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  • Chapter 13Probability

    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 1/3. 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: Sample Spaces

    A sample space of an experiment is the set of all possible outcomes.

    Example 13.1

    If a die is rolled, write a sample space.

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

    {1,2,3,4,5,6}.

    Example 13.2

    A family has three children. Write a sample space.

    The sample space consists of eight possibilities.

    (13.1)
    _autogen-svg2png-0004.png

    The possibility _autogen-svg2png-0005.png, 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.

    graphics1.png
    Figure 13.0

    Example 13.3

    Two dice are rolled. Write the sample space.

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

    Table 13.1.
     Green
    Red123456
    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 S, and an outcome A of S, the following two properties are satisfied.

    1. If A is an outcome of a sample space, then the probability of A, denoted by P(A), is between 0 and 1, inclusive.

      () 0 ≤ P (A) ≤ 1

    2. The sum of the probabilities of all the outcomes in S equals 1.

    Example 13.4

    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.

    Since two dice are rolled, there are 36 possibilities. The probability of each outcome, listed in Example 13.3, is equally likely.

    Since (3, 6) is one such outcome, the probability of obtaining (3, 6) is 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 13.5

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

    Let 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).

    (13.2)E={(1,6),(2,5),(3,4)(4,3),(5,2), and (6,1)}

    and the probability of the event E,

    P(E)=6/36 or 1/6.

    Example 13.6

    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 r1, r2, r3, w1, w2, w3, w4, b1, b2, b3. Let the event C represent that the marble is red or blue.

    The sample space S={r1,r2,r3,w1,w2,w3,w4,b1,b2,b3}

    And the event C={r1,r2,r3,b1,b2,b3}

    Therefore, the probability of C,

    P(C)=6/10 or 3/5.

    Example 13.7

    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.

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

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

    (13.4)F=[(1,3),(3,1)]

    Therefore, the probability of F is

    P(F)=2/6 or 1/3.

    Example 13.8

    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 Example 13.7, consists of the following six possibilities.

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

    Let the event A represent that the sum of the numbers is at least four. Then

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

    Therefore, the probability of F is

    P(F)=4/6 or 2/3.

    Mutually Exclusive Events and the Addition Rule

    In the Section 11.1, 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 and F, EF, is the set of outcomes that are in E or in F or in both.

    The intersection of two events E and F, EF, is the set of outcomes that are in both E and F.

    The complement of an event E, denoted by Ec, is the set of outcomes in the sample space S that are not in E. It is worth noting that P(EC)=1−P(E). This follows from the fact that if the sample space has n elements and E has k elements, then Ec has nk elements. Therefore,

    _autogen-svg2png-0067.png.

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

    Two events E and F are said to be mutually exclusive if they do not intersect. That is, EF=∅.

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

    Example 13.9

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

    (13.7)E={The card drawn is an Ace}
    (13.8)F={The card drawn is a heart}

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

    EF={Ace of hearts}≠∅.

    Therefore, the events E and F are not mutually exclusive.

    Example 13.10

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

    (13.9)G={The sum of the faces is six}
    (13.10)H={One die shows a four}

    For clarity, we list the elements of both sets.

    (13.11)G={(1,5),(2,4),(3,3),(4,2),(5,1)}
    (13.12)H={(2,4),(4,2)}

    Clearly, GH={(2,4),(4,2)}≠Ø.

    Therefore, the two sets are not mutually exclusive.

    Example 13.11

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

    (13.13)M={The family has at least one boy}
    (13.14)N={The family has all girls}

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

    _autogen-svg2png-0083.png and _autogen-svg2png-0084.png

    Clearly, MN

    Therefore, the events M and N are mutually exclusive.

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

    Example 13.12

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

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

    The sample space S={1,2,3,4,5,6}. The event E={2,4,6}, and the event F={5,6}

    We need to find P(EF).

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

    This can be illustrated by a Venn diagram.

    The sample space S, the events E and F, and EF are listed below.

    S={1,2,3,4,5,6}, E={2,4,6}, F={5,6}, and EF={6}.

    graphics2.png
    Figure 13.0

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

    Finding the probability of EF, 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(EF), we remove the duplicity and get the correct answer. So as a rule,

    (13.15)n(EF)=n(E)+n(F)–n(EF)

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

    (13.16)
    _autogen-svg2png-0131.png

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

    (13.17)P(EF)=P(E)+P(F)–P(EF)

    Applying the above for this example, we get

    (13.18)P(EF)=3/6+2/6–1/6=4/6

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

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

    (13.19)P(EF)=P(E)+P(F)–P(EF)

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

    (13.20)P(EF)=P(E)+P(F)
    Example 13.13

    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 be the event that the card is an ace, and H the event that it is a heart.

