(a) The total probability that Kino is late is given by:

\(0.2x + 0.1 \times 2x + 0.7 \times 0.25 = 0.235\)

Simplifying, we have:

\(0.2x + 0.2x + 0.175 = 0.235\)

\(0.4x + 0.175 = 0.235\)

Subtract 0.175 from both sides:

\(0.4x = 0.06\)

Divide by 0.4:

\(x = 0.15\)

(b) We need to find \(P(\text{car} | \text{not late})\).

\(P(\text{car} | \text{not late}) = \frac{P(\text{car and not late})}{P(\text{not late})}\)

\(P(\text{car and not late}) = 0.1 \times (1 - 2x) = 0.1 \times (1 - 0.3) = 0.1 \times 0.7 = 0.07\)

\(P(\text{not late}) = 1 - 0.235 = 0.765\)

Thus,

\(P(\text{car} | \text{not late}) = \frac{0.07}{0.765} = 0.0915\)

Expressed as a fraction, \(\frac{0.07}{0.765} = \frac{70}{765} = \frac{14}{153}\).

(i) The tree diagram is structured as follows:

• First branch: Toast (T) with probability 0.85, Bread roll (B) with probability 0.15.
• Second branch from Toast: Jam (J) with probability 0.8, No Jam (NJ) with probability 0.2.
• Second branch from Bread roll: Jam (J) with probability 0.4, No Jam (NJ) with probability 0.6.

(ii) To find the probability that Maria had toast given she did not have jam, use Bayes' theorem:

\(P(T \mid NJ) = \frac{P(T \text{ and } NJ)}{P(NJ)}\)

Calculate \(P(T \text{ and } NJ)\):

\(P(T \text{ and } NJ) = 0.85 \times 0.2 = 0.17\)

Calculate \(P(NJ)\):

\(P(NJ) = P(T \text{ and } NJ) + P(B \text{ and } NJ) = 0.85 \times 0.2 + 0.15 \times 0.6 = 0.17 + 0.09 = 0.26\)

Substitute back into Bayes' theorem:

\(P(T \mid NJ) = \frac{0.17}{0.26} = \frac{17}{26} \approx 0.654\)

(i) The probability of selecting 2 pears and 1 orange can be calculated using combinations:

\(\frac{{^4C_2 \times ^7C_1}}{{^{11}C_3}} = \frac{6 \times 7}{165} = 0.255\)

Alternatively, using probabilities:

\(\frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} \times 3 = 0.255\)

(ii) The probability that the third fruit is an orange can be calculated by considering different sequences:

\(P(O) = P(P, P, O) + P(P, O, O) + P(O, P, O) + P(O, O, O)\)

\(= \frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9} + \frac{7}{11} \times \frac{4}{10} \times \frac{6}{9} + \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9}\)

\(= \frac{14}{165} + \frac{28}{165} + \frac{28}{165} + \frac{7}{33} = \frac{7}{11}\)

(iii) The probability that the first fruit was a pear, given that the third fruit is an orange, is calculated using conditional probability:

\(P(P | O) = \frac{P(P \cap O)}{P(O)}\)

\(= \frac{P(P, P, O) + P(P, O, O)}{P(O)}\)

\(= \frac{\frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9}}{\frac{7}{11}}\)

\(= \frac{28/110 + 28/70}{7/11} = 0.4\)

(i) The probability of getting 2 heads (HH) is \(\frac{1}{4}\), so \(P(E) = \frac{1}{4}\). The probability of getting 2 tails (TT) is \(\frac{1}{4}\), so \(P(C) = \frac{1}{4}\). The probability of getting 1 head and 1 tail (HT or TH) is \(\frac{1}{2}\), so \(P(JT) = \frac{1}{2}\).

(ii) The tree diagram is as follows:

(iii) The probability that Ravi is frightened is calculated by summing the probabilities of being frightened on each ride:

\(P(F) = \left( \frac{1}{4} \times \frac{6}{10} \right) + \left( \frac{1}{4} \times \frac{7}{10} \right) + \left( \frac{1}{2} \times \frac{8}{10} \right) = \frac{29}{40}\) or 0.725.

(iv) Given that Ravi is not frightened, the probability that he went on the camel ride is:

\(P(C \mid NF) = \frac{P(C \cap NF)}{P(NF)} = \frac{\frac{3}{40}}{1 - \frac{29}{40}} = \frac{3}{11}\) or 0.273.

(i) The tree diagram is drawn with branches for stopping (S) and not stopping (NS) at each set of lights. The probabilities are labeled as follows: first light (S: 0.4, NS: 0.6), second light (S: 0.8, NS: 0.2), third light (S: 0.3, NS: 0.7).

(ii) The probability that Karinne stops at the first two sets but not the third is calculated as:

\(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\) or \(\frac{28}{125}\).

(iii) The probability that Karinne stops at exactly two sets of lights is the sum of the probabilities of the following scenarios:

\(P(S, S, NS) + P(S, NS, S) + P(NS, S, S)\).

Calculating each:

\(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\)

\(P(S, NS, S) = 0.4 \times 0.2 \times 0.3 = 0.024\)

\(P(NS, S, S) = 0.6 \times 0.8 \times 0.3 = 0.144\)

Summing these gives:

\(0.224 + 0.024 + 0.144 = 0.392\) or \(\frac{49}{125}\).

(iv) The probability that Karinne stops at the first set of lights, given that she stops at exactly two sets, is calculated using conditional probability:

\(P(S \text{ at first light} | \text{stops at exactly 2 lights}) = \frac{P(S, NS, S) + P(S, S, NS)}{0.392}\).

\(= \frac{0.024 + 0.224}{0.392} = \frac{0.248}{0.392} = 0.633\) or \(\frac{31}{49}\).