Chapter 5: Probability
5.1 What is probability anyway?
Probability is a branch of mathematics which intrudes on everyday conversation perhaps more than any other (except for just plain arithmetic and counting). A TV weather report might announce that “there is a 70% chance of rain today.” When you hear that statement you automatically infer that there must be a 30% chance that it will not rain, don’t you? When you do that you’re intuitively using one of the basic properties of probability. Other common expressions that implicitly have to do with probability are statements like, “N.C. State is a 2 to 1 favorite to beat Wake Forest today,” or “I think there’s about a 50-50 chance that I’ll pass English this semester.”
Often when we talk about probability in everyday conversation we speak in terms of “percentages” or “odds” or “chances.” When speaking mathematically we usually treat probabilities as fractions (or decimal numbers) rather than percentages, and we usually use the word “probability.” The three statements in the paragraph above are equivalent to these: (1) The probability of rain today is .7. (2) The probability that N.C. State will beat Wake Forest today is 2/3. (3) I think the probability that I will pass English this semester is about 1/2.
Many of the examples which are most useful in developing our understanding of probabilities involve simple experiments. With each experiment there are certain outcomes for the experiment. For instance, if a coin is tossed then the possible outcomes are HEADS and TAILS. If a class holds an election to elect a class president, then the possible outcomes are the individual members of the class, any of whom could be elected president. When talking about probabilities, it’s always a good idea to have a firm understanding of just what the possibilities are. In fact, this idea is so important that there is a special name for the set of all possibilities. It’s called the sample space.
Definition: Sample space
The sample space for an experiment is the set of all possible outcomes.
Example 5.1. A coin is tossed twice. In this case the sample space could be considered to be the set S = {HH,HT,TH,TT}. The expression TH, for example, would
Section 5.1: What Is Probability Anyway? 163 represent the outcome in which a “tails” occurs on the first toss and a “heads” occurs on the second toss.
Example 5.2. A card is drawn from a standard deck of 52 cards. Here the possible outcomes are the 52 cards in the deck, so the sample space is the set
S = {2♣, 2♦,2♥,2♠,3♣,...,K♠,Α♣,A♦,A♥,A♠}.
Don’t get the idea that you have to be an artist, however, in order to represent the elements of the sample space. The important concept is that the sample space consists of the 52 cards. Whatever notation you wish to use to represent the 52 cards is your business.
Example 5.3. Two people are elected from a class of 35 to represent the class on a school committee. What is the sample space for this experiment?
This is a “combinations” question since it is simply a matter of recording which 2 of the 35 are elected. The number of “possibilities” is the number of ways of choosing 2 students from the 35 in the class, and so the number of possibilities is C(35,2). In this example the sample space is obviously too big for us to want to list its elements, or even to have names for all the elements. Nevertheless, it is important to have an understanding for what the sample space is. If we want to verbally describe the sample space we can just say that it is made up of “all the possible pairs of students that could be chosen from the 35 in the class.” From Chapter 4 we know that the number of elements in the sample space is C(35,2) = 595.
In ordinary conversation we sometimes say things like, “Do you think that event will occur?” The words “event” and “occur” used in this sentence are used in probability as mathematical terms, but the way in which they are used is consistent with the way they are used in everyday conversation.
Definition: Event
An event is a subset of the sample space. To say that an event
occurs means that the observed outcome is an element of this subset.
Example 5.4. An ordinary 6-sided die is rolled. The sample space then can be viewed as just the set of numbers S = {1,2,3,4,5,6}. What do we mean when we say something like “an odd number occurs”?
First of all it means we’re talking about the event
A = {1,3,5} = set of all odd numbers in S.
And what we’re saying when we say “an odd number occurs” can be rephrased by saying that “the observed outcome (when the experiment is carried out) belongs to the set A which is the set of odd numbers in the sample space.” You notice that this last way of
saying it is very long and awkward, and furthermore nobody really talks that way. That’s why we like to be able to say just that “an odd number occurs.” But when we talk this way you need to understand conceptually that we mean that the observed outcome belongs to the set A above. If you are ever confused about what a particular event is, you might try writing out the set of elements that make up the event.
Notice also that the event A above is described by listing its elements and also described verbally. In most examples it will be difficult or impossible to list the elements of the sample space, or of events, because there are too many. In these cases we will have to use verbal descriptions of the sample space and of the relevant events in order to convey the necessary information.
Example 5.5. Two dice are rolled, one red and one green. (See Example 4.23) The event
E = {(6,3),(5,4),(4,5),(3,6)}
can easily be described as “the event consisting of all outcomes for which the sum of the numbers appearing on the dice is equal to nine.” Or more briefly we might just say “the event that the sum is nine.”
Example 5.6. Let’s suppose that a class of 35 students has 19 males and 16 females. If two students are elected to represent the class on a committee, then there are C(35,2) different ways of choosing which two students will be on the committee. This means we need to think of a sample space of C(35,2) possible outcomes for the election.
Now what does it mean to talk about the “event that two women are elected”? This event consists of all possible outcomes in which both people elected come from the 16 women in the class. There are C(16,2) ways that two women can be chosen. So when we talk about “the event that two women are elected,” we are talking about an event consisting of C(16,2) elements that is a subset of the sample space of C(35,2) elements.
Sometimes a sample space will need to be based on some kind of existing data. The next example illustrates this.
Example 5.7. A television manufacturer has kept data on all televisions it has sold during the past decade to determine how many complete years of service the TV gave before it had to be brought in for repairs. Here is their data:
Years % of televisions
0 12%
1 26%
2 33%
3 14%
4 9%
5 or more 6%
Notice that the very form in which the data is tabulated forces us to view the sample space in a particular way. For the data as presented, the natural sample space is
Section 5.1: What Is Probability Anyway? 165
S = {0, 1, 2, 3, 4, 5 or more}.
The sample space shows six possibilities because that’s how the information was grouped by the company. If they had kept monthly rather than yearly data, then a much more detailed sample space with many more possible outcomes would be possible.
