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Probability

Module by: Rupinder Sekhon. E-mail the author

Summary: This chapter covers principles of probability. After completing this chapter students should be able to: write sample spaces; determine whether two events are mutually exclusive; use the addition rule; calculate probabilities using tree diagrams and combinations; solve problems involving conditional probability; determine whether two events are independent.

Chapter Overview

In this chapter, you will learn to:

  1. Write sample spaces.
  2. Determine whether two events are mutually exclusive.
  3. Use the Addition Rule.
  4. Calculate probabilities using both tree diagrams and combinations.
  5. Do problems involving conditional probability.
  6. Determine whether two events are independent.

Sample Spaces and Probability

If two coins are tossed, what is the probability that both coins will fall heads? The problem seems simple enough, but it is not uncommon to hear the incorrect answer 1/31/3 size 12{1/3} {}. A student may incorrectly reason that if two coins are tossed there are three possibilities, one head, two heads, or no heads. Therefore, the probability of two heads is one out of three. The answer is wrong because if we toss two coins there are four possibilities and not three. For clarity, assume that one coin is a penny and the other a nickel. Then we have the following four possibilities.

HH HT TH TT

The possibility HT, for example, indicates a head on the penny and a tail on the nickel, while TH represents a tail on the penny and a head on the nickel.

It is for this reason, we emphasize the need for understanding sample spaces.

An act of flipping coins, rolling dice, drawing cards, or surveying people are referred to as an experiment.

Definition 1: Sample Spaces
A sample space of an experiment is the set of all possible outcomes.

Example 1

Problem 1

If a die is rolled, write a sample space.

Solution

A die has six faces each having an equally likely chance of appearing. Therefore, the set of all possible outcomes SS size 12{S} {} is

1,2,3,4,5,61,2,3,4,5,6 size 12{ left lbrace 1,2,3,4,5,6 right rbrace } {}.

Example 2

Problem 1

A family has three children. Write a sample space.

Solution

The sample space consists of eight possibilities.

BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGGBBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG size 12{ left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GBG", ital "GGB", ital "GGG" right rbrace } {}
(1)

The possibility BGBBGB size 12{ ital "BGB"} {}, for example, indicates that the first born is a boy, the second born a girl, and the third a boy.

We illustrate these possibilities with a tree diagram.

Figure 1
The tree diagram illustrates the different possibilities for the gender of three children in a family.

Example 3

Problem 1

Two dice are rolled. Write the sample space.

Solution

We assume one of the dice is red, and the other green. We have the following 36 possibilities.

Table 1
  Green
Red 1 2 3 4 5 6
1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The entry (2, 5), for example, indicates that the red die shows a two, and the green a 5.

Now that we understand the concept of a sample space, we will define probability.

Probability

For a sample space SS size 12{S} {}, and an outcome AA size 12{A} {} of SS size 12{S} {}, the following two properties are satisfied.

  1. If AA size 12{A} {} is an outcome of a sample space, then the probability of AA size 12{A} {}, denoted by PAPA size 12{P left (A right )} {}, is between 0 and 1, inclusive.
    0 ≤ P A ≤ 1 0 ≤ P A ≤ 1 size 12{0 <= P left (A right ) <= 1} {}
    (2)
  2. The sum of the probabilities of all the outcomes in SS size 12{S} {} equals 1.

Example 4

Problem 1

If two dice, one red and one green, are rolled, find the probability that the red die shows a 3 and the green shows a six.

Solution

Since two dice are rolled, there are 36 possibilities. The probability of each outcome, listed in Example 3, is equally likely.

Since (3, 6) is one such outcome, the probability of obtaining (3, 6) is 1/361/36 size 12{1/"36"} {}.

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.

Example 5

Problem 1

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

Solution

Let EE 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=1,6,2,5,3,44,3,5,2, and 6,1E=1,6,2,5,3,44,3,5,2, 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 } {}
(3)

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

PE=6/36PE=6/36 size 12{P left (E right )=6/"36"} {} or 1/61/6 size 12{1/6} {}.

Example 6

Problem 1

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?

Solution

We assume the marbles are r1r1 size 12{r rSub { size 8{1} } } {}, r2r2 size 12{r rSub { size 8{2} } } {}, r3r3 size 12{r rSub { size 8{3} } } {}, w1w1 size 12{w rSub { size 8{1} } } {}, w2w2 size 12{w rSub { size 8{2} } } {}, w3w3 size 12{w rSub { size 8{3} } } {}, w4w4 size 12{w rSub { size 8{4} } } {}, b1b1 size 12{b rSub { size 8{1} } } {}, b2b2 size 12{b rSub { size 8{2} } } {}, b3b3 size 12{b rSub { size 8{3} } } {}. Let the event CC size 12{C} {} represent that the marble is red or blue.

The sample space S=r1,r2,r3,w1,w2,w3,w4,b1,b2,b3S=r1,r2,r3,w1,w2,w3,w4,b1,b2,b3 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=r1,r2,r3,b1,b2,b3C=r1,r2,r3,b1,b2,b3 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 CC size 12{C} {},

PC=6/10PC=6/10 size 12{P left (C right )=6/"10"} {} or 3/53/5 size 12{3/5} {}.

Example 7

Problem 1

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?

Solution

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

S=1,2,1,3,2,3,2,1,3,1,3,2S=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 } {}
(4)

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

F=1,3,3,1F=1,3,3,1 size 12{F= left [ left (1,3 right ), left (3,1 right ) right ]} {}
(5)

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

PF=2/6PF=2/6 size 12{P left (F right )=2/6} {} or 1/31/3 size 12{1/3} {}.

Example 8

Problem 1

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?

Solution

The sample space, as in Example 7, consists of the following six possibilities.

S=1,2,1,3,2,3,2,1,3,1,3,2S=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 } {}
(6)

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

F=1,3,3,1,2,3,3,2F=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 } {}
(7)

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

PF=4/6PF=4/6 size 12{P left (F right )=4/6} {} or 2/32/3 size 12{2/3} {}.

Mutually Exclusive Events and the Addition Rule

In the (Reference), 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 EE size 12{E} {} and FF size 12{F} {}, E∪FE∪F size 12{E union F} {}, is the set of outcomes that are in EE size 12{E} {} or in FF size 12{F} {} or in both.

