Let the event that the pen is in the pencil case be denoted as \(P(C) = 0.7\), and the event that Ted finds the pen given it is in the pencil case be \(P(F|C) = 1\). The event that the pen is not in the pencil case is \(P(C') = 0.3\), and the probability that Ted finds the pen given it is not in the pencil case is \(P(F|C') = 0.2\).

We need to find \(P(C|F)\), the probability that the pen is in the pencil case given that Ted finds it. Using Bayes' theorem:

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

Where:

\(P(C \cap F) = P(C) \cdot P(F|C) = 0.7 \times 1 = 0.7\)

\(P(F) = P(C \cap F) + P(C' \cap F) = 0.7 + (0.3 \times 0.2) = 0.7 + 0.06 = 0.76\)

Thus:

\(P(C|F) = \frac{0.7}{0.76} = 0.921\)

(i) The tree diagram is structured as follows:

• First branch: Probability of using a cell phone (Ph) = 0.68, not using a cell phone (NPh) = 0.32.
• Second branch for Ph: Young (Y) = 0.7, Middle-aged (M) = 0.25, Old (O) = 0.05.
• Second branch for NPh: Young (Y) = 0.26, Middle-aged (M) = 0.10, Old (O) = 0.64.

(ii) To find the probability that a passenger aged 45 (middle-aged) uses a cell phone, use conditional probability:

\(P(\text{Ph} \mid \text{M}) = \frac{P(\text{Ph} \cap \text{M})}{P(\text{M})}\)

Where:

• \(P(\text{Ph} \cap \text{M}) = 0.68 \times 0.25 = 0.17\)
• \(P(\text{M}) = (0.68 \times 0.25) + (0.32 \times 0.10) = 0.17 + 0.032 = 0.202\)

Thus,

\(P(\text{Ph} \mid \text{M}) = \frac{0.17}{0.202} \approx 0.842\)

To find the probability that the spinner landed on an even-numbered side given that Raj's score is 12, we use conditional probability:

Let \(E\) be the event that the spinner lands on an even-numbered side. We need to find \(P(E \mid 12)\).

First, calculate \(P(E \text{ and } 12)\):

The probability of landing on an even-numbered side is \(\frac{2}{5}\) (since there are 2 even numbers: 2 and 4).

The possible dice outcomes that multiply to 12 are: (2,6), (6,2), (3,4), (4,3). Each pair has a probability of \(\frac{1}{36}\) (since there are 36 possible outcomes when rolling two dice).

Thus, \(P(E \text{ and } 12) = \frac{2}{5} \times \frac{4}{36} = \frac{8}{180} = \frac{2}{45}\).

Next, calculate \(P(12)\):

For an odd-numbered side, the possible dice outcomes that add to 12 are: (6,6). The probability of this is \(\frac{1}{36}\).

The probability of landing on an odd-numbered side is \(\frac{3}{5}\) (since there are 3 odd numbers: 1, 3, 5).

Thus, \(P(12) = \frac{3}{5} \times \frac{1}{36} + \frac{8}{180} = \frac{11}{180}\).

Finally, use the conditional probability formula:

\(P(E \mid 12) = \frac{P(E \text{ and } 12)}{P(12)} = \frac{\frac{8}{180}}{\frac{11}{180}} = \frac{8}{11}\).

Let \(D\) be the event that the dog chases ducks, and \(G\) be the event that the dog chases geese. Let \(A\) be the event that the dog is attacked, and \(NA\) be the event that the dog is not attacked.

We need to find \(P(G \mid NA)\), which is given by:

\(P(G \mid NA) = \frac{P(G \cap NA)}{P(NA)}\)

First, calculate \(P(G \cap NA)\):

\(P(G \cap NA) = P(G) \times P(NA \mid G) = \frac{2}{5} \times \left(1 - \frac{3}{4}\right) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)

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

\(P(NA) = P(D \cap NA) + P(G \cap NA)\)

\(P(D \cap NA) = P(D) \times P(NA \mid D) = \frac{3}{5} \times \left(1 - \frac{1}{10}\right) = \frac{3}{5} \times \frac{9}{10} = \frac{27}{50}\)

Thus,

\(P(NA) = \frac{27}{50} + \frac{1}{10} = \frac{27}{50} + \frac{5}{50} = \frac{32}{50}\)

Finally, substitute back to find \(P(G \mid NA)\):

\(P(G \mid NA) = \frac{\frac{1}{10}}{\frac{32}{50}} = \frac{1}{10} \times \frac{50}{32} = \frac{5}{32}\)

(i) The probability that the first question is answered correctly is calculated as follows:

\(P(\text{1st correct}) = P(\text{Peter correct}) + P(\text{Peter asks for help and audience correct})\)

\(= 0.7 + 0.2 \times 0.95 = 0.89\)

(ii) To find the probability that the first two questions are both answered correctly, consider the following scenarios:

• Peter answers both questions correctly: \(P(CC) = 0.7 \times 0.7 = 0.49\)
• Peter answers the first correctly, asks for help, and the audience answers correctly: \(P(CHA) = 0.7 \times 0.2 \times 0.95 = 0.133\)
• Peter asks for help on the first, audience answers correctly, and Peter answers the second correctly: \(P(HAC) = 0.2 \times 0.95 \times 0.7 = 0.133\)
• Peter asks for help on the first, audience answers correctly, Peter asks for help on the second, and friend answers correctly: \(P(HAHP) = 0.2 \times 0.95 \times 0.2 \times 0.65 = 0.0247\)

Summing these probabilities gives:

\(P(\text{both correctly answered}) = 0.49 + 0.133 + 0.133 + 0.0247 = 0.781\)

(iii) Given that the first two questions were both answered correctly, the probability that Peter asked the audience is:

\(P(\text{audience} \mid \text{both correct}) = \frac{P(CHA) + P(HAC) + P(HAHP)}{P(\text{both correctly answered})}\)

\(= \frac{0.133 + 0.133 + 0.0247}{0.781} = \frac{0.2907}{0.781} = 0.372\)