Example 5.8. Here is an example to show you that sometimes you have to make a choice in deciding what to consider your sample space to be.
In a game like monopoly, you roll a pair of dice but what you’re interested in is not the individual numbers that appear on the two dice but the sum of the two numbers that appear. Should you think of the sample space as all 36 ordered pairs that can be formed from the numbers, 1,...,6, or should you think of the sample space as all possible sums that could appear, in which case we would just think of S as being
S = {2,3,4,5,6,7,8,9,10,11,12}?
You should understand that the question here is really just the question “how much information should I record when I roll the dice?” The two sample spaces are not in conflict, they just record different amounts of information. For example the outcome “4” in the latter choice for sample space corresponds to all the outcomes (1,3), (2,2) and (3,1) if the numbers on the individual dice are recorded. The best rule is to err on the side of caution and record all the information that you think you might need. It’s better to record too much information than too little. For example, if you chose the sample space that just shows the sum, you would be unable to talk about the event “at least one die shows a five” because your sample space doesn’t catalog that kind of information. In the sample space of 36 ordered pairs, the event “at least one die shows a five” consists of 11 of the 36 pairs. (Count them to convince yourself that this is correct.)
Example 5.9. Sue and Ann are playing a tennis match. The winner will be the first player to win two sets. What could we use for the sample space in this situation?
Solution: This is the kind of problem for which we used a tree diagram in Chapter 4. A tree diagram for the tennis match looks something like the following one:
Ann Ann
Sue Sue
Sue Ann
Sue Ann Sue
Ann
tree diagram. So if we like we can think of the sample space as consisting of the six outcomes shown in this tree.
Example 5.10. A company is going to build three stores, and there are 10 locations available. What is the sample space corresponding to the possibilities for the sites they select?
Solution: Since they will be choosing 3 of the 10 sites, there are C(10,3) different possibilities for the sites that will be selected. This is obviously a sample space too large for us to list all the elements, but from Chapter 4 we know that there are C(10,3) = 120 possible choices, i.e. 120 “possible outcomes” in the sample space.
Problems
1. The grades that may be obtained in a class are A, B, C, D, and F.
(a) If a student takes this class, describe the sample space that would represent the possible grades the student might get in the class.
(b) Describe the event “the student makes a passing grade.”
(c) Describe the event “the student does not make an A in the class.”
2. A game show contestant who loses has his choice of 2 out of 5 consolation prizes: a microwave oven, a food processor, a tape recorder, a camera, and a blender.
(a) List the 10 elements that comprise the sample space of all possible choices the contestant might make.
(b) List the elements of the event “the contestant chooses the camera.”
(c) List the elements of the event “the contestant chooses the camera but not the blender.”
(d) List the elements of the event “the contestant chooses either the food processor or the tape recorder.” (Remember that “or” allows for the possibility of both.) 3. A TV station news manager is trying to decide whether to send a reporter to cover a
local chemical spill. If she does decide to send someone, it will be Ed, Ann, or Sue. What might a sample space for this situation look like?
4. A club is going to choose its president and vice-president from among these people: Ralph, Mark, Ruth, and Abraham. Give a sample space for this process.
5. A coin is to be tossed 3 times. State the sample space. How many different elements are there for this sample space? How many different events are there? 6. A pair of dice is rolled, one red and one green. Using the usual sample space of 36
possible outcomes, list the elements of each of these events: (a) the event that the sum of numbers on the two dice is 8.
Section 5.1: What Is Probability Anyway? 167 (c) the event that the two dice show the same number.
(d) the event that the number on the red die is greater than the number on the green one.
7. Carl and Hubert are going to play a tennis match. To win a match in men’s tennis a player must win 3 sets. Draw a tree diagram to represent all the possibilities for the match (see Example 5.9). How many different possible outcomes does your tree show for the tennis match?
8. Three days after Halloween, Andy’s bag of candy contains 2 Milky Ways, 1 Snickers, and 1 Almond Joy. He pulls out candy bars one at a time without replacement until he has the Snickers.
(a) Draw a tree diagram representing this situation. (b) Write out the elements of the sample space.
(c) List the elements of the event E = set of all outcomes in which no more than 1 Milky Way is selected.
9. A company is monitoring use of its parking lot. They have collected data on the number of cars that enter the lot during a one-hour time interval by observing the parking lot during 50 different one-hour time intervals. The data they have accumulated is shown below.
n = number of cars observed frequency
0 ≤n < 100 5
100 ≤n < 200 7
200 ≤n < 300 14
300 ≤n < 400 12
400 ≤n < 500 7
n≥ 500 5
(a) Give an appropriate sample space for this data.
(b) Suppose the company wants to “compress” the data into 3 cases: 0 ≤n < 200, or 200 ≤n < 400, or n≥ 400. What sample space would be chosen then?
5.2 Properties of Probability
The properties of probabilities are based on a few simple ideas: (1) Each element of the sample space is assigned a probability which is a non-negative number. (2) The sum of all these probabilities must be equal to one. (3) If we want to know the probability of an event, we compute it by simply adding up the probabilities of the outcomes that make up the event.
Example 5.11. An ordinary 6-sided die is rolled, so that the sample space is S = {1,2,3,4,5,6}.
Generally dice are made so that the various numbers have about the same probability of occurring. If all the probabilities are going to be equal, and if they are going to add up to one, then each outcome must be assigned probability 1/6. (A die that has this property is called a fair die.)
The event
E = {2,4,6} = set of even outcomes
is an event in this sample space. The probability of the event E is denoted by P(E) and its probability is computed by adding the probabilities of the outcomes that make up E:
P(E) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 1/2.
Example 5.12: Suppose on a winter day there are three possibilities for the state of the weather: rain, snow, and fair. We can think then of a sample space consisting of these three possibilities.
