First, determine the probabilities for each person choosing each entrance:

• Rick: \(P(A) = \frac{1}{3}\), \(P(B) = P(C) = P(D) = \frac{2}{9}\)
• Brenda: \(P(A) = P(B) = P(C) = P(D) = \frac{1}{4}\)
• Ali: \(P(C) = \frac{2}{7}\), \(P(D) = \frac{3}{5}\), \(P(A) = P(B) = \frac{2}{35}\)

(i) To find the probability that at least 2 friends choose entrance B:

Calculate the probability for each scenario where at least 2 choose B:

• \(P(\text{Rick B, Brenda B, Ali not B}) = \frac{2}{9} \times \frac{1}{4} \times \left(1 - \frac{2}{35}\right) = \frac{11}{210}\)
• \(P(\text{Rick B, Brenda not B, Ali B}) = \frac{2}{9} \times \left(1 - \frac{1}{4}\right) \times \frac{2}{35} = \frac{2}{210}\)
• \(P(\text{Rick not B, Brenda B, Ali B}) = \left(1 - \frac{2}{9}\right) \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{90}\)
• \(P(\text{Rick B, Brenda B, Ali B}) = \frac{2}{9} \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{315}\)

Sum these probabilities: \(\frac{11}{210} + \frac{2}{210} + \frac{1}{90} + \frac{1}{315} = \frac{24}{315} = \frac{8}{105}\)

(ii) To find the probability that all three choose the same entrance:

• \(P(\text{entrance A}) = \frac{1}{3} \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{210}\)
• \(P(\text{entrance B}) = \frac{2}{9} \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{315}\)
• \(P(\text{entrance C}) = \frac{2}{9} \times \frac{1}{4} \times \frac{2}{7} = \frac{1}{63}\)
• \(P(\text{entrance D}) = \frac{2}{9} \times \frac{1}{4} \times \frac{3}{5} = \frac{1}{30}\)

Sum these probabilities: \(\frac{1}{210} + \frac{1}{315} + \frac{1}{63} + \frac{1}{30} = \frac{2}{35}\)

The probability that the dog barks can be found by considering two scenarios: when they go to the park and when they do not.

1. Probability they go to the park and the dog barks:

\(P(\text{park, bark}) = 0.6 \times 0.35 = 0.21\)

2. Probability they do not go to the park and the dog barks:

\(P(\text{not park, bark}) = 0.4 \times 0.75 = 0.3\)

3. Total probability that the dog barks:

\(P(\text{bark}) = P(\text{park, bark}) + P(\text{not park, bark}) = 0.21 + 0.3 = 0.51\)

Let the probability that it is not raining be \(1 - x\).

The probability that Aran does not wear a hat when it is not raining is \(1 - 0.3 = 0.7\).

The probability that Aran does not wear a scarf when he does not wear a hat is \(1 - 0.1 = 0.9\).

The probability that it is not raining and Aran is not wearing a hat or a scarf is given by:

\((1-x) \times 0.7 \times 0.9 = 0.36\)

Solving for \(x\):

\((1-x) \times 0.63 = 0.36\)

\(0.63 - 0.63x = 0.36\)

\(0.63x = 0.63 - 0.36\)

\(0.63x = 0.27\)

\(x = \frac{0.27}{0.63} = \frac{3}{7}\)

The total number of sweets is 52 (13 red + 13 blue + 13 green + 13 yellow). We need to find the probability of selecting exactly 3 red sweets out of 7.

Using combinations, the number of ways to choose 3 red sweets from 13 is given by:

\(\binom{13}{3}\)

The number of ways to choose the remaining 4 sweets from the 39 non-red sweets is:

\(\binom{39}{4}\)

The total number of ways to choose 7 sweets from 52 is:

\(\binom{52}{7}\)

Thus, the probability is:

\(\frac{\binom{13}{3} \times \binom{39}{4}}{\binom{52}{7}}\)

Calculating these values:

\(\binom{13}{3} = 286\)

\(\binom{39}{4} = 82,251\)

\(\binom{52}{7} = 133,784,560\)

Therefore, the probability is:

\(\frac{286 \times 82,251}{133,784,560} \approx 0.176\)

Each die has 4 faces, so there are a total of 16 possible outcomes when two dice are thrown together.

To find the probability of the sum being 7 or more, we consider the following outcomes:

• Sum of 7: Possible outcomes are (3,4) and (4,3), giving a probability of \(\frac{2}{16}\).
• Sum of 8: The only possible outcome is (4,4), giving a probability of \(\frac{1}{16}\).

Thus, the probability of the sum being 7 or more is:

\(P(7 \text{ or more}) = \frac{2}{16} + \frac{1}{16} = \frac{3}{16}\)

The expected number of occasions when the sum is 7 or more in 200 throws is:

\(200 \times \frac{3}{16} = 37.5\)