The probability of the union of two events is the sum of their probability, minus the probability of their interserction:
[tex] P(A \cup B) = P(A) + P(B) - P(A \cap B) [/tex]
If we plug the known values into this formula, we have
[tex] \dfrac{3}{5} = \dfrac{1}{3} + \dfrac{2}{5} - P(A\cap B) [/tex]
From which we can deduce
[tex] P(A\cap B) = \dfrac{1}{3} + \dfrac{2}{5} - \dfrac{3}{5} = \dfrac{2}{15} [/tex]
So, the probability of [tex] A \setminus B [/tex] is a bit less than [tex] P(A) [/tex], we have to take away all events that belong to B as well:
[tex] P(A \setminus B) = P(A) - P(A\cap B) = \dfrac{1}{3} - \dfrac{2}{15} = \dfrac{1}{5}[/tex]