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

    We need to find P(AH).

    P(AH)=P(A)+P(H)–P(AH)=4/52+13/52−1/52=16/52.

    Example 13.14

    Two dice are rolled, and the events F and T are as follows:

    F={The sum of the dice is four} and T={At least one die shows a three}

    Find P(FT).

    We list F and T, and FT as follows:

    (13.21)F={(1,3),(2,2),(3,1)}
    (13.22)T={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(1,3),(2,3),(4,3),(5,3),(6,3)}
    (13.23)FT={(1,3),(3,1)}

    Since P(FT)=P(F)+P(T)−P(FT)

    We have P(FT)=3/36+11/36−2/36=12/36.

    Example 13.15

    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?

    Let A be the event that the board approves the position, and S be the event that Mr. Washington gets selected. We have,

    P(A)=.80, P(S)=.70, and P(AS)=.90.

    We need to find, P(AS).

    The addition formula states that,

    (13.24)P(AS)=P(A)+P(S)−P(AS)

    Substituting the known values, we get

    (13.25).90=.80+.70−P(AS)

    Therefore, P(AS)=.60.

    Example 13.16

    The probability that this weekend will be cold is .6, the probability that it will be rainy is .7, and probability that it will be both cold and rainy is .5. What is the probability that it will be neither cold nor rainy?

    Let C be the event that the weekend will be cold, and R be event that it will be rainy. We are given that

    P(C)=.6, P(R)=.7, P(CR)=.5

    (13.26)P(CR)=P(C)+P(R)−P(CR)=.6+.7−.5=.8

    We want to find P((CR)c).

    (13.27)P((CR)c)=1−P(CR)=1−.8=.2

    We summarize this section by listing the important rules.

    Example 13.17

    The Addition Rule

    For Two Events E and F, P(EF)=P(E)+P(F)−P(EF)

    The Addition Rule for Mutually Exclusive Events

    If Two Events E and F are Mutually Exclusive, then P(EF)=P(E)+P(F)

    The Complement Rule

    If Ec is the Complement of Event E, then P(Ec)=1−P(E)

    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 13.18

    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?

    Let E be the event that the first marble drawn is red, and let F be the event that the second marble drawn is red.

    We need to find P(EF).

    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 S consists of 49 ordered pairs. Of the 49 ordered pairs, there are 3×3=9 ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

    (13.28)
    _autogen-svg2png-0195.png

    Further note that in this particular case

    (13.29)P(EF)=P(E)⋅P(F)
    Example 13.19

    If in the Example 13.18, the two marbles are drawn without replacement, then what is the probability that both marbles are red?

    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 P(EF).

    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 S consists of 42 ordered pairs. Of the 42 ordered pairs, there are 3×2=6 ordered pairs that show red on the first draw and red on the second draw. Therefore,

    (13.30)
    _autogen-svg2png-0200.png

    Here 3/7 represents P(E), and 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 P(Red on the second∣red on first) or P(FE). The "|" represents the word "given." Therefore,

    (13.31)P(EF)=P(E)⋅P(FE)

    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 13.20

    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.

    Let R be the event that the marble drawn is red, and let W be the event that the marble drawn is white.

    We draw the following tree diagram.

    graphics3.png
    Figure 13.0

    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 Section 11.1. 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 13.21

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

    1. P(Two red and one white)

    2. P(One of each color)

    3. P(None blue)

    4. P(At least one blue)

    Let us suppose the marbles are labeled as R1, R2, R3, W1, W2, B1, B2, B3.

    1. P(Two red and one white)

      We analyze the problem in the following manner.

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

      _autogen-svg2png-0224.png.

    2. P(One of each color)

      Again, there are 8C3=56 possible combinations. Of these 56 combinations, there are 3C1×2C1×3C1=18 combinations consisting of one red, one white, and one blue. Therefore,

      _autogen-svg2png-0228.png.

    3. P(None blue)

      There are 5 non-blue marbles, therefore

      _autogen-svg2png-0230.png.

    4. P(At least one blue)

      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.

      (13.32)P(At least one blue)=P(one blue, two non-blue)+P(two blue, one non-blue)+P(three blue)
      (13.33)
      _autogen-svg2png-0233.png

      P(At least one blue)=30/56+15/56+1/56=46/56=23/28.

      Alternately,

      we use the fact that P(E)=1−P(Ec).

      If the event E=At least one blue, then Ec=None blue.

      But from part c of this example, we have (Ec)=5/28

      Therefore, P(E)=1−5/28=23/28.

    Example 13.22

    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.

