurts himself while riding his skateboard. We have:

\(P(BH) = 0.6 \times 0.05 = 0.03\)

\(P(SH) = 0.4 \times 0.75 = 0.3\)

The probability that Khalid hurts himself on any particular day is:

\(P(H) = P(BH) + P(SH) = 0.03 + 0.3 = 0.33\)

(ii) We need to find \(P(S|H)\), the probability that Khalid is riding on his skateboard given that he hurts himself. Using the formula for conditional probability:

\(P(S|H) = \frac{P(S \cap H)}{P(H)}\)

\(P(S \cap H) = P(SH) = 0.3\)

\(P(H) = 0.33\)

Thus, \(P(S|H) = \frac{0.3}{0.33} = \frac{10}{11} \approx 0.909\)

(i) The tree diagram is drawn with the following branches:

• First match: Win (\(\frac{3}{5}\)), Draw (\(\frac{1}{5}\)), Lose (\(\frac{1}{5}\))
• Second match if Win first: Win (\(\frac{7}{10}\)), Draw (\(\frac{2}{10}\)), Lose (\(\frac{1}{10}\))
• Second match if Draw first: Win (\(\frac{1}{3}\)), Draw (\(\frac{1}{3}\)), Lose (\(\frac{1}{3}\))
• Second match if Lose first: Win (\(\frac{3}{10}\)), Draw (\(\frac{1}{20}\)), Lose (\(\frac{13}{20}\))

(ii) To find the probability that they lose the first match given they win the second match, use Bayes' theorem:

\(P(L_1 \mid W_2) = \frac{P(L_1 \cap W_2)}{P(W_2)}\)

Calculate \(P(L_1 \cap W_2)\):

\(P(L_1 \cap W_2) = \frac{1}{5} \times \frac{3}{10} = \frac{3}{50}\)

Calculate \(P(W_2)\):

\(P(W_2) = \frac{3}{5} \times \frac{7}{10} + \frac{1}{5} \times \frac{1}{3} + \frac{1}{5} \times \frac{3}{10}\)

\(P(W_2) = \frac{21}{50} + \frac{1}{15} + \frac{3}{50}\)

\(P(W_2) = \frac{41}{75}\)

Thus,

\(P(L_1 \mid W_2) = \frac{3/50}{41/75} = \frac{9}{82}\)

Let \(W\), \(C\), and \(B\) represent the events that Anya walks, cycles, and takes the bus, respectively. Let \(L\) represent the event that Anya is late.

The probability that Anya cycles and is late is \(P(C \cap L) = P(C) \times P(L \mid C) = 0.65 \times 0.1 = 0.065\).

The total probability that Anya is late, \(P(L)\), is given by:

\(P(L) = P(W \cap L) + P(C \cap L) + P(B \cap L)\)

\(= (0.3 \times 0.15) + (0.65 \times 0.1) + (0.05 \times 0.6)\)

\(= 0.045 + 0.065 + 0.03 = 0.14\)

The probability that Anya cycles given that she is late is:

\(P(C \mid L) = \frac{P(C \cap L)}{P(L)} = \frac{0.065}{0.14} = 0.464\)

Thus, the probability that Anya cycles given that she is late is \(0.464\) or \(\frac{13}{28}\).

(i) The probability of taking a blue pen from the left pocket is \(\frac{1}{4}\). After moving this pen to the right pocket, the probability of taking a blue pen from the right pocket is \(\frac{2}{5}\). Therefore, the probability is \(\frac{1}{4} \times \frac{2}{5} = \frac{1}{10}\).

(ii) To have 1 blue pen in the left pocket after operation T, either a red pen is moved from left to right and a red pen is moved back, or a blue pen is moved from left to right and a blue pen is moved back. The probability of moving a red pen from left to right is \(\frac{3}{4}\) and then moving a red pen back is \(\frac{4}{5}\). The probability of moving a blue pen from left to right and then moving a blue pen back is \(\frac{1}{10}\). Thus, \(P(X = 1) = \frac{3}{4} \times \frac{4}{5} + \frac{1}{10} = \frac{14}{20} = \frac{7}{10}\).

(iii) Given that a blue pen is taken from the right pocket, the probability that a blue pen was initially taken from the left pocket is \(\frac{P(B \cap B)}{P(B)} = \frac{1/10}{3/4 \times 1/5 + 1/4 \times 2/5} = \frac{1/10}{1/5} = \frac{2}{5}\).

(i) The tree diagram is constructed as follows:

• First branch: Tea (T) with probability 0.65, Coffee (C) with probability 0.28, Hot Chocolate (HC) with probability 0.07.
• Second branch for Tea: Milk (M) with probability 0.8, No Milk (NM) with probability 0.2.
• Second branch for Coffee: Milk (M) with probability 0.5, No Milk (NM) with probability 0.5.
• Second branch for Hot Chocolate: Milk (M) with probability 1, No Milk (NM) with probability 0.

(ii) To find the probability that Ayman’s breakfast drink is coffee, given that his drink has milk in it, we use Bayes' theorem:

\(P(C \mid \text{milk}) = \frac{P(C \cap \text{milk})}{P(\text{milk})}\)

Calculate \(P(C \cap \text{milk})\):

\(P(C \cap \text{milk}) = 0.28 \times 0.5 = 0.14\)

Calculate \(P(\text{milk})\):

\(P(\text{milk}) = (0.65 \times 0.8) + (0.28 \times 0.5) + (0.07 \times 1) = 0.52 + 0.14 + 0.07 = 0.73\)

Substitute these into the formula:

\(P(C \mid \text{milk}) = \frac{0.14}{0.73} \approx 0.192\)