S = {rain, snow, fair}
Let’s suppose that there is a 30% chance of rain, a 20% chance of snow, and a 50% chance that the weather will be fair. This means that we will assign probabilities to the possible outcomes in S as follows: P(rain) = .3, P(snow) = .2, and P(fair) = .5. In this context the event
E = event that precipitation occurs is simply the event
E = {rain, snow}. So the probability of E is
P(E) = P(rain) + P(snow) = .3 + .2 = .5.
Sometimes students wonder things like, “Where do probabilities come from?” That’s a good question, and the answer is not always the same.
Let’s think about one of the simplest of all situations: tossing a coin and observing whether it comes up heads or tails. So our sample space is clearly something like S = {H,T}. But where do we get the probabilities?
Most people’s instinctive reaction is to say P(H) = 1/2 and P(T) = 1/2. One reason might be the symmetry of the coin. Assuming that the coin isn’t bent and that the two faces of the coin are basically flat, it seems reasonable to think that the coin shouldn’t have any more preference for coming up tails than for coming up heads, and vice versa. For that reason we would assign probability 1/2 to each outcome, since this is the only way that the probabilities can be assigned so that heads and tails are equally probable. Similarly, if we are rolling an ordinary six-sided die, we might well be inclined to assign probability 1/6 to each outcome because of the symmetry of the die.
But there’s another intuitive idea that’s important too. It is that the probability of an event represents the relative frequency with which an event occurs. For example, in tossing an unbiased coin where the probabilities of heads and tails are 1/2, the fraction of tosses that result in heads should be close to 1/2 if we toss the coin a large number of times. Similarly, if we say that the probability of getting a “5” when we roll a die is 1/6, we mean that in a large number of rolls the relative frequency with which a “5” should
Section 5.2: Properties of Probability 169 occur is about 1/6 of the time. And if we say that the probability that Sue beats Ann in tennis is 2/3, we mean that in the long run Sue should win about 2/3 of the time when they play.
So you can think of probability as a theoretical relative frequency. This of course doesn’t mean that if you toss a coin 10 times you will get exactly 5 heads. Nor does it mean that in 100 tosses you will get 50 heads. In fact it doesn’t even say that the nickel that might turn up in your pocket tomorrow really will have exactly a .5 probability of coming up heads. When we talk about a “fair” coin with exactly the same probabilities for heads and tails, we are talking about a mathematical model for a coin. Whether the coin in your pocket actually has that property or not isn’t even a question for mathematics. By tossing the coin 100 times and seeing if the number of heads is near 50, for example, you might gather some evidence as to whether the coin is “fair” or not. Testing such as this, and the significance of such tests, is studied in the branch of mathematics known as statistics. But there is no test that will prove once and for all that the coin in your pocket is a perfectly fair coin.
This should not be a great concern for you though. How would you judge the value of a mathematical model for an airplane wing? The question isn’t whether the model is theoretically perfect, it’s whether the airplane will fly. In the same way we get lots of useful information about the way the real world behaves by examining simple mathematical models for things like coins, dice, or tennis matches without having to know that our models offer an absolutely perfect description of the coin in your pocket, the dice you bought at K-Mart, or the tennis players who live down the street.
Here are some properties that will be useful in all our discussions of probability. These properties are all logical consequences of the fact that the probability of an event is simply the sum of the probabilities of the individual elements that make up the event.
Basic Properties of Probabilities
1. For any event E, P(E) ≥ 0. The empty set Ø is assigned probability zero, i.e. P(Ø) = 0.
2. The probability of the entire sample space S is equal to 1,i.e.
P(S) = 1.
3. If A and B are disjoint events then P(A ∪ B) = P(A) + P(B). (Remember that A and B are disjoint if A ∩ B = Ø. Another way of saying A and B are disjoint is to say that they are mutually exclusive.)
4. For any events A and B, P(A∪B) = P(A) + P(B) – P(A∩B).
5. For any event A, P(Ac) = 1 – P(A).
event by adding up the probabilities of the individual elements that make up the event. Since these probabilities are non-negative numbers, their sum is non-negative. Assigning probability zero to the empty set is logically consistent with this.
Property (2) is just a restatement of the fact that the probabilities of all outcomes have to add up to one.
Property (3) is also a consequence of the fact that the probability of an event is the sum of the probabilities of the outcomes that make up the event. The reason is that A ∪ B includes all the elements of A and of B. So all their probabilities get included when we compute P(A ∪ B). Furthermore, since A ∩ B = Ø, none of the probabilities gets included more than once when we add P(A) + P(B).
Property (4) includes an extra term on the right side of the equation that is not included in the equation of Property (3). The extra term P(A ∩ B) is needed when A and B are not disjoint. The following simple example illustrates why this extra term is needed in this case.
Example 5.13. When a fair die is rolled, the sample space is S = {1,2,3,4,5,6}. Think about the two events A = {1,2,3} and B = {3,4}. In this case A ∪ B = {1,2,3,4}, and A ∩ B = {3}.
The die is fair, so each outcome has probability 1/6. This means that P(A ∪ B) = 46 = P(A) + P(B) – P(A ∩ B) = 36 + 26 – 16 .
A glance at this equation shows that the reason that it is necessary to subtract off the term appearing as P(A ∩ B) on the right side is because otherwise the probability of the outcome “3” in A ∩ B would be included twice on the right side of the equation and only once on the left side. By including the term “– P(A ∩ B)” on the right side of the equation in Property (4), the proper “adjusting” is performed to compensate for the elements in A ∩ B.
Notice that Property (5) is really just a special instance of Properties (2) and (3). This is because
A ∪ Ac = S. Furthermore, since A and Ac are disjoint,
P(A) + P(Ac) = P(A ∪ Ac) = P(S) = 1, so P(Ac) = 1 – P(A).
Example 5.14. John thinks that there is a 90% chance that he will pass English, and only a 20% chance that he will make worse than a C. What is the probability that he will make a D?
Solution: Let’s try to describe all the events of interest in terms of the two events F1 = event that John passes English, and
F2 = event that John makes worse than a C. What are the things we know to start with? First we know that
Section 5.2: Properties of Probability 171
Next convince yourself that F1 ∪ F2 includes all possible grades. The reason is that F1 = {A , B , C , D} and F2 = {D , F}.