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

The complement of an event EE size 12{E} {}, denoted by EcEc size 12{E rSup { size 8{c} } } {}, is the set of outcomes in the sample space SS size 12{S} {} that are not in EE size 12{E} {}. It is worth noting that PEC=1−PEPEC=1−PE 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 nn size 12{n} {} elements and EE size 12{E} {} has kk size 12{k} {} elements, then EcEc size 12{E rSup { size 8{c} } } {} has n−kn−k size 12{n - k} {} elements. Therefore,

PEC=n−kn=1−kn=1−PEPEC=n−kn=1−kn=1−PE 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 EE size 12{E} {} and FF size 12{F} {} are said to be mutually exclusive if they do not intersect. That is, E∩F=∅E∩F=∅ size 12{E intersection F=" 00000"} {}.

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

Example 9

Problem 1

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 AceE=The card drawn is an Ace size 12{E= left lbrace "The card drawn is an Ace" right rbrace } {}
(8)
F=The card drawn is a heartF=The card drawn is a heart size 12{F= left lbrace "The card drawn is a heart" right rbrace } {}
(9)
Solution

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

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

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

Example 10

Problem 1

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

G=The sum of the faces is sixG=The sum of the faces is six size 12{G= left lbrace "The sum of the faces is six" right rbrace } {}
(10)
H=One die shows a fourH=One die shows a four size 12{H= left lbrace "One die shows a four" right rbrace } {}
(11)
Solution

For clarity, we list the elements of both sets.

G=1,5,2,4,3,3,4,2,5,1G=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 } {}
(12)
H=2,4,4,2H=2,4,4,2 size 12{H= left lbrace left (2,4 right ), left (4,2 right ) right rbrace } {}
(13)

Clearly, G∩H=2,4,4,2≠ØG∩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.

Example 11

Problem 1

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

M=The family has at least one boyM=The family has at least one boy size 12{M= left lbrace "The family has at least one boy" right rbrace } {}
(14)
N=The family has all girlsN=The family has all girls size 12{N= left lbrace "The family has all girls" right rbrace } {}
(15)
Solution

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

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

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

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

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

Example 12

Problem 1

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

Solution

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

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

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

Since PE=3/6PE=3/6 size 12{P left (E right )=3/6} {}, and PF=2/6PF=2/6 size 12{P left (F right )=2/6} {}, a student may say PE∪F=3/6+2/6PE∪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 EE size 12{E} {} and FF size 12{F} {} has been counted twice, once as an element of EE size 12{E} {} and once as an element of FF size 12{F} {}. In other words, the set E∪FE∪F size 12{E union F} {} has only four elements and not five. Therefore, PE∪F=4/6PE∪F=4/6 size 12{P left (E union F right )=4/6} {} and not 5/65/6 size 12{5/6} {} .

This can be illustrated by a Venn diagram.

The sample space SS size 12{S} {}, the events EE size 12{E} {} and FF size 12{F} {}, and E∩FE∩F size 12{E intersection F} {} are listed below.

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

Figure 2
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 SS, EE, FF, and E∩FE∩F.

Finding the probability of E∪FE∪F, is the same as finding the probability that EE will happen, or FF will happen, or both will happen. If we count the number of elements n(E)n(E) in EE, and add to it the number of elements n(F)n(F) in FF, the points in both EE and FF are counted twice, once as elements of EE and once as elements of FF. Now if we subtract from the sum, n(E)+ n(F)n(E)+n(F), the number n(E∩F)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)n(E∪F)=n(E)+n(F)–n(E∩F)
(16)

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

n(E∪F)n(S)=n(E)n(S)+n(F)n(S)–n(E∩F)n(S)n(E∪F)n(S)=n(E)n(S)+n(F)n(S)–n(E∩F)n(S)
(17)

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)P(E∪F)=P(E)+P(F)–P(E∩F)
(18)

Applying the above for this example, we get

P(E∪F)=3/6+2/6-1/6=4/6P(E∪F)=3/6+2/6-1/6=4/6
(19)

This is because, when we add P(E)P(E) and P(F)P(F), we have added P(E∩F)P(E∩F) twice. Therefore, we must subtract P(E∩F)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)P(E∪F)=P(E)+P(F)–P(E∩F)
(20)

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

P(E∪F)=P(E)+P(F)P(E∪F)=P(E)+P(F)
(21)

Example 13

Problem 1

If a card is drawn from a deck, use the addition rule to find the probability of obtaining an ace or a heart.

Solution

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

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

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

PA∪H=PA+PH–PA∩H=4/52+13/52−1/52=16/52PA∪H=PA+PH–PA∩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"} {}.

Example 14

Problem 1

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

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

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

Solution

We list FF size 12{F} {} and TT size 12{T} {}, and F∩TF∩T size 12{F intersection T} {} as follows:

F=1,3,2,2,3,1F=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 } {}
(22)
T=3,1,3,2,3,3,3,4,3,5,3,6,1,3,2,3,4,3,5,3,6,3T=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 } {}
(23)
F∩T=1,3,3,1F∩T=1,3,3,1 size 12{F intersection T= left lbrace left (1,3 right ), left (3,1 right ) right rbrace } {}
(24)

Since PF∪T=PF+PT−PF∩TPF∪T=PF+PT−PF∩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 PF∪T=3/36+11/36−2/36=12/36PF∪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"} {}.

Example 15

Problem 1

Mr. Washington is seeking a mathematics instructor's position at his favorite community college in Cupertino. His employment depends on two conditions: whether the board approves the position, and whether the hiring committee selects him. There is a 80% chance that the board will approve the position, and there is a 70% chance that the hiring committee will select him. If there is a 90% chance that at least one of the two conditions, the board approval or his selection, will be met, what is the probability that Mr. Washington will be hired?

Solution

Let AA size 12{A} {} be the event that the board approves the position, and S be the event that Mr. Washington gets selected. We have,

PA=.80PA=.80 size 12{P left (A right )= "." "80"} {}, PS=.70PS=.70 size 12{P left (S right )= "." "70"} {}, and PA∪S=.90PA∪S=.90 size 12{P left (A union S right )= "." "90"} {}.

We need to find, PA∩SPA∩S size 12{P left (A intersection S right )} {}.

The addition formula states that,

PA∪S=PA+PS−PA∩SPA∪S=PA+PS−PA∩S size 12{P left (A union S right )=P left (A right )+P left (S right ) - P left (A intersection S right )} {}
(25)

Substituting the known values, we get

.90=.80+.70−PA∩S.90=.80+.70−PA∩S size 12{ "." "90"= "." "80"+ "." "70" - P left (A intersection S right )} {}
(26)

Therefore, PA∩S=.60PA∩S=.60 size 12{P left (A intersection S right )= "." "60"} {}.