    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

    (13.34)
    _autogen-svg2png-0240.png

    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 13C2 different combinations of pairs.

    (13.35)
    _autogen-svg2png-0242.png

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

    Example 13.23

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

    Let event 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 36525 possible birthdays for 25 people. Therefore, the sample space has 36525 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,

    (13.36)
    _autogen-svg2png-0246.png

    Since P(at least two people have the same birthday)=1−P(No two have the same birthday),

    (13.37)
    _autogen-svg2png-0248.png

    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 4/52 or 1/13. And, probability of not obtaining a king is 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 4/12 or 1/3. So your chance of winning is 1/3 and of losing 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 P(EF) denotes the problem of finding the probability of E given that F has occurred. We read P(EF) as "the probability of E, given F."

    Example 13.24

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

    Let event E be that the family has two boys and a girl, and F the event that the first born is a boy.

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

    (13.38)
    _autogen-svg2png-0264.png

    Since we know that the first born is a boy, our possibilities narrow down to four outcomes, _autogen-svg2png-0265.png, _autogen-svg2png-0266.png, _autogen-svg2png-0267.png, and _autogen-svg2png-0268.png.

    Among the four, _autogen-svg2png-0269.png and _autogen-svg2png-0270.png represent two boys and a girl.

    Therefore, _autogen-svg2png-0271.png or 1/2.

    Let us now develop a formula for the conditional probability P(EF).

    Suppose an experiment consists of n equally likely events. Further suppose that there are m elements in F, and c elements in EF, as shown in the following Venn diagram.

    graphics4.png
    Figure 13.1

    If the event F has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset F. Therefore, we only look at the set F and at nothing outside of F. Since F has m elements, the denominator in the calculation of P(EF) is m. We may think that the numerator for our conditional probability is the number of elements in E. But clearly we cannot consider the elements of E that are not in F. We can only count the elements of E that are in F, that is, the elements in EF. Therefore,

    _autogen-svg2png-0292.png

    Dividing both the numerator and the denominator by n, we get

    (13.39)
    _autogen-svg2png-0294.png

    But c/n=P(EF), and m/n=P(F).

    Substituting, we derive the following formula for P(EF).

    Example 13.25

    For Two Events E and F, the Probability of E Given F is

    (13.40)
    _autogen-svg2png-0302.png
    Example 13.26

    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.

    Let E be the event that an even number shows, and F be the event that a number greater than three shows. We want P(EF).

    E={2,4,6} and F={4,5,6}. Which implies, EF={4,6}

    Therefore, P(F)=3/6, and P(EF)=2/6

    _autogen-svg2png-0311.png.

    Example 13.27

    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 13.2.
     Male(M)Female(F)Total
    Public Transportation(P)81321
    Drive(D)394079
    Total4753100

    The events M, F, P, and D are self explanatory. Find the following probabilities.

    1. P(DM)

    2. P(FD)

    3. P(MP)

    We use the conditional probability formula _autogen-svg2png-0319.png.

    1. _autogen-svg2png-0320.png.

    2. _autogen-svg2png-0321.png.

    3. _autogen-svg2png-0322.png.

    Example 13.28

    Given P(E)=.5, P(F)=.7, and P(EF)=.3. Find the following.

    1. P(EF)

    2. P(FE).

    We use the conditional probability formula _autogen-svg2png-0328.png.

    1. _autogen-svg2png-0329.png.

    2. P(FE)=.3/.5=3/5.

    Example 13.29

    Given two mutually exclusive events E and F such that P(E)=.4, P(F)=.9. Find P(EF).

    Since E and F are mutually exclusive, P(EF)=0. Therefore,

    _autogen-svg2png-0339.png.

    Example 13.30

    Given P(FE)=.5, and P(EF)=.3. Find P(E).

    Using the conditional probability formula _autogen-svg2png-0343.png, we get

    (13.41)
    _autogen-svg2png-0344.png

    Substituting,

    _autogen-svg2png-0345.png or P(E)=3/5

    Example 13.31

    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.

    Let event E be that the family has two boys and a girl, and let F be the probability that the family has at least two boys. We want P(EF).

    We list the sample space along with the events E and F.

    (13.42)
    _autogen-svg2png-0352.png

    _autogen-svg2png-0353.png and _autogen-svg2png-0354.png

    (13.43)
    _autogen-svg2png-0355.png

    Therefore, P(F)=4/8, and P(EF)=3/8.

    And

    _autogen-svg2png-0358.png.

    Example 13.32

    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.

    Let event I be that the student uses an IBM computer, and M the probability that he uses a Macintosh.