Since F1 ∪ F2 is the whole sample space, P(F1 ∪ F2) = 1.
Now notice that F1 ∩ F2 = {D}. If we go back to Property (4) and make use of P(F1 ∪ F2) = P(F1) + P(F2) – P(F1 ∩ F2), this tells us
1 = .9 + .2 – P(F1 ∩ F2). So P(F1 ∩ F2) = .1
Example 5.15. A pair of dice is rolled. What is the probability that the sum of the numbers appearing on the two dice is different from 7?
Solution. Of the 36 equally probable outcomes, the event that the sum is equal to 7 is the event
E = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}.
So P(E) = 6/36 = 1/6. Since the event that the sum is different from 7 is simply Ec, we can obtain its probability by a simple subtraction from 1 [using Property (5)]:
P(Ec) = 1 – P(E) = 1 – 16 = 56 .
There are many situations in which all possible outcomes should be treated as equally probable. Often we deliberately devise situations so that outcomes will be equally probable. For example, the whole purpose in shuffling a deck of cards before playing a card game is to mix up the deck and destroy any “pattern” that might be present in the arrangement of cards in the deck. It is natural to assume that a card dealt from a well-shuffled deck would have equal probability of being any of the 52 cards in the deck. Similarly, when an opinion poll is conducted an attempt is made to “randomize” the people questioned. For instance to determine the opinion of students in a school you might question 30 students chosen “at random.” This means that you would like one group of 30 students to be just as likely to be questioned as any other group of thirty students. If there happened to be 400 students in the school, there are C(400,30) different possible ways to choose 30 students. For “random sampling,” this means that each group of thirty should be just as likely to be chosen as any other group of 30, so since there are C(400,30) possibilities, each would be assigned a probability
1 C(400,30) .
The terminology that is used to describe a situation in which all outcomes are assumed equally probable is to describe the sample space as a uniform sample space.
Definition: Uniform sample space
A uniform sample space is one in which all outcomes are assigned the same probability. So if there are n elements in the sample space, this means that each element is assigned probability 1/n.
As a reminder, however, that real data often doesn’t lead to a uniform sample space, let’s return to a situation that illustrates this fact.
Example 5.16. The distribution of grades in a class is as follows Grade Number of students
A 4
B 9
C 17
D 6
F 3
If a student is randomly picked from the class what is the probability that the student is a C student? What is the probability that the student has a grade higher than C? Solution: Since there are 39 students in the class and 4 have A’s, the probability assigned to the outcome “A” should be 4/39. The reason is that if a student is randomly chosen from the class, there are 4 chances in 39 that the student is an A student, and this converts to a probability of 4/39.
In the same way P(C student) = 17/39. And P(grade higher than C) = P(A student) + P(B student) = 4/39 + 9/39 = 13/39.
The assignment of probabilities to the various outcomes such as A, B, C, D, and F in the above example is frequently referred to as specification of a probability distribution. The point we have just been observing is that depending on the situation the correct probability distribution may be quite different from a uniform probability distribution in which all outcomes are assigned the same probability.
Depending upon the sample space that is used, a uniform probability distribution may or may not be correct. For instance, if two coins are tossed we can use the sample space given by S = {HH, HT, TH, TT} and assign probability 1/4 to each outcome. On the other hand, if it is only the number of heads we are interested in, we might be tempted to use the sample space S = {two heads, one head, zero heads}. If we want this latter viewpoint to be consistent with the former, we would have to assign probability 1/2 to the outcome “one head” and probability 1/4 to each of the other two outcomes. So in this case we would not be dealing with a uniform sample space. Often for ease of computation it is easiest to choose a uniform sample. This will be illustrated in the next
Section 5.2: Properties of Probability 173 section.
Probabilities and Venn diagrams
Sometimes Venn diagrams are useful in determining probabilities. Suppose for example that A and B are events with P(A ∩ B) = .2, P(A ∩ Bc) = .3, and P(B) = .4. If you need to find the probabilities of some other events that are described in terms of A and B, a good procedure is to place the probabilities into a Venn diagram in much the same manner we used with counting problems in Chapter 4. You can immediately place the information P(A ∩ B) = .2 and P(A ∩ Bc) = .3 in the Venn Diagram. Then knowing
that P(B) = .4 you can fill in the probability of all the regions and end up with a picture looking like this one.
A B
.2
.3 .2
.3
Now if you need to know the probability of an event such as A ∪ Bc, for example,
you can get it by finding the appropriate region in the Venn diagram. Thus P(A ∪ Bc) =
.8.
Problems
1. Suppose A and B are mutually exclusive events with P(A) = 1/3 and P(B) = 1/6. (a) Find P(A ∪ B)
(b) Find P(A ∩ B)
2. 50% of the students in a school weigh more than 140 pounds, but 70% weigh less than 170. If a student is randomly selected, what is the probability the student will weigh between 140 and 170?
3. Suppose P(E) = .4, P(F) = .3, and P(E ∩ F) = .2. Find (a) P(E ∩ Fc) (c) P(Ec ∪ Fc)
(b) P(E ∪ F) (d) P(Ec ∪ F)
4. A pair of dice is rolled, one red and one green. Suppose
E = event consisting of all outcomes in which the red die shows the number 4 F = event consisting of all outcomes in which the sum of the numbers on the
red and green dice is equal to 8
(a) Find P(E) (b) Find P(E ∩F) (c) Find P(E ∪ F)
probability that his parents let him go to the movie is .5, and the probability that they let him go to the concert is .7. However, he suspects that the probability that they let him go to both is only .3. What is the probability that he gets to go nowhere at all? 6. P(A ∪ B) = .8, P(Ac ∩ B) = .1, and P(A ∩ Bc) = .2.