Example 16

Problem 1

The probability that this weekend will be cold is .6.6 size 12{ "." 6} {}, the probability that it will be rainy is .7.7 size 12{ "." 7} {}, and probability that it will be both cold and rainy is .5.5 size 12{ "." 5} {}. What is the probability that it will be neither cold nor rainy?

Solution

Let CC size 12{C} {} be the event that the weekend will be cold, and RR size 12{R} {} be event that it will be rainy. We are given that

PC=.6PC=.6 size 12{P left (C right )= "." 6} {}, PR=.7PR=.7 size 12{P left (R right )= "." 7} {}, PC∩R=.5PC∩R=.5 size 12{P left (C intersection R right )= "." 5} {}

PC∪R=PC+PR−PC∩R=.6+.7−.5=.8PC∪R=PC+PR−PC∩R=.6+.7−.5=.8 size 12{P left (C union R right )=P left (C right )+P left (R right ) - P left (C intersection R right )= "." 6+ "." 7 - "." 5= "." 8} {}
(27)

We want to find PC∪RcPC∪Rc size 12{P left ( left (C union R right ) rSup { size 8{c} } right )} {}.

PC∪Rc=1−PC∪R=1−.8=.2PC∪Rc=1−PC∪R=1−.8=.2 size 12{P left ( left (C union R right ) rSup { size 8{c} } right )=1 - P left (C union R right )=1 - "." 8= "." 2} {}
(28)

We summarize this section by listing the important rules.

The Addition Rule

For Two Events EE size 12{E} {} and FF size 12{F} {}, PE∪F=PE+PF−PE∩FPE∪F=PE+PF−PE∩F size 12{P left (E union F right )=P left (E right )+P left (F right ) - P left (E intersection F right )} {}

The Addition Rule for Mutually Exclusive Events

If Two Events EE size 12{E} {} and FF size 12{F} {} are Mutually Exclusive, then PE∪F=PE+PFPE∪F=PE+PF size 12{P left (E union F right )=P left (E right )+P left (F right )} {}

The Complement Rule

If EcEc size 12{E rSup { size 8{c} } } {} is the Complement of Event EE size 12{E} {}, then PEc=1−PEPEc=1−PE size 12{P left (E rSup { size 8{c} } right )=1 - P left (E right )} {}

Probability Using Tree Diagrams and Combinations

In this section, we will apply previously learnt counting techniques in calculating probabilities, and use tree diagrams to help us gain a better understanding of what is involved.

We begin with an example.

Example 18

Problem 1

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn with replacement, what is the probability that both marbles are red?

Solution

Let EE size 12{E} {} be the event that the first marble drawn is red, and let FF size 12{F} {} be the event that the second marble drawn is red.

We need to find PE∩FPE∩F size 12{P left (E intersection F right )} {}.

By the statement, "two marbles are drawn with replacement," we mean that the first marble is replaced before the second marble is drawn.

There are 7 choices for the first draw. And since the first marble is replaced before the second is drawn, there are, again, seven choices for the second draw. Using the multiplication axiom, we conclude that the sample space SS size 12{S} {} consists of 49 ordered pairs. Of the 49 ordered pairs, there are 3×3=93×3=9 size 12{3 times 3=9} {} ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

PE∩F=949=37⋅37PE∩F=949=37⋅37 size 12{P left (E intersection F right )= { {9} over {"49"} } = { {3} over {7} } cdot { {3} over {7} } } {}
(29)

Further note that in this particular case

PE∩F=PE⋅PFPE∩F=PE⋅PF size 12{P left (E intersection F right )=P left (E right ) cdot P left (F right )} {}
(30)

Example 19

Problem 1

If in the Example 18, the two marbles are drawn without replacement, then what is the probability that both marbles are red?

Solution

By the statement, "two marbles are drawn without replacement," we mean that the first marble is not replaced before the second marble is drawn.

Again, we need to find PE∩FPE∩F size 12{P left (E intersection F right )} {}.

There are, again, 7 choices for the first draw. And since the first marble is not replaced before the second is drawn, there are only six choices for the second draw. Using the multiplication axiom, we conclude that the sample space SS size 12{S} {} consists of 42 ordered pairs. Of the 42 ordered pairs, there are 3×2=63×2=6 size 12{3 times 2=6} {} ordered pairs that show red on the first draw and red on the second draw. Therefore,

PE∩F=642=37⋅26PE∩F=642=37⋅26 size 12{P left (E intersection F right )= { {6} over {"42"} } = { {3} over {7} } cdot { {2} over {6} } } {}
(31)

Here 3/73/7 size 12{3/7} {} represents PEPE size 12{P left (E right )} {}, and 2/62/6 size 12{2/6} {} represents the probability of drawing a red on the second draw, given that the first draw resulted in a red. We write the latter as PRed on the second∣red on firstPRed on the second∣red on first size 12{P left ("Red on the second" \lline "red on first" right )} {} or PF∣EPF∣E size 12{P left (F \lline E right )} {}. The "|" represents the word "given." Therefore,

PE∩F=PE⋅PF∣EPE∩F=PE⋅PF∣E size 12{P left (F intersection E right )=P left (E right ) cdot P left (E \lline F right )} {}
(32)

The above result is an important one and will appear again in later sections.

We now demonstrate the above results with a tree diagram.

Example 20

Problem 1

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn without replacement, find the following probabilities using a tree diagram.

  1. The probability that both marbles are white.
  2. The probability that the first marble is red and the second white.
  3. The probability that one marble is red and the other white.
Solution

Let RR size 12{R} {} be the event that the marble drawn is red, and let WW size 12{W} {} be the event that the marble drawn is white.

We draw the following tree diagram.

Figure 3
The tree diagram illustrates the different probability of drawing a red marble or a white marble.

Although the tree diagrams give us better insight into a problem, they are not practical for problems where more than two or three things are chosen. In such cases, we use the concept of combinations that we learned in (Reference). This method is best suited for problems where the order in which the objects are chosen is not important, and the objects are chosen without replacement.

Example 21

Problem 1

Suppose a jar contains 3 red, 2 white, and 3 blue marbles. If three marbles are drawn without replacement, find the following probabilities.

