# Adding Probabilities - Not Mutually Exclusive

Reference > Mathematics > ProbabilityIn the previous section, we showed you a formula for calculating the the probability of two (or more) mutually exclusive outcomes. The formula was pretty simple:

P(A or B) = P(A) + P(B)

But this only works if the outcomes are mutually exclusive. What happens if they aren't mutually exclusive?

Well, the formula gets just a little bit more complicated. Now it looks like this:

P(A or B) = P(A) + P(B) - P(A and B)

That's right - now we don't just need to figure out the probability of A and the probability of B - we need to figure out the probability that *both *happen. Why? Well, suppose I asked you to find the probability that a card was either a spade or a king. You can add the probability of a spade, and the probability of a king, but what have you done? You've counted one desired outcome *twice*! You counted the king of spades because it's a king, and you also counted it because it's a spade. So we need to subtract that one outcome back out of the result. That's what the last term of our formula is: subtract out the probability of it being both a king and a spade.

So let's go ahead and calculate that probability. We'll use S for spade, and K for king:

P(S or K) = P(S) + P(K) - P(S and K)

P(S) =P(K) =

P(S and K) =

**Random Letter Example**

Suppose I pick a letter from the alphabet. What is the probability that it is either in the word "PUG" or the word "DOGS?"

**Solution**

These are not mutually exclusive, because the letter G is in both words.

Let P(P) be the probability that the letter is in PUG.

Let P(D) be the probability that the letter is in DOGS.

P(P or D) = P(P) + P(D) - P(P and D)

P(P) =P(D) =

P(P and D) =

**Die Example**

If I roll a twelve sided die, what is the probability that the result will either be an odd number or a perfect square?

**Solution**

These are not mutually exclusive, because some outcomes are both odd and perfect squares (1 and 9).

Let P(O) be the probability of an odd number.

Let P(S) be the probability of a perfect square.

P(O or S) = P(O) + P(S) - P(O and S)

P(O) =P(S) =

P(O and S) =