Combinatorics For $4$ Pool Balls There lie $4$ pool balls on a pool table: two striped and two plain. Two of the pool balls are selected at the same time, at random. Given that one of the selected balls is striped, what's the probability that the other is also striped?
This problem seems very hard to me? Need Help>Please?
 A: There are four different ways that we can select two pool balls:


*

*Both striped: $\big(\frac{1}{2}\big)\big(\frac{1}{3}\big) = \frac{1}{6}$

*First striped, second solid: $\big(\frac{1}{2}\big)\big(\frac{2}{3}\big) = \frac{2}{6}$

*First solid, second striped: $\big(\frac{1}{2}\big)\big(\frac{2}{3}\big) = \frac{2}{6}$

*Both solid: $\big(\frac{1}{2}\big)\big(\frac{1}{3}\big) = \frac{1}{6}$
Since we are given that one of the balls was striped, we ignore scenario $4$. The total probability is then the probability of the scenario we want (scenario $1$) divided by the sum of the probabilities of all possible scenarios ($1$, $2$, and $3$):
$$
\frac{\big(\frac{1}{6}\big)}{\big(\frac{1}{6}\big)+\big(\frac{2}{6}\big)+\big(\frac{2}{6}\big)} = \frac{1}{5}
$$
A: Am I under thinking this by assuming that there is a $\frac 12$ chance of picking a striped ball at the beginning. Which leave 2 solids and 1 striped ball. So now the chance of getting that striped ball is $\frac 13$?
I'm assuming the question is asking P(2nd ball striped|1st ball striped)
Edit
Okay, let me think about this a little more. I used this hypergeometric formula from probability and statistics. $$P(X=x)=\frac {{N_1\choose x}{N_2\choose {n-x}}}{N\choose n}$$ where N is the total number of objects $N_2$ is solid ball $N_1$ is striped ball $n$ is the number of balls we are choosing and $x$ is the probability we are looking for.
So our $$N=3, N_1=1, N_2=2,n=3,x=1$$
So our total number of balls is 3 since we've already picked on that is striped. Now we have 1 striped ball left and 2 solid balls left and we only want 1 more striped ball.
Which gives us:
$$P(X=1)=\frac {{1\choose 1}{2\choose 0}}{3\choose 1}=\frac 13$$