  1. PTwo red and one whitePTwo red and one white size 12{P left ("Two red and one white" right )} {}
  2. POne of each colorPOne of each color size 12{P left ("One of each color" right )} {}
  3. PNone bluePNone blue size 12{P left ("None blue" right )} {}
  4. PAt least one bluePAt least one blue size 12{P left ("At least one blue" right )} {}
Solution

Let us suppose the marbles are labeled as R1R1 size 12{R rSub { size 8{1} } } {}, R2R2 size 12{R rSub { size 8{2} } } {}, R3R3 size 12{R rSub { size 8{3} } } {}, W1W1 size 12{W rSub { size 8{1} } } {}, W2W2 size 12{W rSub { size 8{2} } } {}, B1B1 size 12{B rSub { size 8{1} } } {}, B2B2 size 12{B rSub { size 8{2} } } {}, B3B3 size 12{B rSub { size 8{3} } } {}.

  1. PTwo red and one whitePTwo red and one white size 12{P left ("Two red and one white" right )} {}

    We analyze the problem in the following manner.

    Since we are choosing 3 marbles from a total of 8, there are 8C3=568C3=56 size 12{8C3="56"} {} possible combinations. Of these 56 combinations, there are 3C2×2C1=63C2×2C1=6 size 12{3C2 times 2C1=6} {} combinations consisting of 2 red and one white. Therefore,

    PTwo red and one white=3C2×2C18C3=656PTwo red and one white=3C2×2C18C3=656 size 12{P left ("Two red and one white" right )= { {3C2 times 2C1} over {8C3} } = { {6} over {"56"} } } {}.

  2. POne of each colorPOne of each color size 12{P left ("One of each color" right )} {}

    Again, there are 8C3=568C3=56 size 12{8C3="56"} {} possible combinations. Of these 56 combinations, there are 3C1×2C1×3C1=183C1×2C1×3C1=18 size 12{3C1 times 2C1 times 3C1="18"} {} combinations consisting of one red, one white, and one blue. Therefore,

    POne of each color=3C1×2C1×3C18C3=1856POne of each color=3C1×2C1×3C18C3=1856 size 12{P left ("One of each color" right )= { {3C1 times 2C1 times 3C1} over {8C3} } = { {"18"} over {"56"} } } {}.

  3. PNone bluePNone blue size 12{P left ("None blue" right )} {}

    There are 5 non-blue marbles, therefore

    PNone blue=5C38C3=1056=528PNone blue=5C38C3=1056=528 size 12{P left ("None blue" right )= { {5C3} over {8C3} } = { {"10"} over {"56"} } = { {5} over {"28"} } } {}.

  4. PAt least one bluePAt least one blue size 12{P left ("At least one blue" right )} {}

    By "at least one blue marble," we mean the following: one blue marble and two non-blue marbles, or two blue marbles and one non-blue marble, or all three blue marbles. So we have to find the sum of the probabilities of all three cases.

    PAt least one blue=Pone blue, two non-blue+Ptwo blue, one non-blue+Pthree bluePAt least one blue=Pone blue, two non-blue+Ptwo blue, one non-blue+Pthree blue size 12{P left ("At least one blue" right )=P left ("one blue, two non-blue" right )+P left ("two blue, one non-blue" right )+P left ("three blue" right )} {}
    (33)
    PAt least one blue=3C1×5C28C3+3C2×5C18C3+3C38C3PAt least one blue=3C1×5C28C3+3C2×5C18C3+3C38C3 size 12{P left ("At least one blue" right )= { {3C1 times 5C2} over {8C3} } + { {3C2 times 5C1} over {8C3} } + { {3C3} over {8C3} } } {}
    (34)

    PAt least one blue=30/56+15/56+1/56=46/56=23/28PAt least one blue=30/56+15/56+1/56=46/56=23/28 size 12{P left ("At least one blue" right )="30"/"56"+"15"/"56"+1/"56"="46"/"56"="23"/"28"} {}.

    Alternately,

    we use the fact that PE=1−PEcPE=1−PEc size 12{P left (E right )=1 - P left (E rSup { size 8{c} } right )} {}.

    If the event E=At least one blueE=At least one blue size 12{E="At least one blue"} {}, then Ec=None blueEc=None blue size 12{E rSup { size 8{c} } ="None blue"} {}.

    But from part c of this example, we have Ec=5/28Ec=5/28 size 12{ left (E rSup { size 8{c} } right )=5/"28"} {}

    Therefore, PE=1−5/28=23/28PE=1−5/28=23/28 size 12{P left (E right )=1 - 5/"28"="23"/"28"} {}.

Example 22

Problem 1

Five cards are drawn from a deck. Find the probability of obtaining two pairs, that is, two cards of one value, two of another value, and one other card.

Solution

Let us first do an easier problem–the probability of obtaining a pair of kings and queens.

Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is

PA pair of kings and queens=4C2×4C2×44C152C5PA pair of kings and queens=4C2×4C2×44C152C5 size 12{P left ("A pair of kings and queens" right )= { {4C2 times 4C2 times "44"C1} over {"52"C5} } } {}
(35)

To find the probability of obtaining two pairs, we have to consider all possible pairs.

Since there are altogether 13 values, that is, aces, deuces, and so on, there are 13C213C2 size 12{"13"C2} {} different combinations of pairs.

PTwo pairs=13C2⋅4C2×4C2×44C152C5=.04754PTwo pairs=13C2⋅4C2×4C2×44C152C5=.04754 size 12{P left ("Two pairs" right )="13"C2 cdot { {4C2 times 4C2 times "44"C1} over {"52"C5} } = "." "04754"} {}
(36)

We end the section by solving a problem called the Birthday Problem.

Example 23

Problem 1

If there are 25 people in a room, what is the probability that at least two people have the same birthday?

Solution

Let event EE size 12{E} {} represent that at least two people have the same birthday.

We first find the probability that no two people have the same birthday.

We analyze as follows.

