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A family has three children. Determine whether the following pair of events are mutually exclusive.

M = The family has at least one boy size 12{M= left lbrace "The family has at least one boy" right rbrace } {}
N = The family has all girls size 12{N= left lbrace "The family has all girls" right rbrace } {}

Although the answer may be clear, we list both the sets.

M = BBB , BBG , BGB , BGG , GBB , GBG , GGB size 12{M= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GBG", ital "GGB" right rbrace } {} and N = GGG size 12{N= left lbrace ital "GGG" right rbrace } {}

Clearly, M N = Ø size 12{M intersection N="Ø"} {}

Therefore, the events M size 12{M} {} and N size 12{N} {} are mutually exclusive.

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We will now consider problems that involve the union of two events.

If a die is rolled, what is the probability of obtaining an even number or a number greater than four?

Let E size 12{E} {} be the event that the number shown on the die is an even number, and let F size 12{F} {} be the event that the number shown is greater than four.

The sample space S = 1,2,3,4,5,6 size 12{S= left lbrace 1,2,3,4,5,6 right rbrace } {} . The event E = 2,4,6 size 12{E= left lbrace 2,4,6 right rbrace } {} , and the event F = 5,6 size 12{F= left lbrace 5,6 right rbrace } {}

We need to find P E F size 12{P left (E union F right )} {} .

Since P E = 3 / 6 size 12{P left (E right )=3/6} {} , and P F = 2 / 6 size 12{P left (F right )=2/6} {} , a student may say P E F = 3 / 6 + 2 / 6 size 12{P left (E union F right )=3/6+2/6} {} . This will be incorrect because the element 6, which is in both E size 12{E} {} and F size 12{F} {} has been counted twice, once as an element of E size 12{E} {} and once as an element of F size 12{F} {} . In other words, the set E F size 12{E union F} {} has only four elements and not five. Therefore, P E F = 4 / 6 size 12{P left (E union F right )=4/6} {} and not 5 / 6 size 12{5/6} {} .

This can be illustrated by a Venn diagram.

The sample space S size 12{S} {} , the events E size 12{E} {} and F size 12{F} {} , and E F size 12{E intersection F} {} are listed below.

S = 1,2,3,4,5,6 size 12{S= left lbrace 1,2,3,4,5,6 right rbrace } {} , E = 2,4,6 size 12{E= left lbrace 2,4,6 right rbrace } {} , F = 5,6 size 12{F= left lbrace 5,6 right rbrace } {} , and E F = 6 size 12{E intersection F= left lbrace 6 right rbrace } {} .

The figure shows that everything in the square equals S. In the Venn Diagram 2 and 4 equal E, while 5 equals F and 6 equals both E and F.

The above figure shows S , E , F , and E F .

Finding the probability of E F , is the same as finding the probability that E will happen, or F will happen, or both will happen. If we count the number of elements n ( E ) in E , and add to it the number of elements n ( F ) in F , the points in both E and F are counted twice, once as elements of E and once as elements of F . Now if we subtract from the sum, n ( E ) + n ( F ) , the number n ( E F ) , we remove the duplicity and get the correct answer. So as a rule,

n ( E F ) = n ( E ) + n ( F ) n ( E F )

By dividing the entire equation by n ( S ) , we get

n ( E F ) n ( S ) = n ( E ) n ( S ) + n ( F ) n ( S ) n ( E F ) n ( S )

Since the probability of an event is the number of elements in that event divided by the number of all possible outcomes, we have

P ( E F ) = P ( E ) + P ( F ) P ( E F )

Applying the above for this example, we get

P ( E F ) = 3 / 6 + 2 / 6 - 1 / 6 = 4 / 6

This is because, when we add P ( E ) and P ( F ) , we have added P ( E F ) twice. Therefore, we must subtract P ( E F ) , once.

This gives us the general formula, called the Addition Rule , for finding the probability of the union of two events. It states

P ( E F ) = P ( E ) + P ( F ) P ( E F )

If two events E and F are mutually exclusive, then E F = and P ( E F ) = 0 , and we get

P ( E F ) = P ( E ) + P ( F )
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If a card is drawn from a deck, use the addition rule to find the probability of obtaining an ace or a heart.

Let A size 12{A} {} be the event that the card is an ace, and H size 12{H} {} the event that it is a heart.

Since there are four aces, and thirteen hearts in the deck, P A = 4 / 52 size 12{P left (A right )=4/"52"} {} and P H = 13 / 52 size 12{P left (H right )="13"/"52"} {} . Furthermore, since the intersection of two events is an ace of hearts, P A H = 1 / 52 size 12{P left (A intersection H right )=1/"52"} {}

We need to find P A H size 12{P left (A union H right )} {} .

P A H = P A + P H P A H = 4 / 52 + 13 / 52 1 / 52 = 16 / 52 size 12{P left (A union H right )=P left (A right )+P left (H right )–P left (A intersection H right )=4/"52"+"13"/"52" - 1/"52"="16"/"52"} {} .

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Two dice are rolled, and the events F size 12{F} {} and T size 12{T} {} are as follows:

F = The sum of the dice is four size 12{F= left lbrace "The sum of the dice is four" right rbrace } {} and T = At least one die shows a three size 12{T= left lbrace "At least one die shows a three" right rbrace } {}

Find P F T size 12{P left (F union T right )} {} .

We list F size 12{F} {} and T size 12{T} {} , and F T size 12{F intersection T} {} as follows:

F = 1,3 , 2,2 , 3,1 size 12{F= left lbrace left (1,3 right ), left (2,2 right ), left (3,1 right ) right rbrace } {}
T = 3,1 , 3,2 , 3,3 , 3,4 , 3,5 , 3,6 , 1,3 , 2,3 , 4,3 , 5,3 , 6,3 size 12{T= left lbrace left (3,1 right ), left (3,2 right ), left (3,3 right ), left (3,4 right ), left (3,5 right ), left (3,6 right ), left (1,3 right ), left (2,3 right ), left (4,3 right ), left (5,3 right ), left (6,3 right ) right rbrace } {}
F T = 1,3 , 3,1 size 12{F intersection T= left lbrace left (1,3 right ), left (3,1 right ) right rbrace } {}

Since P F T = P F + P T P F T size 12{P left (F union T right )=P left (F right )+P left (T right ) - P left (F intersection T right )} {}

We have P F T = 3 / 36 + 11 / 36 2 / 36 = 12 / 36 size 12{P left (F union T right )=3/"36"+"11"/"36" - 2/"36"="12"/"36"} {} .

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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