Probability picking certain balls out of bin An bin has 5 red and 4 green balls. Pick 6 without replacement. Find the probability that the second ball is green and the fifth ball is red.
I'm struggling to see how to find the sample space and even then, I'm confused what to do from there. I thought that there are $\binom{9}{6}$ outcomes, but the balls are indistinguishable. Any advice/solutions would be much appreciated!
 A: The sample space is
$$\{R_2 \cap R_5, R_2\cap G_5, G_2\cap R_5, G_2 \cap G_5   \}$$
Their respective probabilities are
$$\begin{align}
\Pr(R_2 \cap R_5) &= \frac 5 9 \frac 4 8\\
\Pr(R_2\cap G_5) &= \frac 5 9 \frac 4 8 \\
\Pr(G_2\cap R_5) &= \frac 4 9 \frac 5 8 \\
\Pr(G_2 \cap G_5) &= \frac 4 9 \frac 3 8 
\end{align}$$
This adds up to $1,$ and
$$\Pr(G_2\cap R_5) = \frac 4 9 \frac 5 8 =0.278$$ can be numerically verified in R:
balls <- c(rep(1,5), rep(0,4))

set.seed(561)
n <- 1e6
m <- matrix(,n,6)
for(i in 1:n) m[i,] <- sample(balls, 6, replace = F)
s <- m[m[,2]==0 & m[,5]==1,,drop=F]
nrow(s)/n
0.277662


Thank you for accepting my answer. I was thinking that any combination of whichever two balls (second and fifth, first and third, etc) being of different color will have the same probability. So this can be solved with the expectation of the indicator variable $X_{ij}$ with $ij$ being the $i$-th and $j$-th balls, and $X_{ij}$ the random variable that is $1$ if $G_i R_j,$ and $0$ otherwise. Being that this is an indicator variable, the expectation equals the probability, and can be calculated as $\Pr(G_1,R_2)=\frac 4 9 \frac 5 8.$
A: Sample space:
\begin{aligned}
0 \ green \ balls&, & 0 \ combinations & \\
1 \ green \ balls&, & \binom 6 1  \ combinations\\
2 \ green \ balls&, & \binom 6 2  \ combinations\\
3 \ green \ balls&, & \binom 6 3  \ combinations \\
4 \ green \ balls&, & \binom 6 4  \ combinations \\
5 \ green \ balls&, & \binom 6 5  \ combinations \\
\end{aligned}
Total sample space: $2^6-2=62$ cases. Now we need to find how many of them are favorable.  
When we have one green ball, we have one favorable case - it was picked second. When we have two green balls, one of them have to be in the second slot and the other one can go in any of the remaining 4 slots (it cannot go into the fifth one). So, four cases. When we have three green balls, one goes into the second slot, the other two are distributed into four available slots, etc.