Suppose there are 365 days to every year. According to the multiplication axiom, there are 3652536525 size 12{"365" rSup { size 8{"25"} } } {} possible birthdays for 25 people. Therefore, the sample space has 3652536525 size 12{"365" rSup { size 8{"25"} } } {} elements. We are interested in the probability that no two people have the same birthday. There are 365 possible choices for the first person and since the second person must have a different birthday, there are 364 choices for the second, 363 for the third, and so on. Therefore,

PNo two have the same birthday=365⋅364⋅363⋯34136525=365P2536525PNo two have the same birthday=365⋅364⋅363⋯34136525=365P2536525 size 12{P left ("No two have the same birthday" right )= { {"365" cdot "364" cdot "363" dotsaxis "341"} over {"365" rSup { size 8{"25"} } } } = { {"365"P"25"} over {"365" rSup { size 8{"25"} } } } } {}
(37)

Since Pat least two people have the same birthday=1−PNo two have the same birthday,Pat least two people have the same birthday=1−PNo two have the same birthday, size 12{P left ("at least two people have the same birthday" right )=1 - P left ("No two have the same birthday" right ),} {}

Pat least two people have the same birthday=1−365P2536525=.5687Pat least two people have the same birthday=1−365P2536525=.5687 size 12{P left ("at least two people have the same birthday" right )=1 - { {"365"P"25"} over {"365" rSup { size 8{"25"} } } } = "." "5687"} {}
(38)

Conditional Probability

Suppose you and a friend wish to play a game that involves choosing a single card from a well-shuffled deck. Your friend deals you one card, face down, from the deck and offers you the following deal: If the card is a king, he will pay you $5, otherwise, you pay him $1. Should you play the game?

You reason in the following manner. Since there are four kings in the deck, the probability of obtaining a king is 4/524/52 size 12{4/"52"} {} or 1/131/13 size 12{1/"13"} {}. And, probability of not obtaining a king is 12/1312/13 size 12{"12"/"13"} {}. This implies that the ratio of your winning to losing is 1 to 12, while the payoff ratio is only $1 to $5. Therefore, you determine that you should not play.

Now consider the following scenario. While your friend was dealing the card, you happened to get a glance of it and noticed that the card was a face card. Should you, now, play the game?

Since there are 12 face cards in the deck, the total elements in the sample space are no longer 52, but just 12. This means the chance of obtaining a king is 4/124/12 size 12{4/"12"} {} or 1/31/3 size 12{1/3} {}. So your chance of winning is 1/31/3 size 12{1/3} {} and of losing 2/32/3 size 12{2/3} {}. This makes your winning to losing ratio 1 to 2 which fares much better with the payoff ratio of $1 to $5. This time, you determine that you should play.

In the second part of the above example, we were finding the probability of obtaining a king knowing that a face card had shown. This is an example of conditional probability. Whenever we are finding the probability of an event E under the condition that another event F has happened, we are finding conditional probability.

The symbol PE∣FPE∣F size 12{P left (E \lline F right )} {} denotes the problem of finding the probability of EE size 12{E} {} given that FF size 12{F} {} has occurred. We read PE∣FPE∣F size 12{P left (E \lline F right )} {} as "the probability of EE size 12{E} {}, given FF size 12{F} {}."

Example 24

Problem 1

A family has three children. Find the conditional probability of having two boys and a girl given that the first born is a boy.

Solution

Let event EE size 12{E} {} be that the family has two boys and a girl, and FF size 12{F} {} the event that the first born is a boy.

First, we list the sample space for a family of three children as follows.

S=BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGGS=BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG size 12{S= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GBG", ital "GGB", ital "GGG" right rbrace } {}
(39)

Since we know that the first born is a boy, our possibilities narrow down to four outcomes, BBBBBB size 12{ ital "BBB"} {}, BBGBBG size 12{ ital "BBG"} {}, BGBBGB size 12{ ital "BGB"} {}, and BGGBGG size 12{ ital "BGG"} {}.

Among the four, BBGBBG size 12{ ital "BBG"} {} and BGBBGB size 12{ ital "BGB"} {} represent two boys and a girl.

Therefore, P(E∣F=2/4P(E∣F=2/4 size 12{P \( E \lline F rbrace =2/4} {} or 1/21/2 size 12{1/2} {}.

Let us now develop a formula for the conditional probability PE∣FPE∣F size 12{P left (E \lline F right )} {}.

Suppose an experiment consists of nn size 12{n} {} equally likely events. Further suppose that there are mm size 12{m} {} elements in FF size 12{F} {}, and cc size 12{c} {} elements in E∩FE∩F size 12{E intersection F} {}, as shown in the following Venn diagram.

Figure 4
The figure shows that everything within the square is equal to S. In the Venn diagram m-c is equal to F where c is equal to both E and F.

If the event FF size 12{F} {} has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset FF size 12{F} {}. Therefore, we only look at the set FF size 12{F} {} and at nothing outside of FF size 12{F} {}. Since FF size 12{F} {} has mm size 12{m} {} elements, the denominator in the calculation of PE∣FPE∣F size 12{P left (E \lline F right )} {} is m. We may think that the numerator for our conditional probability is the number of elements in EE size 12{E} {}. But clearly we cannot consider the elements of EE size 12{E} {} that are not in FF size 12{F} {}. We can only count the elements of EE size 12{E} {} that are in FF size 12{F} {}, that is, the elements in E∩FE∩F size 12{E intersection F} {}. Therefore,

P E ∣ F = c m P E ∣ F = c m size 12{P left (E \lline F right )= { {c} over {m} } } {}

Dividing both the numerator and the denominator by nn size 12{n} {}, we get

P E ∣ F = c / n m / n P E ∣ F = c / n m / n size 12{P left (E \lline F right )= { {c/n} over {m/n} } } {}
(40)

But c/n=PE∩Fc/n=PE∩F size 12{c/n=P left (E intersection F right )} {}, and m/n=PFm/n=PF size 12{m/n=P left (F right )} {}.

Substituting, we derive the following formula for PE∣FPE∣F size 12{P left (E \lline F right )} {}.

For Two Events EE size 12{E} {} and FF size 12{F} {}, the Probability of E E size 12{E} {} Given FF size 12{F} {} is

P E ∣ F = P E ∩ F P F P E ∣ F = P E ∩ F P F size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}
(41)

Example 26

Problem 1

A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.

Solution

Let EE size 12{E} {} be the event that an even number shows, and FF size 12{F} {} be the event that a number greater than three shows. We want PE∣FPE∣F size 12{P left (E \lline F right )} {}.

E=2,4,6E=2,4,6 size 12{E= left lbrace 2,4,6 right rbrace } {} and F=4,5,6F=4,5,6 size 12{F= left lbrace 4,5,6 right rbrace } {}. Which implies, E∩F=4,6E∩F=4,6 size 12{E intersection F= left lbrace 4,6 right rbrace } {}

Therefore, PF=3/6PF=3/6 size 12{P left (F right )=3/6} {}, and PE∩F=2/6PE∩F=2/6 size 12{P left (E intersection F right )=2/6} {}

PE∣F=PE∩FPF=2/63/6=23PE∣F=PE∩FPF=2/63/6=23 size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } = { {2/6} over {3/6} } = { {2} over {3} } } {}.

