Is the probability of a chain of dependent events, independent of the order in which they occur?

Question

So I have a bag of 6 balls: 3 red, 2 blue and 1 green.

Events:

A: I draw a red ball

B: I draw a blue ball

C: I draw a green ball

I do not replace the balls (thus, resulting in conditional probability).

Is the probability that I remove balls in the order: R, B, G, the same as the probability that I remove them in the order B, G, R? Or any order?

Intuitively, the probability of order R-B-G is $\frac{3}{6}.\frac{2}{5}.\frac{1}{4}$ and B-G-R gives $\frac{2}{6}.\frac{1}{5}.\frac{3}{4}$, both of which are $\frac{1}{20}$.

Thus, is p(A and-then B and-then C) = p(A and-then-C and-then B)? If so, why?

Answer 1

1. Order does not matter. You are right.

2. I think the rule for which you are looking is called the chain rule (of probability).

http://en.wikipedia.org/wiki/Chain_rule_%28probability%29

If you want to compute

$P(A_1, A_2, A_3)$, you can do so like this:

$$P(A_1, A_2, A_3) = P(A_3 | A_1, A_2) P(A_2 | A_1) P(A_1)$$

But since $P(A_1, A_2, A_3) = P(A_1, A_3, A_2)$, we can do:

$$P(A_1, A_2, A_3) = P(A_2 | A_1, A_3) P(A_1 | A_3) P(A_3)$$

Oh, when I say $P(A_1, A_2, A_3)$, I mean $P(A_1, A_2, A_3) = P(A_1 \cap A_2 \cap A_3)$

Edit: Um, $A_1$ can be removing the green ball, $A_2$ for red and $A_3$ for blue.

Answer 2

I think your own explanation is already a mathematical proof that both probabilities are indeed equal. Your products of probabilities are correct and hence the result follows. Intuitively, this is due to the fact that you are only drawing $1$ ball per color. Things would be more complicated if you would consider the sequence RRB for example.

Answer 3

Yes, for the mentioned problem the result is independent on the order. Let's say you have $n$ balls and their distribution by colors: $(n_1, n_2, n_3 = n-n_1-n_2)$. Then you extract $i$ balls and their distribution by colors: $(i_1, i_2, i_3 = i-i_1-i_2)$, then the answer:

$$\frac{i!}{n!}\cdot \frac{n_1!}{(n_1-i_1)!}\cdot \frac{n_2!}{(n_2-i_2)!}\cdot \frac{n_3!}{(n_3-i_3)!},$$ regardless of the order.

Answer 4

Let $E$ denote the event that the first $3$ balls that are drawn have distinct color.

Do you understand that $P(BGR\mid E)=P(RBG\mid E)$?

(Here $BGR$ stands for the event that the first ball removed is blue, the second is green and the third is red.)

This because working under condition $E$ is the same as drawing $3$ balls from a bag containing exactly $3$ balls that have distinct color.

But here we have also $BGR\cap E=BGR$ and $RBG\cap E=RBG$ which leads to:$$P(BGR)=P(BGR\cap E)=P(BGR\mid E)P(E)=P(RBG\mid E)P(E)=P(RGB\cap E)=P(RBG)$$

Answer 5

After a few years and learning more probability theory, I think the answer I was looking for is twofold:

1. If you assume three events happen together, then the probability is $p(A, B, C)$, which just means $p(A \text{ and } B \text{ and } C)$. Since "and" (conjunction) is independent of order, $p(A \text{ and } B \text{ and } C) = p(B \text{ and } C \text{ and } A)$, or any other permutation.
2. Another way to look at three events is in terms of information: what information does the occurrence of event A have on the probability of B? Subsequently, what does the occurrence of events (A and B) have on the probability of event C? If we assume that the chronological order of events was A, B, C, we can instead look at the conditional probability, i.e. $p(A)p(B|A)p(C|B,A)$. This view assumes that event A occurs first (causing us to update our probability that B occurs, given that A has occurred), then B occurs (causing us to update our probability that C occurs, given that A and B have occurred).

The chain rule just equates the above probabilities. It says that you can view a set of events in both ways: as occurring all at once, or one after the other, and you end up with the same probability value. This is quite convenient, as in certain problems, we might have different probability values available.

Another thing worth noting about the chain rule is that it is actually n! rules in one: since $p(A \text{ and } B \text{ and } C) = p(B \text{ and } C \text{ and } A) = \dots$, you are free to choose any permutation of events in the chain rule's RHS. You should pick the one which allows you to make use of the probability values available to you in your problem.