D. Mutually Exclusive
Events
Another
type of probability deals with mutually exclusive events. What
do we mean by mutually exclusive events? And what does it mean
for two events not to be mutually exclusive? Consider the following
example of drawing cards:
Example 8
What
is the probability that a card selected from a deck will be either
an ace or a spade?
a)2/52
b)2/13
c)7/26
d)4/13
e)17/52
Solution
First, identify events A and B and notice the "or"
in between them. That means a greater probability than either
A or B individually. Therefore, we expect the answer to be greater
than 4/52(ace)=1/13 or 13/52(spade)= 1/4. Eliminate a and b.
The tricky part of this question lies in the fact that when we
figure probability, we are really just counting, and sometimes,
we count twice. In this case we have counted the ace of spades
twice. If you don't see this, consider what the 4 in 4/52 stands
for: ace of hearts, ace of diamonds, ace of clubs, ace of spades.
The 13 in 13/52 stands for all the spades: 1,2,3…King, Ace(of
spades). Therefore if we just combined the probabilities by the
rule for P(A or B) = P(A) + P(B) we would be over counting. We
have to subtract 1/52, the ace of spades that was counted twice.
Our answer becomes 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
Another way to think about the question is to just count aces
and spades; that is, use simple probability. There are 13 spades
in a deck and 3 aces other than the ace of spades already included
in the 13 spades. Therefore, there are 16 desired outcomes out
of a total of 52 possible outcomes, or 16/52 = 4/13.
In the above example, events A and B are not mutually exclusive.
Figuring the probability for event A includes part of the probability
of event B, and we must therefore subtract out this "over-counted"
probability to get the correct answer.
The following example illustrates mutually exclusive events:
Example 9
What
is the probability that a card selected from a deck will be either
an ace or a king?
a)1/169
b)1/26
c)2/13
d)4/13
e)8/13
Solution
The question asks for either an ace or a king. Since there are
four kings and four aces in a deck, the probabilities for event
A and event B are the same, 4/52 = 1/13. Our answer must be more
than this, so eliminate a and b. Do kings and aces have anything
to do with each other? Is there such a thing as an ace of kings
or a king of aces? No, so we don't have to worry about having
over-counted; the events are mutually exclusive. The probability
is straightforward: P(A or B) = P(A) + P(B) = 1/13 + 1/13 = 2/13.
C is correct.
Again we could have used simple probability. Count the total
number of kings and aces (4+4) and divide by the total number
of cards in a deck: 8/52 = 2/13.
w E. Conditional Probabilities
If you have any more questions or suggestions, email 24hourtutor@800score.com
<<
go back to table of contents