(a) Find P(B) (b) Find P(A ∩ B) (c) Find P(Ac ∩ Bc)
7. The probability that Ann will go to the beach this summer is .6 and the probability that she will go to the mountains is .5. The probability she does both is .25. What is the probability that
(a) she goes to the beach and not the mountains? (b) she doesn’t go to either place?
8. Suppose S = {s1, s2, s3, s4, s5} is a sample space, and suppose the probability distribution for S is given by
P(s1) = 1/3, P(s2) = 1/6, P(s3) = 1/4, P(s4) = 1/12, and P(s5) = 1/6. If A = {s1, s2, s3} and B = {s1, s3, s5}, find
(a) P(A) (b) P(A ∩ B) (c) P(Ac ∪ Bc) 9. Answer TRUE or FALSE.
(a) P(A ∩ Bc) = P(A) – P(A ∩ B). (b) If A!B then P(A ∩ B) = P(A). (c) P(A ∪ B) ≥P(A).
(d) If A and B are mutually exclusive then P(A ∩ Bc) = P(A).
10. An experiment is conducted in which the possible outcomes are the counting numbers from 1 to 10. Assign probability 1/10 to each possible outcome. In this sample space let A = {1,2,3}, B = {5,7,8}, C = {2,3,7,8}, and D = {1,2,3,4,7,8,9}. (a) Does P(A ∩ Cc) = P(A) – P(A ∩ C)? [See 9(a)]
(b) Does P(A ∩ D) = P(A)? [see 9(b)] (c) Is P(A ∪ C) ≥P(A)? [see 9(c)] (d) Does P(A ∩ Bc) = P(A)? [see 9(d)]
11. To say that the odds in favor of an event are m to n means that the probability of the event is m/(m+n). For instance if the odds are 3 to 4 it means that the probability of the event is 3/7. As another illustration, if a die is rolled the odds of getting a “4” are 1 to 5, since there is 1 way to get a “4” and 5 ways not to get a “4”. The probability of getting a “4” is 1/6. Give the odds of each of the following:
(a) drawing a heart from a deck of cards. (b) getting a “heads” when a coin is tossed. (c) not getting a “4” when a die is rolled.
12. To say that the odds against an event are m to n means that the probability is m/(m+n) that the event will not occur. Find the odds against each of the following
Section 5.2: Properties of Probability 175 things happening:
(a) getting a “4” when a die is rolled.
(b) drawing a heart when a card is drawn from a deck.
(c) Syracuse beating Stanford if Stanford is a 2 to 1 favorite. [Hint: to say that Stanford is a 2 to 1 favorite means that the odds of Stanford beating Syracuse are 2 to 1.]
(d) Duke beating N.C. State if the probability is 2/3 that N.C. State will win.
5.3 Uniform Probability Distributions
We have already seen numerous examples in which the probabilities of the various outcomes are equal. We describe the situation in which all outcomes have the same probability as a uniform probability distribution.
Not only is this situation very common, there are some special ideas that apply to uniform probability distributions that don’t apply to other situations.
Example 5.17. A president and vice-president are to be elected in a club which has 14 members including 6 men and 8 women. Assuming that all members are equally likely to be elected, what is the probability that both officers will be women?
Solution: The first observation here is that the sample space is made up of all possible choices of 2 people from 14. There are C(14,2) different ways of specifying which two people are selected as officers. (For the moment we will ignore any distinction between the two offices, i.e. we will consider only which two people are selected and not worry about which one is president and which is vice-president.)
Since C(14,2) = 91, we are visualizing a sample space of 91 possible outcomes. Furthermore, we are going to treat all of these 91 outcomes as equally likely. This means that we assign probability 1/91 to each one.
Now the event we are interested in is the event that both officers are women: E = event that two women are elected
Since there are 8 women, this can happen in C(8,2) = 28 different ways, and so E consists of 28 of the 91 possible outcomes. But remember that each outcome has probability 1/91. So the probability of E is just the sum of the probabilities of these 28 outcomes, or
P(E) = 28 × 91 = 1 13 4
Alternate Solution: Suppose we decide that we’d like to keep track of who is elected president and who is elected vice-president. If we consider the offices one at a time (with president being considered first), there are 14 choices for president, and then after the president is selected that leaves 13 possibilities for whoever is elected vice-president. If you think about it you should realize that this is simply a matter of selecting 2 people from 14 keeping track of the order of selection. So the number of ways of doing
this is P(14,2) = 14 × 13 = 182. Conclusion: We now have a sample space of 182 equally likely possible outcomes, which means we assign probability 1/182 to each.
Still the event we are interested in is
E = event that two women are elected
Now, however, since we are considering the offices separately this means that there would be 8 choices for which woman is elected president and then 7 choices for which woman is elected vice-president. This gives a total of P(8,2) = 8 × 7 = 56 ways in which two women could be elected. Since each of these 56 possibilities now has probability 1/182,
P(E) = 56 × 182 = 1 182 = 56 13 4
Notice that the answers for these two solutions agree. What is this telling us? It points out that in this problem (as will be the case in many others) we have some freedom of choice about what kind of information to keep track of and what information to ignore. Here it simply doesn’t matter whether we keep track of who fills what office. This is equivalent to saying that we can either take into consideration or ignore the order in which the two people are selected. In other words, we have a choice as to whether we want to consider the sample space as “combinations” or “permutations.”
This “freedom of choice” may be bothersome to you at first, because some students would rather be told exactly what they have to do. There’s not a simple rule to follow here, but a general guideline might be to keep track of order only when necessary. Following that guideline here means choosing the first solution over the second. But both are correct.
What is important about both solutions, however, is to realize that the solution really reduces to two counting problems. In both instances
P(E) = number of elements of the sample space = number of elements of E nn((ES) . )
Uniform probability distributions
P(E) = number of elements of the sample space = number of elements of E nn((ES)) .
Keep in mind however, that this is true only when all elements of the sample space are equally probable, i.e. when we have a uniform probability distribution.
