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I. Probability
wA. Simple Probability
wB. Probability of Multiple Events
wC. Independent and Dependent Events
wD. Mutually Exclusive Events
wE. Conditional Probabilities
wF. Combinations

     

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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
D. 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.


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w E. Conditional Probabilities





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