Expert: Scott A Wilson - 9/21/2009

Question
First of all, thank you for your time. I'm trying to follow my class notes on this, but none of them "match" this type of probability, and being a person who cannot seem to comprehend math very well, I'm having trouble writing the answer to this question out in detail (the steps):

Let us flip a coin and roll a die. Find the probability that the coin will be heads or that the die will be even, given the following:  A is the event that the coin is heads and B is the event that the die is even, and
A = {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}
B = {(H,2),(T,2),(H,4),(T,4),(H,6),(T,6)}

All the other "examples" we did in class were things like, they are BOTH dice, and what is the probability of rolling a 7 or 11, etc. I know that the formula for this is P(A or B) = P(A) - P(B) - P(A and B) ... and I know this sounds dumb, but I'm having trouble "filling in" for the steps ... my brain seems to keep scrambling over it ... I know it's probably so simple and I just can't see it...  Thank you for your help!

Answer
By AnB, I mean "A intersect B".

The entire set has the first one as H or T and the second one as 1,2,3,4,5 or 6.  Thus, the entire set is {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}.

The set A is listed correctly and has 6 elements.
The set B is listed correctly and has 6 elements.
The set AnB is the set of elements in A and B.
They are {(H,2),(H,4), and (H,6)}.  There are 3 elements here.

It can now be said that the P(A) = 6/12 = 1/2, P(B) = 6/12 = 1/2, and P(AnB) = 3/12 = 1/4.
Thus, it is now seen that P(AnB) = P(A)•P(B), which says that they are independent.

A complement would be just like A with all of the H changed to T.
B complement would be just like B with 2,4,6 changed to 1,3,5, respectively.