# What is the probability of drawing a blue ball on the 3rd trial (without replacement)?

I am self-studying Professor Ery Arias-Castro's Principles of Statistical Analysis (it's a pretty great book so far), and I could use some help on this problem.

Suppose we draw without replacement from an urn with $$r$$ red balls and $$b$$ blue balls. At each stage, every ball remaining in the urn is equally likely to be picked. Use (1.15) to derive the probability of drawing a blue ball on the 3rd trial.

Equation 1.15 is

$$\mathbb{P}(A) = \mathbb{P}(A | B) \mathbb{P}(B) + \mathbb{P}(A | B^\complement) \mathbb{P}(B^\complement).$$

Using the notation $$B_n$$/$$R_n$$ for the event of drawing a blue/red ball on the $$n$$th trial, I calculated

\begin{align} \mathbb{P}(B_2) &= \mathbb{P}(B_2 | B_1) \mathbb{P}(B_1) + \mathbb{P}(B_2 | R_1) \mathbb{P}(R_1)\\ &= \frac{b - 1}{b + r -1} \frac{b}{b + r} + \frac{b}{b + r - 1} \frac{r}{b + r}\\ &= \frac{b}{b + r}. \end{align}

Similarly, we get $$\mathbb{P}(R_2) = r / (b + r)$$, but I am having a hard time setting up the equation for $$\mathbb{P}(B_3)$$. I would greatly appreciate some help.

I'd suppose that, at some point, you'll need the following extension of the equation you wrote: $$P(A|B) = P(A|B,C)P(C|B) + P(A|B,C^c)P(C^c|B).$$ You can prove this the same way that you proved the original statement.

With this in hand, $$P(B_3) = P(B_3|R_1) P(R_1) + P(B_3|B_1) P(B_1).$$ But $$P(B_3|R_1) = P(B_3|R_1, R_2) P(R_2|R_1) + P(B_3|R_1,B_2) P(B_2|R_1),$$ and everything in the right hand side should be straightforward to compute. Similarly, you can expand $$P(B_3|B_1)$$, and then merge the answers.

Incidentally, for sampling without replacement, there's an easier way to compute things: think of the sample space as the entire sequence of $$N = b+r$$ length strings composed of either $$B$$ or $$R$$. Then the sampling without replacement process will end up producing one of these strings, where each string is equally likely.

Now, the total number of such strings is $$\binom{N}{b}$$ - take $$N$$ slots, and pick $$b$$ of them to be blue. The number of strings where the third symbol is a $$B$$ is instead $$\binom{N-1}{b-1}$$: place one $$B$$ in the third slot, and choose $$b-1$$ out of the remaining $$N-1$$ slots to put $$B$$s in. So the probability is $$\binom{N-1}{b-1}/\binom{N}{b} = \frac{b}{N} = \frac{b}{b+r}.$$ Note that the answer doesn't change if you replace "third entry is $$B$$" by "$$k$$th entry is $$B$$" for any $$k \le n+r$$.

• Thanks for the tip and the alternative way of looking at the problem! I think I got it now. I will post my solution below. Commented Mar 27 at 2:03

My detailed solution to the problem. Thank you to @stochasticboy321 for helping me clarify the ideas.

First, we note the general law of total probability in the discrete case.

Theorem (Law of Total Probability): Let $$(\Omega, \Sigma, \mathbb{P})$$ be a probability space, and suppose $$\{B_n\}_{n=1}^\infty \subseteq \Sigma$$ is a countable collection of mutually exclusive ($$B_i \cap B_j = \varnothing$$ for $$i \neq j$$) and collectively exhaustive ($$\cup_{n = 1}^\infty B_n = \Omega$$) events. Then, for any $$A \in \Sigma$$, we have

\begin{align} \mathbb{P}(A) &= \sum_{n=1}^\infty(A \cap B_n)\\ &= \sum_{n=1}^\infty\mathbb{P}(A | B_n)\mathbb{P}(B_n). \end{align}

In the problem, our experiment is drawing balls out of an urn without replacement until we empty the urn. Our sample space $$\Omega$$ consists of words of length $$b+r$$ in two letters $$B$$ and $$R$$. Let $$B_k$$ and $$R_k$$ represent the event of choosing a blue or red ball on the $$k$$th draw respectively.

We note that

$$(B_2 \cap B_1) \cup (B_2 \cap R_1) \cup (R_2 \cap B_1) \cup (R_2 \cap R_1) = \Omega$$

and that any two of these events are disjoint (their intersection is empty). Thus, by the law of total probability, we have

\begin{align} \mathbb{P}(B_3) &= \mathbb{P}(B_3 | B_2 \cap B_1) \mathbb{P}(B_2 \cap B_1) + \mathbb{P}(B_3 | B_2 \cap R_1) \mathbb{P}(B_2 \cap R_1) + \mathbb{P}(B_3 | R_2 \cap B_1) \mathbb{P}(R_2 \cap B_1) + \mathbb{P}(B_3 | R_2 \cap R_1) \mathbb{P}(R_2 \cap R_1)\\ &= \mathbb{P}(B_3 | B_2 \cap B_1) \mathbb{P}(B_2 | B_1) \mathbb{P}(B_1) + \mathbb{P}(B_3 | B_2 \cap R_1) \mathbb{P}(B_2 | R_1) \mathbb{P}(R_1) + \mathbb{P}(B_3 | R_2 \cap B_1) \mathbb{P}(R_2 | B_1) \mathbb{P}(B_1) + \mathbb{P}(B_3 | R_2 \cap R_1) \mathbb{P}(R_2 | R_1) \mathbb{P}(R_1)\\ &= \frac{b - 2}{b + r - 2} \frac{b - 1}{b + r -1} \frac{b}{b + r} + \frac{b - 1}{b + r - 2} \frac{b}{b + r -1} \frac{r}{b + r} + \frac{b - 1}{b + r - 2} \frac{r}{b + r -1} \frac{b}{b + r} + \frac{b}{b + r - 2} \frac{r - 1}{b + r -1} \frac{r}{b + r}\\ &= \frac{b(b^2 + r^2 + 2br - 3b -3r + 2)}{(b+r)(b^2 + r^2 + 2br - 3b -3r + 2)}\\ &= \frac{b}{b + r}. \end{align}

Remark: As noted by @stochasticboy321, you could solve this problem in a combinatorial fashion as shown in their solution, but this solution uses the law of total probability as requested.