    1. _autogen-svg2png-0361.png

    2. _autogen-svg2png-0362.png.

    Independent Events

    In the section called “Conditional Probability”, 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 4/52 to 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 13.33

    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.

    1. Clearly, P(The card is a king )=4/52=1/13.

    2. To find P(The card is a king∣ A red card has shown), 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,

      P(The card is a king ∣ A red card has shown)=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)

    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 E is not affected by the occurrence of another event F, and vice versa, we say that the two events E and F are independent. This leads to the following definition.

    Two Events E and F are independent if and only if at least one of the following two conditions is true.

    1. P(EF)=P(E) or

    2. () P ( FE ) = P (F)


    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.

    (13.44)
    _autogen-svg2png-0377.png

    Multiplying both sides by P(F), we get

    (13.45)P(EF)=P(EF)P(F)

    Now if the two events are independent, then by definition

    (13.46)P(EF)=P(E)

    Substituting, P(EF)=P(E)P(F)

    We state it formally as follows.

    Test for Independence

    Two Events E and F are independent if and only if

    ()P(EF)=P(E)P(F)
    Example 13.34

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

    Table 13.3.
     Male(M)Female(F)Total
    Color-Blind(C)617
    Not Color-Blind (N)464793
    Total5248100

    Where M represents male, F represents female, C represents color-blind, and N not color-blind. Use the independence test to determine whether the events color-blind and male are independent.

    According to the test, C and M are independent if and only if P(CM)=P(C)P(M).

    (13.47)P(C)=7/100,   P(M)=52/100   and   P(CM)=6/100
    (13.48)P(C)P(M)=(7/100)(52/100)=.0364

    and P(CM)=.06

    Clearly .0364≠.06

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

    Example 13.35

    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?

    Let M be the event that a woman wears makeup, and L the event that a woman has a low self-image. We have

    P(ML)=9/100, P(M)=45/100 and P(L)=20/100

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

    (13.49)P(ML)=P(M)P(L)

    Since 9/100=(45/100)(20/100)

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

    Example 13.36

    A coin is tossed three times, and the events E, F and G are defined as follows:

    E: The coin shows a head on the first toss.

    F: At least two heads appear.

    G: Heads appear in two successive tosses.

    Determine whether the following events are independent.

    1. E and F

    2. F and G

    3. E and G

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

    (13.50)
    _autogen-svg2png-0415.png

    _autogen-svg2png-0416.png, P(E)=4/8 or 1/2

    _autogen-svg2png-0419.png, P(F)=4/8 or 1/2

    _autogen-svg2png-0422.png, P(G)=2/8 or 1/4

    _autogen-svg2png-0425.png, P(EF)=3/8

    _autogen-svg2png-0427.png, P(FG)=2/8 or 1/4

    _autogen-svg2png-0430.png P ( EG ) = 1 / 8

    1. In order for E and F to be independent, we must have

      P(EF)=P(E)P(F).

      But 3/8≠1/2⋅1/2

      Therefore, E and F are not independent.

    2. F and G will be independent if

      P(FG)=P(F)P(G).

      Since 1/4≠1/2⋅1/4

      F and G are not independent.

    3. We look at

      (13.51)P(EG)=P(E)P(G)
      (13.52)1/8=1/2⋅1/4

      Therefore, E and G are independent events.

    Example 13.37

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

    Let A be the event that Jaime will visit his aunt this year, and R be the event that he will go river rafting.

    We are given P(A)=.30 and P(R)=.50, and we want to find P(AR).

    Since we are told that the events A and R are independent,

    (13.53)P(AR)=P(A)P(R)=(.30)(.50)=.15.
    Example 13.38

    Given P(BA)=.4. If A and B are independent, find P(B).

    If A and B are independent, then by definition P(BA)=P(B)

    Therefore, P(B)=.4

    Example 13.39

    Given P(A)=.7, P(BA)=.5. Find P(AB).

    By definition _autogen-svg2png-0469.png

    Substituting, we have

    (13.54)
    _autogen-svg2png-0470.png

    Therefore, P(AB)=.35

    Example 13.40

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

    The addition rule states that

    (13.55)P(AB)=P(A)+P(B)–P(AB)

    Since A and B are independent, P(AB)=P(A)P(B)

    We substitute for P(AB) in the addition formula and get

    (13.56)P(AB)=P(A)+P(B)–P(A)P(B)

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

    (13.57).7=.5+x–.5x
    (13.58).7=.5+.5x
    (13.59).2=.5x
    (13.60).4=x

    Therefore, P(B)=.4.

    Glossary

    Definition: Sample Spaces

    A sample space of an experiment is the set of all possible outcomes.

    Solutions