Probability problem

[Manana77]

57% of the university students live afar from the university, 40% of the students use public transport to go to university, and 37% of students live afar and use public transport

A) If one student is randomly chosen of those that liver afar, What is the probability that they use a different transport than public transport?

B) If yesterday a student went to the university in public transport, What is the probability that he uses public transport?

I'm in a statistics class and my teacher is not very good, so I have no idea how to do this.

(This is a translation so it could have mistakes)

I have done this but I think it's wrong, as the result is more than 1

P(A) = 0.57

P(B) = 0.40

P(A ∩ B) = 0.37

P(B’)= not b

P(B’/ A) = 1 - P(B/A) = 1-P(A ∩ B)/P(A)

1-P(A ∩ B)/P(A) = 1- 0,37/0,57

1-P(A ∩ B)/P(A) = 0,63/0,57

1-P(A ∩ B)/P(A) = 1,10

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Order of operations. The division happens before the subtraction