Example 5.18. Each student at a school is assigned an ID number which is a 5-digit number with the first digit not equal to zero. What is the probability that John’s ID will consist of 5 different digits?
Section 5.3: Uniform Probability Distributions 177 Solution: How many possible ID numbers are there? The first digit must be one of the digits 1,2,3,4,5,6,7,8,9, and all the others may be any of these digits or a 0. So there are 9 possibilities for the first digit and 10 for each of the other four digits. The multipli-cation principle says then that there are 9 × 104 = 90,000 possible ID numbers. We will
treat these possibilities as equally likely, i.e. we will assume a uniform probability distribution.
If E denotes the event
E = event that John’s ID has no repeated digits,
then there are 9 possibilities for the first digit in John’s ID. Then there are 9 possibilities for the second digit (any of the other 8 non-zero digits plus 0), 8 possibilities for the third (any of the 8 not used as the first two digits), 7 for the fourth, and 6 for the fifth. The multiplication principle therefore gives n(E) = 9 × 9 × 8 × 7 × 6 = 27,216.
So P(E) = nn((ES) = ) 90,000 27,216 ≈ .302.
In other words, there is about a 30% chance that John’s ID will consist of 5 different digits.
Example 5.19. Two cards are dealt from a well-shuffled standard deck of 52 cards. What is the probability that both cards are clubs?
Solution: There are C(52,2) = 1326 ways to deal two cards from the deck. There are C(13,2) = 78 ways to deal two clubs from the deck. Therefore the probability of getting two clubs is
C(13,2)
C(52,2) = 1326 = 78 17 . 1
It would be a worthwhile exercise for you to redo this example keeping track of the order in which the cards are dealt. In other words, treat the possible outcomes as permutations rather than combinations.
Example 5.20. Charlie has 4 pairs of white socks, 2 pairs of black socks, and 3 pairs of blue socks. They are all mixed up in a drawer in his dresser. One morning he gets out of bed in the dark and reaches into the drawer and pulls out two socks. What is the probability that the socks will be of the same color?
Solution: Altogether Charlie has 9 pairs of socks, for a total of 18 socks. So there are C(18,2) = 153 possible ways to choose two socks from this jumble of socks.
How many matching pairs are there, i.e. how many pairs of the same color? Since he has 8 white socks, that makes C(8,2) = 28 ways to pick a pair of white socks. Similarly there are C(4,2) = 6 ways to pick a pair of black socks and C(6,2) = 15 ways to pick a pair of blue socks.
Altogether this gives 28 + 6 + 15 = 49 ways to pick a pair of matching socks. This is 49 ways out of the total of 153 possible ways, giving a probability
P(socks match) = C(8,2) + CC(18,2)(4,2) + C(6,2) = 153 . 49
Make sure you understand why the combinations terms in the numerator of this fraction are added rather than multiplied.
Example 5.21. Eric and Edith have applied for jobs at Sadlacks. There are 6 job openings available and 25 people have applied.
(a) What is the probability that Edith gets a job?
(b) What is the probability that both of them get jobs? (Assume that the 6 people to get jobs are to be randomly chosen from among the 25 applicants.)
Solution: (a) Since 6 of the 25 applicants will get jobs, there are C(25,6) = 177100 ways that the applicants can be selected. The other counting problem that we have to solve is this: In how many of these cases will Edith be one of the applicants chosen? If Edith is chosen, this means that there are 5 remaining jobs to be distributed among the other 24 applicants. This can be done in C(24,5) = 42504 ways. So the probability that Edith gets a job is
P(Edith gets a job) = CC(24,5)(25,6) = 177100 = 42504 25 = .24. 6
A moment’s thought should convince you that this answer is intuitively correct. Since 6 out of 25 applicants (or 24% of the applicants) get jobs, Edith’s chances are 6 out of 25 = .24.
(b) Now what about the probability that they both get jobs? If both get jobs, this leaves only 4 other jobs to be distributed among the other 23 applicants. And this can be done in C(23,4) = 8855 ways. Therefore the probability that they both get jobs is
P(both Eric and Edith get jobs) = CC(23,4)(25,6) = 25 ∞ 24 = .05. 6 ∞ 5
Example 5.22. Joyce is a junior in a club with 18 juniors and 14 seniors, and Pat is a senior in the club.
(a) If 10 students are to be randomly chosen from the club to go on a trip, what is the probability that Joyce and Pat will both be chosen?
(b) If it is decided that 5 juniors and 5 seniors are to go, what is the probability that Joyce and Pat will both be chosen?
Solution: The first question is exactly like the previous example.
P(Joyce and Pat are both chosen) = CC(32,10) = (30,8) 32 ∞ 31 = 10 ∞ 9 496 45 ≈ .0907.
In the second question, we need to consider the juniors and seniors separately. There are C(18,5) ways to choose the juniors that go on the trip and C(14,5) ways to choose the seniors. The multiplication principle therefore gives a total of C(18,5) × C(14,5) ways to choose all the trip participants. If Joyce is to be one of the juniors, this leaves C(17,4) ways to choose the other 4 juniors. And if Pat is to be one of the seniors
Section 5.3: Uniform Probability Distributions 179 chosen, then there are C(13,4) ways to choose the other seniors. So the number of ways to fill out the list in such a way that Pat and Joyce are both included is C(17,4) × C(13,4). This means that the probability that Joyce and Pat get to go is
C(17,4) ∞ C(13,4) C(18,5) ∞ C(14,5) =
5 ∞ 5
18 ∞ 14 ≈ .099.
Example 5.23. Pud is playing 5-card draw (a poker game). When he picks up the hand that is dealt to him, what is the probability that it will be a “full house”?
Solution: In this game each player is dealt 5 cards from the standard 52-card deck. A “full house” is a hand of 5 cards in which 3 of the cards are of one rank and the remaining 2 are of some other rank. For example, 3 sevens and 2 fives would be a full house.
Since the 5 cards are dealt from a standard 52-card deck, there are C(52,5) different hands possible.
