(a) Let the probability that Alexa arrives late be \(P(\text{late})\). We know that \(P(\text{not late}) = 0.48\).

Using the total probability theorem, \(P(\text{not late}) = 0.4 \times 0.45 + 0.35 \times 0.3 + 0.25 \times (1-x) = 0.48\).

Calculating, \(0.18 + 0.105 + 0.25(1-x) = 0.48\).

\(0.285 + 0.25 - 0.25x = 0.48\).

\(0.535 - 0.25x = 0.48\).

\(0.25x = 0.535 - 0.48\).

\(0.25x = 0.055\).

\(x = \frac{0.055}{0.25} = 0.22\).

(b) To find \(P(\text{train} | \text{late})\), use Bayes' theorem:

\(P(\text{train} | \text{late}) = \frac{P(\text{train} \cap \text{late})}{P(\text{late})}\).

\(P(\text{train} \cap \text{late}) = 0.35 \times 0.7 = 0.245\).

\(P(\text{late}) = 1 - P(\text{not late}) = 0.52\).

Thus, \(P(\text{train} | \text{late}) = \frac{0.245}{0.52} = \frac{49}{104}\) or 0.471.

(a) The tree diagram is constructed as follows:

• First branch: Pass first written test (W1P) with probability 0.8, Fail first written test (W1F) with probability 0.2.
• Second branch (if W1F): Pass second written test (W2P) with probability 0.6, Fail second written test (W2F) with probability 0.4.
• Third branch (if W1P or W2P): Pass practical test (PP) with probability 0.3, Fail practical test (PF) with probability 0.7.

(b) The probability that a student will succeed in gaining a place is calculated by considering the successful paths:

\(P(W1P \cap PP) + P(W1F \cap W2P \cap PP) = (0.8 \times 0.3) + (0.2 \times 0.6 \times 0.3) = 0.24 + 0.036 = 0.276\)

(c) The probability that a student passes the written test at the first attempt given that they succeed in gaining a place is:

\(P(W1P \mid \text{getting place}) = \frac{P(W1P \cap PP)}{P(\text{getting place})} = \frac{0.24}{0.276} = \frac{20}{23} \approx 0.87\)

(a) To find the probability that Georgie wears a hat, we calculate the total probability of wearing a hat with each scarf:

\(P(H) = 0.2 \times 1 + 0.45 \times 0.4 + 0.35 \times 0.3\)

\(= 0.2 + 0.18 + 0.105\)

\(= 0.485\)

Thus, the probability that Georgie wears a hat is \(\frac{97}{200}\) or 0.485.

(b) To find the probability that Georgie wears a yellow scarf given that she does not wear a hat, we use conditional probability:

\(P(Y | \overline{H}) = \frac{P(Y \cap \overline{H})}{P(\overline{H})}\)

\(P(Y \cap \overline{H}) = 0.35 \times 0.7 = 0.245\)

\(P(\overline{H}) = 1 - P(H) = 1 - 0.485 = 0.515\)

\(P(Y | \overline{H}) = \frac{0.245}{0.515} \approx 0.476\)

Thus, the probability that Georgie wears a yellow scarf given that she does not wear a hat is \(\frac{49}{103}\) or 0.476.

(a) The tree diagram is completed with the following probabilities:

• 1 April Fine to 2 April Fine: 0.75
• 1 April Fine to 2 April Rainy: 0.25
• 1 April Rainy to 2 April Fine: 0.4
• 1 April Rainy to 2 April Rainy: 0.6

(b) The probability that 2 April is fine is calculated as follows:

\(P(\text{2 April Fine}) = 0.8 \times 0.75 + 0.2 \times 0.4 = 0.6 + 0.08 = 0.68\)

Thus, \(P(\text{2 April Fine}) = \frac{17}{25}\).

(c) To find \(P(X \cap Y)\), we calculate:

\(P(X \cap Y) = 0.8 \times 0.25 + 0.8 \times 0.6 = 0.2 + 0.12 = 0.27\)

(d) To find the probability that 1 April is fine given that 3 April is rainy, we use:

\(P(Y) = 0.2 \times 0.4 \times 0.25 + 0.2 \times 0.6 \times 0.6 = 0.362\)

\(P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0.27}{0.362} = 0.746\)

(a) The tree diagram is constructed as follows:

• First level: Modes of transport with probabilities: Car (0.2), Bus (0.45), Walk (0.35).
• Second level: Arrival times for each mode: Early (E) or Not Early (NE).
• Probabilities for Car: E (0.6), NE (0.4).
• Probabilities for Bus: E (0.1), NE (0.9).
• Probabilities for Walk: E (1), NE (0).

(b) To find the probability that Juan goes to college by car given that he arrives early, we use Bayes' theorem:

\(P(C|E) = \frac{P(C \cap E)}{P(E)}\)

Where:

• \(P(C \cap E) = 0.2 \times 0.6 = 0.12\)
• \(P(E) = (0.2 \times 0.6) + (0.45 \times 0.1) + (0.35 \times 1) = 0.12 + 0.045 + 0.35 = 0.515\)

Thus,

\(P(C|E) = \frac{0.12}{0.515} = \frac{12}{515} \approx 0.233\)