We are given the following information:
The probability that event B occurs is:

And the probability that event A occurs given that event B occurs is:

And we need to find the probability that both A and B occur.
To solve this problem, we have to use the conditional probability formula:

Where
P(A|B) is the probability of A given that B occurred.
P(B) is the probability of B.
And P(A∩B) is the probability of A and B occuring.
Thus, we solve for P(A∩B) in the previous equation:

And substitute the known values:

We multiply the fractions and get the following result:

Finally, we simplify the fraction by dividing both numbers in the fraction by 15:

Answer: 1/2