# Homework Help: Multiplication rule for conditional probabilities

1. Sep 9, 2015

### TheMathNoob

1. The problem statement, all variables and given/known data
Selecting Two Balls. Suppose that two balls are to be selected at random, without replacement, from a box containing r red balls and b blue balls. We shall determine the probability p that the ﬁrst ball will be red and the second ball will be blue

I am confusing Pr(A|B) and Pr(A∩B). In some cases they might look the same to me, so how can I recognize the difference between them.

2. Relevant equations

Pr(A∩B)=Pr(A) Pr(B|A)
Pr(A∩B)=Pr(B) Pr(A|B).

3. The attempt at a solution
This is an example from the book, so the solution is already there.
Let A be the event that the ﬁrst ball is red,and let B be the event that the second ball is blue. Obviously
, Pr(A)=r/(r +b)

Furthermore, if the event A has occurred, then one red ball has been removed from the box on the ﬁrst draw. Therefore, the probability of obtaining a blue ball on the second draw will be

Pr(B|A)= b/( r +b−1)

It follows that
Pr(A∩B)= r/( r +b)* (b/( r +b−1))

So the line " Furthermore, if the event A has occurred, then one red ball has been removed from the box on the ﬁrst draw. Therefore, the probability of obtaining a blue ball on the second draw will be " looks the same to me as Pr(AnB) because the two events have to have occurred.

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2. Sep 9, 2015

### Ray Vickson

No, you are confused: the joint probability $P(A \cap B)$ is the probability that BOTH $A$ AND $B$ will occur. The conditional probability $P(B|A)$ is the probability that $B$ will occur, GIVEN THAT $A$ is known to have occurred already.

One of these refers to a probability of $B$ only (of course, in the new "sample space" $A$!); the other refers to both occurring.

3. Sep 12, 2015

### haruspex

If you are still confused, it may help to consider A=B. P(B|B)=1, but $P(B \cap B)=P(B)$.

4. Sep 12, 2015

### Ray Vickson

To expand on my initial answer, suppose r = b = 5, and A is an event which talks about what happened in the first 9 draws, while B = {last ball is blue}. What might be surprising is that P(B) = 1/2 (if you are told nothing at all about what happened in the first 9 draws---only that 9 balls have already been drawn and one is left in the urn). [More on this later.]

Clearly, for some A we have P(B|A) = 0 (because there are no blue balls left), while for other A we have P(B|A) = 1 (because there is only a blue ball left). However, if A and B are not incompatible, then $0 < P(A \cap B) < 1$; this will be the case for any A such that $A \cap B \neq \emptyset$.

Why $P(B) = 1/2$? Number the balls from 1--10, and say balls 1--5 are blue. Look at all 10! permutations of the numbers 1--10, and look at those that have numbers 1--5 last. Exactly half of the permutations have that property, so $P(B) = (1/2) 10! / 10! = 1/2$.