I'm working through a practice question in combinatorics and my first attempt at counting through a particular problem has yielded the wrong answer (I have a copy of the actual solutions, for reference) but I'm not quite sure why.
The problem I'm doing is
An urn contains $4$ red balls and $3$ white balls, and we draw three balls from the urn without replacement. What is the conditional probability that all $3$ balls are red, given that the third ball drawn is red?
The worked answers I have do it in the standard way by letting $Y$ be a $\{R,W\}^3$-valued random variable and calculating
$$\frac{\mathbb{P}\left(Y = (R,R,R)\right)}{\mathbb{P}\left(Y = (R,R,R)\right) + \mathbb{P}\left(Y = (R,W,R)\right) + \mathbb{P}\left(Y=(W,R,R)\right) + \mathbb{P}\left(Y=(W,W,R) \right)}=\frac{1}{5}.$$
Now, I suppose my question is: isn't this equivalent to just asking what the probability is of drawing two red balls out of the urn without replacement? The way I intuitively wanted to treat this question is by designating a random variable $X\sim \mathcal{H}_{2,(4,3)}$, where this denotes the hypergeometric distribution modelling drawing $2$ items out of a population with $4$ successes and $3$ failures, without replacement.
Then is this problem not the same as simply calculating $\mathbb{P}\left(X = (2,0)\right) = \frac{2}{7}$ as the probability that the first two balls we draw are red? Clearly, we get different answers, but I do not understand why. What part of my intuitions were off about trying to use the hypergeometric distribution this way? Is there a subtlety to conditional probability that I am not picking up on?
Best Answer
A route that comes close to your second approach is the following:
Let there be $7$ spots (one for each ball) and on spot $3$ place a red ball. Then from the remaining balls ($3$ reds and $3$ whites) select $2$ balls that are placed on spot $1$ and $2$.
The probability that the selected balls are both red is:$$\frac{\binom32\binom30}{\binom62}=\frac15$$
Here identify the balls landing on the spots $1,2,3,$ as the $3$ balls that were drawn from the urn.