# Possibility Problem

1. Sep 12, 2008

### k1point618

I don't remember how exactly to do the following kind of possibility problems...

for example... there are x amount of object A, and y amount of object B. How do you find the possibility of choosing 2 numbers such that they are either both A or both B?

more specifically, ex) You have 12 black cards and 5 red cards, what is the possibility of randoming selecting 2 of these cards such that they are both red or both black?

Thanks for the help!

2. Sep 12, 2008

### sutupidmath

First what is the possibility of getting one black card? then again after you have gotten it, what is the possibility of getting another black card, given that the first card was black?

Second, what is the possibility of getting a red card out of the whole pile of cards? Then again what is the possibility of getting another red card given that the first card drawn was red?

Now after you come up with these, you need to find the possibility that the first happens or the second situation happens.

Let the probbability of the first situation happening be A, and the second B

then you need to find P(A or B) =P(A) +P(B)-P(A and B)

can u go from here?

3. Sep 12, 2008

### k1point618

Oh, i see.

So ... P(A)= 12/17+11/16 and P(B) = 5/17 + 4/16?
and does P(A and B) = P(A)*P(B) ? I'm not sure

4. Sep 12, 2008

### sutupidmath

How is P(A and B) defined?

P(A and B)=P(A)*P(B/A)

P(B/A) means the probbability of B given that A has already happened. What is this in your case? 0 right? since if we have already drawn two red balls, then what is the probbability that one or both of them will be black? Zero!
Here we are assuming that once we have drawn a ball, we are not returning it back, so these are called dependent events, since one affects the other.
P(A and B)=P(A)*P(B) if the two events A,B are independent, that is if after we have drawn two balls, we return them back to the pile, and make another draw.

5. Sep 12, 2008

### k1point618

ah...

so in this case, P(A or B) =P(A) +P(B), well that makes sense.

6. Sep 12, 2008

### sutupidmath

Another way of doing this is listing all possible samples, and then looking how many of them are 2 red and how many 2 black, but listing all samples in this case is painful, and the chances to make a mistake are very very high.

7. Sep 12, 2008

Thank you!