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I have this question:

Suppose you have an urn containing 2 red and 2 black balls. We will consider the situation where we select two balls from the urn without replacement. Now suppose you select one ball, observe its color, do NOT replace it, and draw a second ball and observe its color. What are the probabilities that you selected

1   First a red ball and then a black ball, i.e., a sequence ( R, B )?
2   Two balls of different colors, i.e., a set { R, B }?


Answer:
1) $\frac{1}{2}\frac{2}{3}=\frac{1}{3}$
2) $(2)(\frac{1}{3}) = \frac{2}{3}$

The answer is prety intuative but my question is can we solve the same problem using the General Multiplication Rule? i.e: $P(B|A) = \frac{P(A∩B)}{P(A)}$
The problem is I dont know how to compute $P(A∩B)$

Any help?

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  • $\begingroup$ You can compute $P(B | A)$ and you can compute $P(A)$. Using the formula, you can compute $P(A\cap B)$. $\endgroup$ – Braindead Feb 14 '14 at 0:01
  • $\begingroup$ Hi. I want to compute the probability of B given A has occurred. In other words the probability of a black ball given that I already picked a red ball. $\endgroup$ – Krimson Feb 14 '14 at 0:18
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The conditional probability here is natural to compute, so is the best approach. However, we can do it as follows:

Let $A$ be the event first is red, and $B$ be the event second is blue. Imagine the balls all have distinct ID labels. There are $4!$ equally likely ways to draw $4$ balls. There are $(2)(2)$ ways to draw a red then a blue. Thus $\Pr(A\cap B)=\frac{(2)(2)}{4!}$.

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  • $\begingroup$ Awesome! Yes I agree that it is easier to compute naturally but I was just curious to learn how to solve it the other way :) $\endgroup$ – Krimson Feb 14 '14 at 1:21

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