Now how many ways are there to get a full house? This question has already been answered in Example 4.30. We know we must have 3 cards from one of the 13 ranks (like “sevens,” for example). And then the pair must be from one of the remaining 12 ranks (such as “fives”). So we can analyze the process like this:
C(13,1) = 13 ways to choose the rank that the 3-of-a-kind come from C(12,1) = 12 ways to choose the rank that the pair comes from C(4,3) = 4 ways to choose the 3-of-a-kind from the designated rank C(4,2) = 6 ways to choose the pair from the designated rank
So from the multiplication principle this gives 13 × 12 × 4 × 6 = 3744 ways to get a full house. The probability of a full house is therefore
C(13,1) ∞ C(12,1)∞ C(4,3) ∞ C(4,2)
C(52,5) =
3744
2598960 ≈ .00144.
Problems
1. If 3 people are randomly lined up in a row, what is the probability that the shortest is first and the tallest is last in the line?
2. A license plate is made up of 3 letters followed by three digits, for example ZTP-284. If they are made and given out randomly, what is the probability that your license plate will be ABC-123?
3. In a group of 17 girls and 15 boys, what is the probability that the smartest member of the group will be a girl?
4. Mary wrote postcards to 4 friends in California and 6 friends in South Carolina. 4 of the 10 cards fell out of her purse and were lost.
(a) What is the probability that all of the cards that were lost were cards to South Carolina?
(b) What is the probability that exactly 3 of the lost cards were addressed to California?
5. A company has 7 stores in Raleigh and 4 in Durham. If 3 of the stores are randomly selected as a site for a special display, what is the probability that all three selected will be in Raleigh?
6. Ed, Carol, Ann, and Pinky are going to pair up randomly to play tennis. What is the probability that Ann and Pinky will be partners?
7. If five friends (with different names) randomly seat themselves in a row of five theater seats, what is the probability that they will be seated alphabetically by first name?
8. Sue bought 3 raffle tickets. 100 tickets were sold.
(a) What is the probability she holds the winning ticket?
(b) If there are 2 prizes and 2 winning tickets, what is the probability she wins at least one of the prizes?
9. John has 4 red greeting cards, 2 blue ones, and 3 green ones. If he randomly picks 2 of the cards, what is the probability that they will be the same color?
10. A school teacher is going to assign four children to safety patrol duty on four different streets. The children are Eddie, Bob, Mike, and Alice. The streets are Elm, Park, Van Dyke, and Oakwood.
(a) What is the probability that Mike will be assigned duty at Elm Street? (b) What is the probability that Mike gets Elm Street and Alice gets Oakwood? (c) What is the probability that Mike gets Elm, Alice gets Oakwood, and Bob gets
Van Dyke?
(d) What is the probability that Mike gets Elm Street, Alice gets Oakwood, Bob is assigned Van Dyke, and Eddie gets Park Street?
11. A survey of 100 college students revealed the following:
40 read TIME 15 read TIME and NEWSWEEK
30 read NEWSWEEK 12 read TIME and SOJOURNERS
25 read SOJOURNERS 10 read NEWSWEEK and SOJOURNERS 4 read all three
(a) What is the probability that a randomly chosen student will read exactly one magazine?
(b) What is the probability that a randomly chosen student will read exactly two magazines?
Section 5.3: Uniform Probability Distributions 181 (c) What is the probability that a randomly chosen student will read Sojourners and
exactly one other magazine?
12. While talking about birthdays, Ann, Bob, and Terry discovered that they were all born on a Tuesday.
(a) Assuming that the seven days of the week are equally likely as birthdays, what is the probability that 3 people would be born on the same day of the week? (b) What is the probability that 3 people would all be born on different days of the
week?
13. If the letters of the word LITTLE are written in a random order, what is the probability that they will appear in alphabetical order?
14. Six married couples are attending a party where two door prizes are awarded to two different people.
(a) What is the probability that a married couple wins both prizes?
(b) What is the probability that one prize goes to a man and the other to a woman? 15. Teledex Corporation is going to hire 5 programmers from a pool of 10 female and 8
male applicants. Sam and Martha are two of the applicants. If the 5 to get jobs are randomly chosen,
(a) what is the probability Sam and Martha both get jobs?
(b) what is the probability they both get jobs if Teledex has decided to hire 3 women and 2 men?
16. Ten people are dividing up to play basketball. Two of the ten are brothers.
(a) What is the probability the brothers will be on the same team if the players divide up arbitrarily?
(b) What is the probability the brothers and their friend Max will all be on the same team?
17. In a lottery a player can win a prize if any of the digits in a 4-digit number chosen by the player matches the corresponding digit in the winning 4-digit number. Thus a prize can be won if all 4 digits match, if 3 corresponding digits match, if 2 match, or if only 1 matches. Assume that any or all of the digits may be zero. What is the probability that
(a) all 4 digits match? (b) exactly 3 digits match? (c) exactly 2 digits match? (d) at least 1 digit matches?
18. The letters of the word MAMMAL are scrambled and written in a row. (a) What is the probability that the 3 M’s come first?
(b) What is the probability that none of the M’s appear side by side?
themselves randomly,
(a) What is the probability the men sit together and the women sit together?
(b) What is the probability the men and the women sit alternately (no two people of the same sex sitting beside each other)?
(c) What is the probability that Bill, Carla, and Ernie sit in 3 adjacent seats?
20. A class has 23 students. What is the probability that there are two people in the class who have the same birthday? [Assume that the 365 days of the year are equally likely for each person’s birthday. This means that we’ll assume that none of the students was born on Feb. 29 in a leap year.]
Section 5.4: Conditional Probability 183
5.4 Conditional Probability
Sometimes additional information becomes available that forces us to modify the probabilities of events that we might be interested in.
Example 5.24. Suppose you roll a pair of dice (one red and one green). The probability that the sum of the two numbers appearing is equal to 9 is 4/36 = 1/9 because 4 of the 36 possible outcomes show a sum of 9.
