That A and B pass/fail are independent events so the probability that

they both fail:

P(A fails and B fails)=P(A fails) P(B fails) = (1-3/5) x 4/7 = 8/35

P(only one will pass) =P(A pass and B fails) + P(A fails and B pass)=(3/5)x(4/7) + (2/5)x(3/7)=...b) calculate the probability that only one will pass

This is a bit vaguely worded, but given that B has failed the probability that shec) if Brenda fails the probability that she will pass next time is 2/5

(i) calculate the probability that Brenda fails again

fails the second time is 1 minus the probability that she passes the second time = 3/5.

The wording of this is even more ambiguous, the best I can make of what is asked is 2/5,(ii) calculate the probability that Brenda passes having taken the examination for a second time

but I don't think that is what is intended.

RonL