There is a bag with $n$ marbles. Of these $n$ marbles, $k$ of them are black, and the remaining are red.
You draw 3 marbles from the bag. What is the probability that 2 of the marbles are red?
I understand how this would work if it were an event such as flipping a coin, that is independent each time, but I am confused by this since the events of drawing in succession appear to be dependent. Can anyone help me understand this?
Indeed this is the difference between sampling with replacement (independent) and without replacement (dependent). These correspond to the binomial distribution and the hypergeometric distribution respectively, so if you know the hypergeometric distribution you can answer your problem.
Here is a direct way to answer your question (which contains a derivation of the pmf of the hypergeometric distribution).
\begin{align} \frac{\#\{\text{ways to choose 2 reds and one black}\}}{\#\{\text{ways to choose 3 marbles}\}} &= \frac{\binom{n-k}{2} \binom{k}{1}}{\binom{n}{3}}. \end{align} (There are $\binom{n-k}{2}$ ways to choose 2 marbles from the $n-k$ red marbles, and $\binom{k}{1}$ ways to choose one marble from the $k$ black ones.)
I understand how this would work if it were an event such as flipping a coin, that is independent each time, but I am confused by this since the events of drawing in succession appear to be dependent. Can anyone help me understand this?
Indeed; if we replaced the marbles each time the probability for each successive marble drawn being green would be independent. $$\frac kn \frac{(n-k)}{n}\frac{(n-k)}{n}+\frac{(n-k)}{n}\frac kn \frac{(n-k)}{n}+\frac{(n-k)}{n}\frac{(n-k)}{n}\frac kn ~=~ \binom 32\frac{(n-k)^2k}{n^3}$$
However, as we are not replacing the marbles every time we pick one, the probability for each successive marble being green is dependent on the previous selections: $$\frac kn \frac{(n-k)}{n-1}\frac{(n-k-1)}{n-2}+\frac{(n-k)}{n}\frac k{n-1} \frac{(n-k-1)}{n-2}+\frac{(n-k)}{n}\frac{(n-k-1)}{n-1}\frac k{n-2} ~=~ \left.{\binom k1\binom{n-k}2}\middle/{\binom n 3}\right.$$
You have $4$ possibilities for the drawings, RRR, RRB, RBR and BRR.
Calculate each probability separately:
$$RRR: \frac{n-k}{n}\cdot\frac{n-k-1}{n-1}\cdot\frac{n-k-2}{n-2}$$
$$RRB: \frac{n-k}{n}\cdot\frac{n-k-1}{n-1}\cdot\frac{k}{n-2}$$
$$RBR: \frac{n-k}{n}\cdot\frac{k}{n-1}\cdot\frac{n-k-1}{n-2}$$
$$BRR: \frac{k}{n}\cdot\frac{n-k}{n-1}\cdot\frac{n-k-1}{n-2}$$
The denominator in each case is the same, $\dfrac1{n(n-1)(n-2)}$, and we can see that each numerator contains a common factor of $(n-k)(n-k-1)$, so that we can write the sum of these probabilities as:
$$\dfrac{(n-k)(n-k-1)}{n(n-1)(n-2)}(n-k-2+3k)=\dfrac{(n-k)(n-k-1)(n+2k-2)}{n(n-1)(n-2)}$$
which is the probability that you want.
We can prove this by considering the probabilities that we don't get at least two reds, and sum them. This gives:
$$\dfrac{k(k-1)}{n(n-1)(n-2)}(3(n-k)+k-2)=\dfrac{k(k-1)(3n-2k-2)}{n(n-1)(n-2)}$$
If we add the two probabilities together, we get $1$.
These events themselves are not independent, as $P(RRR\cap RRB)\ne P(RRR)P(RRB)$, but whether we pull a red or black ball next is only dependent on the proportion of red to black balls left.