Example 27

Problem 1

The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.

Table 2
  Male(M) Female(F) Total
Public Transportation(P) 8 13 21
Drive(D) 39 40 79
Total 47 53 100

The events MM size 12{M} {}, FF size 12{F} {}, PP size 12{P} {}, and DD size 12{D} {} are self explanatory. Find the following probabilities.

  1. PD∣MPD∣M size 12{P left (D \lline M right )} {}
  2. PF∣DPF∣D size 12{P left (F \lline D right )} {}
  3. PM∣PPM∣P size 12{P left (M \lline P right )} {}
Solution

We use the conditional probability formula PE∣F=PE∩FPFPE∣F=PE∩FPF size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}.

  1. PD∣M=PD∩MPM=39/10047/100=3947PD∣M=PD∩MPM=39/10047/100=3947 size 12{P left (D \lline M right )= { {P left (D intersection M right )} over {P left (M right )} } = { {"39"/"100"} over {"47"/"100"} } = { {"39"} over {"47"} } } {}.
  2. PF∣D=PF∩DPD=40/10079/100=4079PF∣D=PF∩DPD=40/10079/100=4079 size 12{P left (F \lline D right )= { {P left (F intersection D right )} over {P left (D right )} } = { {"40"/"100"} over {"79"/"100"} } = { {"40"} over {"79"} } } {}.
  3. PM∣P=PM∩PPP=8/10021/100=821PM∣P=PM∩PPP=8/10021/100=821 size 12{P left (M \lline P right )= { {P left (M intersection P right )} over {P left (P right )} } = { {8/"100"} over {"21"/"100"} } = { {8} over {"21"} } } {}.

Example 28

Problem 1

Given PE=.5PE=.5 size 12{P left (E right )= "." 5} {}, PF=.7PF=.7 size 12{P left (F right )=/7} {}, and PE∩F=.3PE∩F=.3 size 12{P left (E intersection F right )} {}. Find the following.

  1. PE∣FPE∣F size 12{P left (E \lline F right )} {}
  2. PF∣EPF∣E size 12{P left (F \lline E right )} {}.
Solution

We use the conditional probability formula PE∣F=PE∩FPFPE∣F=PE∩FPF size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}.

  1. PE∣F=.3.7=37PE∣F=.3.7=37 size 12{P left (E \lline F right )= { { "." 3} over { "." 7} } = { {3} over {7} } } {}.
  2. PF∣E=.3/.5=3/5PF∣E=.3/.5=3/5 size 12{P left (F \lline E right )= "." 3/ "." 5=3/5} {}.

Example 29

Problem 1

Given two mutually exclusive events EE size 12{E} {} and FF size 12{F} {} such that PE=.4PE=.4 size 12{P left (E right )= "." 4} {}, PF=.9PF=.9 size 12{P left (F right )= "." 9} {}. Find PE∣FPE∣F size 12{P left (E \lline F right )} {}.

Solution

Since EE size 12{E} {} and FF size 12{F} {} are mutually exclusive, PE∩F=0PE∩F=0 size 12{P left (E intersection F right )=0} {}. Therefore,

PE|F=0 .9=0PE|F=0 .9=0 size 12{P left (E intersection F right )=0} {}.

Example 30

Problem 1

Given PF∣E=.5PF∣E=.5 size 12{P left (F \lline E right )= "." 5} {}, and PE∩F=.3PE∩F=.3 size 12{P left (E intersection F right )= "." 3} {}. Find PEPE size 12{P left (E right )} {}.

Solution

Using the conditional probability formula PE∣F=PE∩FPFPE∣F=PE∩FPF size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}, we get

PF∣E=PE∩FPEPF∣E=PE∩FPE size 12{P left (F \lline E right )= { {P left (E intersection F right )} over {P left (E right )} } } {}
(42)

Substituting,

.5=.3PE.5=.3PE size 12{ "." 5= { { "." 3} over {P left (E right )} } } {} or PE=3/5PE=3/5 size 12{P left (E right )=3/5} {}

Example 31

Problem 1

In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.

Solution

Let event EE size 12{E} {} be that the family has two boys and a girl, and let FF size 12{F} {} be the probability that the family has at least two boys. We want PE∣FPE∣F size 12{P left (E \lline F right )} {}.

We list the sample space along with the events EE size 12{E} {} and FF size 12{F} {}.

S=BBB,BBG,BGB,BGG,GBB,GGB,GGGS=BBB,BBG,BGB,BGG,GBB,GGB,GGG size 12{S= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GGB", ital "GGG" right rbrace } {}
(43)

E=BBG,BGB,GBBE=BBG,BGB,GBB size 12{E= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {} and F=BBB,BBG,BGB,GBBF=BBB,BBG,BGB,GBB size 12{F= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "GBB" right rbrace } {}

E∩F=BBG,BGB,GBBE∩F=BBG,BGB,GBB size 12{E intersection F= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {}
(44)

Therefore, PF=4/8PF=4/8 size 12{P left (F right )=4/8} {}, and PE∩F=3/8PE∩F=3/8 size 12{P left (E intersection F right )=3/8} {}.

And

PE∣F−3/84/8=34PE∣F−3/84/8=34 size 12{P left (E \lline F right ) - { {3/8} over {4/8} } = { {3} over {4} } } {}.

Example 32

Problem 1

At a community college 65% of the students use IBM computers, 50% use Macintosh computers, and 20% use both. If a student is chosen at random, find the following probabilities.

  1. The student uses an IBM given that he uses a Macintosh.
  2. The student uses a Macintosh knowing that he uses an IBM.
Solution

Let event II size 12{I} {} be that the student uses an IBM computer, and MM size 12{M} {} the probability that he uses a Macintosh.

  1. PI∣M=.20.50=25PI∣M=.20.50=25 size 12{P left (I \lline M right )= { { "." "20"} over { "." "50"} } = { {2} over {5} } } {}
  2. PM∣I=.20.65=413PM∣I=.20.65=413 size 12{P left (M \lline I right )= { { "." "20"} over { "." "65"} } = { {4} over {"13"} } } {}.

