Not terribly tricky- it is only necessary to use the fact that the "probability of A or B" is "probability of A" plus "probability of B" as long as A and B are "mutually exclusive".
Suppose, for the moment, that the first three are first type, of which there are 20, and the last two are not. The probability the first card is of that type is 20/(20+ 12+ 8+ 2)= 20/32= 5/8. The probability that the second and third are also of that type are, similarly, 19/31 and 18/30= 3/5 so the probability the first three are of that type is (5/8)(19/31)(3/5). There are then 29 cards left of which 22 are NOT of the first type. That means that the probablity that the last two cards are NOT of the first type is (22/29)(21/28). The probability that the first three cards are of the first type is (20/32)(19/31)(18/30)(22/29)(21/28) (which can, of course, be reduced as I did before). Further there are \(\displaystyle begin{\pmatrix} 5 \\ 3\end{pmatrix}= \frac{5!}{2!3!}= 10\) different orders of three things of one type and 2 of another so the probability of three cards of the first type and 2 cards or any other is 10(20/32)(19/31)(18/30)(22/29)(21/28).
Do the same for "three cards of the second type and two of any other", "three cards of the third type and two of any other", and "three cards of the third type and two of any other" and add the four probabilities.