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# Adding Probabilities - Not Mutually Exclusive

Reference > Mathematics > Probability

In 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) =
13
52
because there are 13 spades out of 52 cards.
P(K) =
4
52
because there are 4 kings out of 52 cards.
P(S and K) =
1
52
because there is only one card out of 52 that is both a king and a spade.

P(S or K) =
13
52
+
4
52
-
1
52

P(S or K) =
9
26

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) =
3
26

P(D) =
4
26

P(P and D) =
1
26
(because there's only one letter that's in both)

P(P or D) =
3
26
+
4
26
-
1
26
=
6
26
=
3
13

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) =
6
12

P(S) =
3
12
because 1,4, and 9 are all perfect squares.
P(O and S) =
2
12
because there are two odd perfect squares.

P(O or S) =
6
12
+
3
12
-
2
12
=
7
12

## Questions

1.
If I roll a twelve-sided die, what is the probability that I will get either a perfect square or a value less than 8?
2.
If I pick a letter from the alphabet, what is the probability that it is a letter in either the word "COW" or "LOW?"
3.
If I pick a letter from the alphabet, what is the probability that it is a letter in either the word "BELLOW" or "FOLLOW?"
4.
If I pick a card from the deck, what is the probability that it is either red, or a Jack?
5.
If I pick a card from the deck, what is the probability that it is either a number (not an ace or face card) or it is a black card?
6.
Ten of my socks are red. Two are green. Eight are red and green. Twelve are white and green. If I pick one sock, what is the probability that it will either have green on it or white?
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Adding Probabilities - Mutually Exclusive Outcomes
Complementary Outcomes

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