Independent Events

In Section 5, we considered conditional probabilities. In some examples, the probability of an event changed when additional information was provided. For instance, the probability of obtaining a king from a deck of cards, changed from 4/524/52 size 12{4/"52"} {} to 4/124/12 size 12{4/"12"} {}, when we were given the condition that a face card had already shown. This is not always the case. The additional information may or may not alter the probability of the event. For example consider the following example.

Example 33

Problem 1

A card is drawn from a deck. Find the following probabilities.

  1. The card is a king.
  2. The card is a king given that a red card has shown.
Solution

  1. Clearly, PThe card is a king =4/52=1/13PThe card is a king =4/52=1/13 size 12{P left ("The card is a king " right )=4/"52"=1/"13"} {}.

  2. To find PThe card is a king∣ A red card has shownPThe card is a king∣ A red card has shown size 12{P left ("The card is a king" \lline " A red card has shown" right )} {}, we reason as follows:

    Since a red card has shown, there are only twenty six possibilities. Of the 26 red cards, there are two kings. Therefore,

    PThe card is a king ∣ A red card has shown=2/26=1/13PThe card is a king ∣ A red card has shown=2/26=1/13 size 12{P left ("The card is a king " \lline " A red card has shown" right )=2/"26"=1/"13"} {}.

The reader should observe that in the above example,

P The card is a king ∣ A red card has shown = P The card is a king P The card is a king ∣ A red card has shown = P The card is a king size 12{P left ("The card is a king" \lline " A red card has shown" right )=P left ("The card is a king" right )} {}

In other words, the additional information, a red card has shown, did not affect the probability of obtaining a king. Whenever the probability of an event EE size 12{E} {} is not affected by the occurrence of another event FF size 12{F} {}, and vice versa, we say that the two events EE size 12{E} {} and FF size 12{F} {} are independent. This leads to the following definition.

Two Events EE size 12{E} {} and FF size 12{F} {} are independent if and only if at least one of the following two conditions is true.

  1. PE∣F=PEPE∣F=PE size 12{P left (E \lline F right )=P left (E right )} {} or
  2. P F ∣ E = P F P F ∣ E = P F size 12{P left (F \lline E right )=P left (F right )} {}
    (45)

If the events are not independent, then they are dependent.

Next, we need to develop a test to determine whether two events are independent.

We recall the conditional probability formula.

PE∣F=PE∩FPFPE∣F=PE∩FPF size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}
(46)

Multiplying both sides by PFPF size 12{P left (F right )} {}, we get

PE∩F=PE∣FPFPE∩F=PE∣FPF size 12{P left (E intersection F right )=P left (E \lline F right )P left (F right )} {}
(47)

Now if the two events are independent, then by definition

PE∣F=PEPE∣F=PE size 12{P left (E \lline F right )=P left (E right )} {}
(48)

Substituting, PE∩F=PEPFPE∩F=PEPF size 12{P left (E intersection F right )=P left (E right )P left (F right )} {}

We state it formally as follows.

Test for Independence

Two Events EE size 12{E} {} and FF size 12{F} {} are independent if and only if

PE∩F=PEPFPE∩F=PEPF size 12{P left (E intersection F right )=P left (E right )P left (F right )} {}
(49)

Example 34

Problem 1

The table below shows the distribution of color-blind people by gender.

Table 3
  Male(M) Female(F) Total
Color-Blind(C) 6 1 7
Not Color-Blind (N) 46 47 93
Total 52 48 100

Where MM size 12{M} {} represents male, FF size 12{F} {} represents female, CC size 12{C} {} represents color-blind, and NN size 12{N} {} not color-blind. Use the independence test to determine whether the events color-blind and male are independent.

Solution

According to the test, CC size 12{C} {} and MM size 12{M} {} are independent if and only if PC∩M=PCPMPC∩M=PCPM size 12{P left (C intersection M right )=P left (C right )P left (M right )} {}.

PC=7/100,   PM=52/100   and   PC∩M=6/100PC=7/100 size 12{P left (C right )=7/"100"} {},  PM=52/100 size 12{P left (M right )="52"/"100"} {}   and   PC∩M=6/100 size 12{P left (C intersection M right )=6/"100"} {}
(50)
PCPM=7/10052/100=.0364PCPM=7/10052/100=.0364 size 12{P left (C right )P left (M right )= left (7/"100" right ) left ("52"/"100" right )= "." "0364"} {}
(51)

and PC∩M=.06PC∩M=.06 size 12{P left (C intersection M right )= "." "06"} {}

Clearly .0364≠.06.0364≠.06 size 12{ "." "0364" <> "." "06"} {}

Therefore, the two events are not independent. We may say they are dependent.

Example 35

Problem 1

In a survey of 100 women, 45 wore makeup, and 55 did not. Of the 45 who wore makeup, 9 had a low self-image, and of the 55 who did not, 11 had a low self-image. Are the events "wearing makeup" and "having a low self-image" independent?

Solution

Let MM size 12{M} {} be the event that a woman wears makeup, and LL size 12{L} {} the event that a woman has a low self-image. We have

PM∩L=9/100PM∩L=9/100 size 12{P left (M intersection L right )=9/"100"} {}, PM=45/100PM=45/100 size 12{P left (M right )="45"/"100"} {} and PL=20/100PL=20/100 size 12{P left (L right )="20"/"100"} {}

In order for two events to be independent, we must have

PM∩L=PMPLPM∩L=PMPL size 12{P left (M intersection L right )=P left (M right )P left (L right )} {}
(52)

Since 9/100=45/10020/1009/100=45/10020/100 size 12{9/"100"= left ("45"/"100" right ) left ("20"/"100" right )} {}

The two events "wearing makeup" and "having a low self-image" are independent.

Example 36

Problem 1

A coin is tossed three times, and the events EE size 12{E} {}, FF size 12{F} {} and GG size 12{G} {} are defined as follows:

EE size 12{E} {}: The coin shows a head on the first toss.

FF size 12{F} {}: At least two heads appear.

GG size 12{G} {}: Heads appear in two successive tosses.

Determine whether the following events are independent.

  1. EE size 12{E} {} and FF size 12{F} {}
  2. FF size 12{F} {} and GG size 12{G} {}
  3. EE size 12{E} {} and GG size 12{G} {}
Solution

To make things easier, we list the sample space, the events, their intersections and the corresponding probabilities.

