You can use a binomial distribution.
p = 1/5 (get it right)
q = 4/5 (get it wrong)
You want the probability you get 2 right (and 8 wrong).
P(k = 2) = C(10,2) p^2 q^8
= (10 x 9) / 2 * (1/5)^2 (4/5)^8
= 0.301989888
Answer:
≈ 30.2%
P.S. You computed the probability of getting 2 *specific* problems correct and the other 8 wrong (say the first two right, the last 8 wrong). But you need to multiply by the number of ways to have 2 correct answers anywhere on the test. That is C(10,2) = 45. So multiply your answer by 45.