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The example we just considered consisted of only one outcome of the sample space. We are often interested in finding probabilities of several outcomes represented by an event.

An event is a subset of a sample space. If an event consists of only one outcome, it is called a simple event .

If two dice are rolled, find the probability that the sum of the faces of the dice is 7.

Let E size 12{E} {} represent the event that the sum of the faces of two dice is 7.

Since the possible cases for the sum to be 7 are: (1, 6), (2,5), (3, 4), (4, 3), (5, 2), and (6, 1).

E = 1,6 , 2,5 , 3,4 4,3 , 5,2 , and 6,1 size 12{E= left lbrace left (1,6 right ), left (2,5 right ), left (3,4 right ) left (4,3 right ), left (5,2 right )", and " left (6,1 right ) right rbrace } {}

and the probability of the event E size 12{E} {} ,

P E = 6 / 36 size 12{P left (E right )=6/"36"} {} or 1 / 6 size 12{1/6} {} .

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A jar contains 3 red, 4 white, and 3 blue marbles. If a marble is chosen at random, what is the probability that the marble is a red marble or a blue marble?

We assume the marbles are r 1 size 12{r rSub { size 8{1} } } {} , r 2 size 12{r rSub { size 8{2} } } {} , r 3 size 12{r rSub { size 8{3} } } {} , w 1 size 12{w rSub { size 8{1} } } {} , w 2 size 12{w rSub { size 8{2} } } {} , w 3 size 12{w rSub { size 8{3} } } {} , w 4 size 12{w rSub { size 8{4} } } {} , b 1 size 12{b rSub { size 8{1} } } {} , b 2 size 12{b rSub { size 8{2} } } {} , b 3 size 12{b rSub { size 8{3} } } {} . Let the event C size 12{C} {} represent that the marble is red or blue.

The sample space S = r 1 , r 2 , r 3, w 1 , w 2 , w 3 , w 4 , b 1 , b 2 , b 3 size 12{S= left lbrace r rSub { size 8{1} } ,r rSub { size 8{2} } ,r rSub { size 8{3,} } w rSub { size 8{1} } ,w rSub { size 8{2} } ,w rSub { size 8{3} } ,w rSub { size 8{4} } ,b rSub { size 8{1} } ,b rSub { size 8{2} } ,b rSub { size 8{3} } right rbrace } {}

And the event C = r 1 , r 2 , r 3 , b 1 , b 2 , b 3 size 12{C= left lbrace r rSub { size 8{1} } ,r rSub { size 8{2} } ,r rSub { size 8{3} } ,b rSub { size 8{1} } ,b rSub { size 8{2} } ,b rSub { size 8{3} } right rbrace } {}

Therefore, the probability of C size 12{C} {} ,

P C = 6 / 10 size 12{P left (C right )=6/"10"} {} or 3 / 5 size 12{3/5} {} .

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A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn, what is the probability that the sum of the numbers is 4?

Since two marbles are drawn, the sample space consists of the following six possibilities.

S = 1,2 , 1,3 , 2,3 , 2,1 , 3,1 , 3,2 size 12{S= left lbrace left (1,2 right ), left (1,3 right ), left (2,3 right ), left (2,1 right ), left (3,1 right ), left (3,2 right ) right rbrace } {}

Let the event F represent that the sum of the numbers is four. Then

F = 1,3 , 3,1 size 12{F= left [ left (1,3 right ), left (3,1 right ) right ]} {}

Therefore, the probability of F size 12{F} {} is

P F = 2 / 6 size 12{P left (F right )=2/6} {} or 1 / 3 size 12{1/3} {} .

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A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn, what is the probability that the sum of the numbers is at least 4?

The sample space, as in [link] , consists of the following six possibilities.

S = 1,2 , 1,3 , 2,3 , 2,1 , 3,1 , 3,2 size 12{S= left lbrace left (1,2 right ), left (1,3 right ), left (2,3 right ), left (2,1 right ), left (3,1 right ), left (3,2 right ) right rbrace } {}

Let the event A size 12{A} {} represent that the sum of the numbers is at least four. Then

F = 1,3 , 3,1 , 2,3 , 3,2 size 12{F= left lbrace left (1,3 right ), left (3,1 right ), left (2,3 right ), left (3,2 right ) right rbrace } {}

Therefore, the probability of F size 12{F} {} is

P F = 4 / 6 size 12{P left (F right )=4/6} {} or 2 / 3 size 12{2/3} {} .

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Mutually exclusive events and the addition rule

In the [link] , we learned to find the union, intersection, and complement of a set. We will now use these set operations to describe events.

The union of two events E size 12{E} {} and F size 12{F} {} , E F size 12{E union F} {} , is the set of outcomes that are in E size 12{E} {} or in F size 12{F} {} or in both.

The intersection of two events E size 12{E} {} and F size 12{F} {} , E F size 12{E intersection F} {} , is the set of outcomes that are in both E size 12{E} {} and F size 12{F} {} .

The complement of an event E size 12{E} {} , denoted by E c size 12{E rSup { size 8{c} } } {} , is the set of outcomes in the sample space S size 12{S} {} that are not in E size 12{E} {} . It is worth noting that P E C = 1 P E size 12{P left (E rSup { size 8{C} } right )=1 - P left (E right )} {} . This follows from the fact that if the sample space has n size 12{n} {} elements and E size 12{E} {} has k size 12{k} {} elements, then E c size 12{E rSup { size 8{c} } } {} has n k size 12{n - k} {} elements. Therefore,

P E C = n k n = 1 k n = 1 P E size 12{P left (E rSup { size 8{C} } right )= { {n - k} over {n} } =1 - { {k} over {n} } =1 - P left (E right )} {} .

Of particular interest to us are the events whose outcomes do not overlap. We call these events mutually exclusive.

Two events E size 12{E} {} and F size 12{F} {} are said to be mutually exclusive if they do not intersect. That is, E F = size 12{E intersection F=" 00000"} {} .

Next we'll determine whether a given pair of events are mutually exclusive.

A card is drawn from a standard deck. Determine whether the pair of events given below is mutually exclusive.

E = The card drawn is an Ace size 12{E= left lbrace "The card drawn is an Ace" right rbrace } {}
F = The card drawn is a heart size 12{F= left lbrace "The card drawn is a heart" right rbrace } {}

Clearly the ace of hearts belongs to both sets. That is

E F = Ace of hearts size 12{E intersection F= left lbrace "Ace of hearts" right rbrace<>"00000"} {} .

Therefore, the events E size 12{E} {} and F size 12{F} {} are not mutually exclusive.

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Two dice are rolled. Determine whether the pair of events given below is mutually exclusive.

G = The sum of the faces is six size 12{G= left lbrace "The sum of the faces is six" right rbrace } {}
H = One die shows a four size 12{H= left lbrace "One die shows a four" right rbrace } {}

For clarity, we list the elements of both sets.

G = 1,5 , 2,4 , 3,3 , 4,2 , 5,1 size 12{G= left lbrace left (1,5 right ), left (2,4 right ), left (3,3 right ), left (4,2 right ), left (5,1 right ) right rbrace } {}
H = 2,4 , 4,2 size 12{H= left lbrace left (2,4 right ), left (4,2 right ) right rbrace } {}

Clearly, G H = 2,4 , 4,2 Ø size 12{G intersection H= left lbrace left (2,4 right ), left (4,2 right ) right rbrace<>"Ø"} {} .

Therefore, the two sets are not mutually exclusive.

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Questions & Answers

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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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