Jackpot probablity There are two parts to a problem:
Part (a);
In a Lottery, three white balls are drawn (at random) from twenty balls numbered 1 through 20, and a blue SuperBall is drawn (at random) from ten balls numbered 21 through 30. When you buy a ticket, you select three numbers from 1-20 and one number from 21-30. To win the jackpot, the numbers on your ticket must match the three white balls and the SuperBall. (You don't need to match the white balls in order).
If you buy a ticket, what is your probability of winning the jackpot?
Part (b):
In a Lottery, three white balls are drawn (at random) from twenty balls numbered 1 through 20, and a blue SuperBall is drawn (at random) from ten balls numbered 21 through 30. When you buy a ticket, you select three numbers from 1-20 and one number from 21-30. To win a prize, the numbers on your ticket must match at least two of the white balls or must match the SuperBall.
If you buy a ticket, what is your probability of winning a prize?
Approach for part (a):
Here is how I am solving the problem.Since 3 number in white balls needs to match during the first pick, the probability is 3/20. For the second one it will be 2/19 and 3rd one 1/18. That combined with probability for blue ball will be 3/20 * 2/19 * 1/18 * 1/10
part (b):
3 cases (1) only blue ball matches and no match in white balls (2) no match in blue ball but 2 out of 3 white ball matches (3) Both matches
Probability for case (a): 
1/10 * 17/20 * 16/19 * 15/18
case (b) :
9/10 * 3/20 * 2/19
case (c) :
1/10 * 3/20 * 2/19 * 1/18
Add case (a) + (b) + (c) because they are mutually exclusive.
Am I correct in my approach?

3 matching white balls are drawn.
B: 2 matching white balls are drawn.
C: The right blue ball is drawn.
Therefore $P(A\cup B \cup C )=P(A)+P(B)+P(C)-P(A \cap C)-P(B \cap C)$
$\left[\frac{3}{20} \cdot \frac{2}{19} \cdot \frac{1}{18}\right]+\left[3 \cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{17}{18}\right]+\left[\frac{1}{10}-\frac{1}{10} \cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{1}{18}\right]-\left[\frac{1}{10}\cdot3\cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{17}{18}\right]\approx 14.1\%$
