To find the probability that a randomly chosen student is at Devar college given that they prefer soccer, we use conditional probability.

The formula for conditional probability is:

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

Where:

• \(P(D|S)\) is the probability that a student is at Devar college given they prefer soccer.
• \(P(D \cap S)\) is the probability that a student is both at Devar college and prefers soccer.
• \(P(S)\) is the probability that a student prefers soccer.

From the table, the number of students who are at Devar college and prefer soccer is 120.

The total number of students who prefer soccer is 280.

Thus, \(P(D|S) = \frac{120}{280}\).

Simplifying \(\frac{120}{280}\) gives \(\frac{3}{7}\).

Therefore, the probability is \(\frac{3}{7}\).

(a) The tree diagram shows the following probabilities:

• Choosing bag X or Y: each with probability \(\frac{1}{2}\).
• For bag X: Removing a red marble \(\frac{7}{10}\), blue marble \(\frac{3}{10}\).
• For bag Y: Removing a red marble \(\frac{4}{5}\), blue marble \(\frac{1}{5}\).
• After adding a red marble, for bag X: Removing a red marble \(\frac{8}{10}\), blue marble \(\frac{2}{10}\).
• After adding a red marble, for bag Y: Removing a red marble \(\frac{5}{6}\), blue marble \(\frac{1}{6}\).

(b) Probability that both marbles are the same colour:

\(P(\text{both same colour}) = P(BB) + P(RR) = P(XBB) + P(XRR) + P(YRR)\)

\(= \frac{1}{2} \times \frac{3}{10} \times \frac{2}{10} + \frac{1}{2} \times \frac{7}{10} \times \frac{7}{10} + \frac{1}{2} \times \frac{4}{5} \times \frac{4}{5}\)

\(= \frac{6}{200} + \frac{49}{200} + \frac{16}{50}\)

\(= 0.03 + 0.245 + 0.32 = 0.595\)

(c) Probability that bag Y is chosen given different colours:

\(P(\text{bag } Y | \text{different colours}) = \frac{P(\text{bag } Y \cap \text{different colours})}{P(\text{different colours})}\)

\(= \frac{\frac{1}{2} \times \frac{4}{5} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{5} \times \frac{5}{6}}{1 - 0.595}\)

\(= \frac{\frac{4}{50} + \frac{1}{12}}{0.405}\)

\(= \frac{9}{81} = 0.444\)

(a) The tree diagram is as follows:

- First branch: Pizza (0.35), Burger (0.44), Curry (0.21)

- Second branch for Pizza: Fruit (0.3), No Fruit (0.7)

- Second branch for Burger: Fruit (0.8), No Fruit (0.2)

- Second branch for Curry: Fruit (0), No Fruit (1)

(b) The probability that Rani has some fruit is calculated as:

\(0.35 \times 0.3 + 0.44 \times 0.8 + 0.21 \times 0 = 0.105 + 0.352 + 0 = 0.457\)

(c) The probability that Rani does not have a burger given that she does not have any fruit is:

\(P(\text{not B} | \text{not fruit}) = \frac{P(\text{not B} \cap \text{not fruit})}{P(\text{not fruit})}\)

Calculate \(P(\text{not B} \cap \text{not fruit}) = 0.35 \times 0.7 + 0.21 \times 1 = 0.245 + 0.21 = 0.455\)

Calculate \(P(\text{not fruit}) = 1 - 0.457 = 0.543\)

Thus, \(P(\text{not B} | \text{not fruit}) = \frac{0.455}{0.543} = 0.838\)

(a) For Box A, the probability of choosing a red ball is \(\frac{7}{8}\) and a blue ball is \(\frac{1}{8}\).

For Box B, if a red ball is added, the probability of choosing a red ball is \(\frac{10}{15}\) and a blue ball is \(\frac{5}{15}\). If a blue ball is added, the probability of choosing a red ball is \(\frac{9}{15}\) and a blue ball is \(\frac{6}{15}\).

(b) The probability that the two balls chosen are not the same colour is calculated as:

\(\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{9}{15} = \frac{44}{120} = \frac{11}{30}\)

(c) The probability that the ball chosen from box A is blue given that the ball chosen from box B is blue is:

\(P(A \text{ blue } | B \text{ blue }) = \frac{P(A \text{ blue } \cap B \text{ blue })}{P(B \text{ blue })}\)

\(= \frac{\frac{1}{8} \times \frac{6}{15}}{\frac{7}{8} \times \frac{5}{15} + \frac{1}{8} \times \frac{6}{15}} = \frac{\frac{1}{20}}{\frac{41}{120}} = \frac{6}{41}\)

(i) The total probability that Benju is late for work is given by:

\(0.4x + 0.6 \times 2x = 0.36\)

Simplifying, we have:

\(0.4x + 1.2x = 0.36\)

\(1.6x = 0.36\)

Solving for \(x\), we get:

\(x = \frac{0.36}{1.6} = 0.225\)

(ii) We need to find the probability that Benju chooses the hilly route given that he is not late for work. The probability that he is not late is \(1 - 0.36 = 0.64\).

The probability that he chooses the hilly route and is not late is \(0.4(1-x)\).

The probability that he chooses the busy route and is not late is \(0.6(1-2x)\).

Thus, the probability that he is not late is:

\(0.4(1-x) + 0.6(1-2x) = 0.64\)

Using \(x = 0.225\), the probability that he chooses the hilly route and is not late is:

\(0.4(1-0.225) = 0.4 \times 0.775 = 0.31\)

The probability that he is not late is:

\(0.4 \times 0.775 + 0.6 \times 0.55 = 0.31 + 0.33 = 0.64\)

Therefore, the probability that he chooses the hilly route given that he is not late is:

\(\frac{0.31}{0.64} = \frac{31}{64} \approx 0.484\)