S=HHH,HHT,HTH,HTT,THH,THT,TTH,TTTS=HHH,HHT,HTH,HTT,THH,THT,TTH,TTT size 12{S= left lbrace ital "HHH", ital "HHT", ital "HTH", ital "HTT", ital "THH", ital "THT", ital "TTH", ital "TTT" right rbrace } {}
(53)

E=HHH,HHT,HTH,HTTE=HHH,HHT,HTH,HTT size 12{E= left lbrace ital "HHH", ital "HHT", ital "HTH", ital "HTT" right rbrace } {}, PE=4/8PE=4/8 size 12{P left (E right )=4/8} {} or 1/21/2 size 12{1/2} {}

F=HHH,HHT,HTH,THHF=HHH,HHT,HTH,THH size 12{F= left lbrace ital "HHH", ital "HHT", ital "HTH", ital "THH" right rbrace } {}, PF=4/8PF=4/8 size 12{P left (F right )=4/8} {} or 1/21/2 size 12{1/2} {}

G=HHT,THHG=HHT,THH size 12{G= left lbrace ital "HHT", ital "THH" right rbrace } {}, PG=2/8PG=2/8 size 12{P left (G right )=2/8} {} or 1/41/4 size 12{1/4} {}

E∩F=HHH,HHT,HTHE∩F=HHH,HHT,HTH size 12{E intersection F= left lbrace ital "HHH", ital "HHT", ital "HTH" right rbrace } {}, PE∩F=3/8PE∩F=3/8 size 12{P left (E intersection F right )=3/8} {}

E∩G=HHT,THHE∩G=HHT,THH size 12{E intersection G= left lbrace ital "HHT", ital "THH" right rbrace } {}, PF∩G=2/8PF∩G=2/8 size 12{P left (F intersection G right )=2/8} {} or 1/41/4 size 12{1/4} {}

E ∩ G = HHT E ∩ G = HHT size 12{E intersection G= left lbrace ital "HHT" right rbrace } {} P E ∩ G = 1 / 8 P E ∩ G = 1 / 8 size 12{P left (E intersection G right )=1/8} {}

  1. In order for EE size 12{E} {} and FF size 12{F} {} to be independent, we must have

    PE∩F=PEPFPE∩F=PEPF size 12{P left (E intersection F right )=P left (E right )P left (F right )} {}.

    But 3/8≠1/2⋅1/23/8≠1/2⋅1/2 size 12{3/8 <> 1/2 cdot 1/2} {}

    Therefore, EE size 12{E} {} and FF size 12{F} {} are not independent.

  2. FF size 12{F} {} and GG size 12{G} {} will be independent if

    PF∩G=PFPGPF∩G=PFPG size 12{P left (F intersection G right )=P left (F right )P left (G right )} {}.

    Since 1/4≠1/2⋅1/41/4≠1/2⋅1/4 size 12{1/4≠1/2 cdot 1/4} {}

    FF size 12{F} {} and GG size 12{G} {} are not independent.

  3. We look at

    PE∩G=PEPGPE∩G=PEPG size 12{P left (E intersection G right )=P left (E right )P left (G right )} {}
    (54)
    1/8=1/2⋅1/41/8=1/2⋅1/4 size 12{1/8=1/2 cdot 1/4} {}
    (55)

    Therefore, EE size 12{E} {} and GG size 12{G} {} are independent events.

Example 37

Problem 1

The probability that Jaime will visit his aunt in Baltimore this year is .30.30 size 12{ "." "30"} {}, and the probability that he will go river rafting on the Colorado river is .50.50 size 12{ "." "50"} {}. If the two events are independent, what is the probability that Jaime will do both?

Solution

Let AA size 12{A} {} be the event that Jaime will visit his aunt this year, and RR size 12{R} {} be the event that he will go river rafting.

We are given PA=.30PA=.30 size 12{P left (A right )= "." "30"} {} and PR=.50PR=.50 size 12{P left (R right )= "." "50"} {}, and we want to find PA∩RPA∩R size 12{P left (A intersection R right )} {}.

Since we are told that the events AA size 12{A} {} and RR size 12{R} {} are independent,

PA∩R=PAPR=.30.50=.15.PA∩R=PAPR=.30.50=.15 size 12{P left (A intersection R right )=P left (A right )P left (R right )= left ( "." "30" right ) left ( "." "50" right )= "." "15"} {}.
(56)

Example 38

Problem 1

Given PB∣A=.4PB∣A=.4 size 12{P left (B \lline A right )= "." 4} {}. If AA size 12{A} {} and BB size 12{B} {} are independent, find PBPB size 12{P left (B right )} {}.

Solution

If AA size 12{A} {} and BB size 12{B} {} are independent, then by definition PB∣A=PBPB∣A=PB size 12{P left (B \lline A right )=P left (B right )} {}

Therefore, PB=.4PB=.4 size 12{P left (B right )= "." 4} {}

Example 39

Problem 1

Given PA=.7PA=.7 size 12{P left (A right )= "." 7} {}, PB∣A=.5PB∣A=.5 size 12{P left (B \lline A right )= "." 5} {}. Find PA∩BPA∩B size 12{P left (A intersection B right )} {}.

Solution

By definition PB∣A=PA∩BPAPB∣A=PA∩BPA size 12{P left (B \lline A right )= { {P left (A intersection B right )} over {P left (A right )} } } {}

Substituting, we have

.5=PA∩B.7.5=PA∩B.7 size 12{ "." 5= { {P left (A intersection B right )} over { "." 7} } } {}
(57)

Therefore, PA∩B=.35PA∩B=.35 size 12{P left (A intersection B right )= "." "35"} {}

Example 40

Problem 1

Given P(A)=.5P(A)=.5, P(A∪B)=.7P(A∪B)=.7, if AA and BB are independent, find P(B)P(B).

Solution

The addition rule states that

P(A∪B)=P(A)+P(B)–P(A∩B)P(A∪B)=P(A)+P(B)–P(A∩B)
(58)

Since AA and BB are independent, P(A∩B)=P(A)P(B)P(A∩B)=P(A)P(B)

We substitute for P(A∩B)P(A∩B) in the addition formula and get

P(A∪B)=P(A)+P(B)–P(A)P(B)P(A∪B)=P(A)+P(B)–P(A)P(B)
(59)

By letting P(B)=xP(B)=x, and substituting values, we get

.7=.5+x–.5x.7=.5+x–.5x
(60)
.7=.5+.5x.7=.5+.5x
(61)
.2=.5x.2=.5x
(62)
.4=x.4=x
(63)

Therefore, P(B)=.4P(B)=.4.

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