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