Probability - marbles without replacement Math is my weakest subject and I'm having a hard time trying to figure out what equation to use in this problem:
A jar contains 5 purple balls, 10 pink balls, and 7 blue balls. If 3 balls are to be drawn successively without replacement. What is the probability of getting 2 purple balls and 1 pink ball?
Ans: I tried doing (5/22)(4/21) & (10/22). But I don't think it's right... Any help is appreciated.
 A: Since you’re drawing without replacement, you are in effect just choosing a $3$-element subset of the set of $22$ balls. All $3$-element subsets are equally likely to be chosen, so a straightforward way to solve the problem is to count the $3$-element subsets containing $2$ purple balls and one pink ball and divide by the total number of $3$-element subsets.
There are $\binom52=10$ different pairs of purple balls, and there are $10$ pink balls, so there are $10\cdot10=100$ possible $3$-element sets consisting of $2$ purple balls and one pink ball. There are 
$$\binom{22}3=\frac{22!}{3!19!}=\frac{22\cdot21\cdot20}{3\cdot2\cdot1}=11\cdot7\cdot20$$
sets of $3$ balls, so the desired probability is
$$\frac{100}{11\cdot7\cdot20}=\frac5{77}\;.$$
You can also work the problem directly in terms of probabilities, but not quite the way you tried. What you calculated is the probability of drawing a purple ball followed by another purple ball followed by a pink ball. However, you can also get the desired outcome by drawing purple-pink-purple or pink-purple-purple. If you do the calculations, you’ll find that each of those outcomes also has probability
$$\frac{5\cdot4\cdot10}{22\cdot21\cdot20}\;,$$
so the actual probability is $3$ times that, or
$$\frac{3\cdot5\cdot4\cdot10}{22\cdot21\cdot20}=\frac5{11\cdot7}=\frac5{77}\;.$$
A: The number of ways to get the desired combination is:
$$\binom{5}{2}\cdot\binom{10}{1}\cdot\binom{7}{0}=100$$
The number of ways to get any combination of $3$ balls is:
$$\binom{5+10+7}{3}=1540$$
Hence the probability of getting the desired combination is:
$$\frac{100}{1540}\approx6.49\%$$