Here are the outcomes where the sum is 9: {(6,3),(5,4),(4,5),(3,6)}
Now what if we happen to see that the red die is showing the number 5, but we still haven’t seen the green die? How do we “process” this information? Since the red die has a 5 showing, we know that the outcome is one of these:
{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
Among these outcomes, there is exactly one in which the sum is equal to 9. So the chances of getting a sum of 9 based on the new information are 1 chance in 6. We say that the conditional probability that the sum is 9 given that the red die shows the number 5 is equal to 1/6.
Example 5.25. A class has the following distribution of grades:
Grade Number of students
A 5
B 14
C 7
D 9
F 8
If a student is randomly picked from this class, what is the probability that the student is an A student if it is known that the student is passing the course?
Solution: The probability that the student is an A student (given no information as to whether the student is passing or not) is 5/43. As soon as we know that the student is passing, the conditional probability that the student is an A student becomes 5/35, since 5 of the 35 passing students have an A average.
Definition: Conditional probability
If A and B are events with P(A) ≠ 0, then the conditional probability of B given A is defined by
Let’s go back and check this definition on Examples 5.24 and 5.25. In Example 5.24 let’s take A to be the event “red die shows the number 5” and B to be the event “sum = 9.” Then P(A) = 1/6, and since A ∩ B consists of the single outcome (5,4), P(A ∩ B) = 1/36. Therefore
P(B|A) = P(AP ↔ (A)B) = 1/361/6 = 16 .
In Example 5.25 let’s take E1 = “A students” and E2 = “students who are passing.” Then P(E2) = 35/43 since 35 of the 43 students are passing.
What is E1 ∩ E2 in this case? It is simply the A students since all A students are passing. So P(E1 ∩ E2) = P(E1) = 5/43. Therefore,
P(E1|E2) = P(EP1( ↔ E E2)
2 ) =
5/43
35/43 = 17 .
One important aspect of the definition of conditional probability above is that it enables you to express a conditional probability as a fraction where both the numerator and denominator are ordinary (unconditional) probabilities.
Example 5.26. Barbara, Carol, Alice, Peggy, and Sabrina are competing for two roles in a play. We will assume that the two to get roles will be randomly chosen from the five girls. What is the conditional probability that Peggy gets a role if we know that Carol does not get a role?
Solution: Let’s name the events that are involved here:
A = event that Carol does not get a role B = event that Peggy does get a role
This problem involves a small number of possible outcomes, and it is helpful to list them all. The ten possibilities for which 2 girls get the roles are as follows (all 10 of these possibilities are assumed equally likely, so each has probability 1/10):
#1 Barbara & Carol #2 Barbara & Alice #3 Barbara & Peggy #4 Barbara & Sabrina #5 Carol & Alice #6 Carol & Peggy #7 Carol & Sabrina #8 Alice & Peggy #9 Alice & Sabrina #10 Peggy & Sabrina
The event A consists of outcomes #2,#3,#4,#8,#9, and #10, so P(A) = 6/10. And the event A ∩ B consists of outcomes #3, #8, and #10, so P(A ∩ B) = 3/10. Therefore
Section 5.4: Conditional Probability 185
P(B|A) = P(BP ↔ (A)A) = 3/106/10 = 12 .
It is not absolutely necessary to list all the 10 outcomes as done here in order to do this problem. What is necessary is to figure out P(B ∩ A) and P(A) in some way, and these can both be determined using the basic counting methods we have practiced. For example
P(A) = CC(4,2)(5,2) = 10 6
because if Carol does not get a role then the two roles must be given out among the other 4 girls. And
P(B ∩ A) = CC(3,1)(5,2) = 10 3
because saying that Carol does not get a role and that Peggy does get a role means that there is one other role that will go to one of the three other girls.
Alternate solution: Let’s look at another approach to the same problem. If we know that Carol does not get a role, then we can just forget about her altogether and reformulate the problem. The question then becomes: If the two roles are assigned among the four other girls (Barbara, Alice, Peggy, and Sabrina), what is the probability that Peggy gets a role? There are C(4,2) = 6 ways to pick 2 of the 4 girls for the roles. Of these 6 possibilities, Peggy gets a role in 3 of the 6 cases. (The 3 cases in which Peggy gets a role are the case in which Peggy and Barbara are chosen, the case in which Peggy and Alice are chosen, and the case in which Peggy and Sabrina are chosen.) So from these considerations we see that
P(Peggy gets a role) = 36 = 12 .
This answer is of course consistent with the answer obtained in the first solution. In this latter solution we used the given information (that Carol does not get a role) in constructing the sample space. So our sample space had C(4,2) = 6 possibilities rather than C(5,2) = 10 as in the first solution. Since our new reduced sample space has already been adjusted for the new information (that Carol does not get a role), the conditional probability question in our original sample space becomes an ordinary (not conditional) probability question in the new reduced sample space.
Example 5.27. If two cards are dealt from a standard 52-card deck, what is the probability that both are hearts if we know that at least one of them is a heart?
Solution: This problem gives us a chance to review a couple of useful ideas. The sample space consists of the C(52,2) = 1326 possible ways to choose 2 cards from 52. The two events involved are
A = event that 2 hearts are dealt
Determining P(A) is easy: P(A) = CC(13,2)(52,2) = 17 . 1
It is easiest to get P(B) by first finding P(Bc) and then using P(B) = 1 – P(Bc). Bc is just the set of outcomes for which it is not true that at least one heart is dealt, meaning that both cards dealt come from the other three suits (with a total of 39 cards in the three suits). The probability of this happening is
P(Bc) = CC(39,2)(52,2) = 1934 . So P(B) = 1 – 1934 = 1534 .
Now what is A ∩ B? In this problem A ∩ B is simply A. (Because any outcome in which two hearts are dealt is certainly an outcome in which at least one heart is dealt.) Therefore P(A ∩ B) = P(A) = 1/17. So
P(A|B) = P(AP ↔ (B) = B) 15/34 = 1/17 15 . 2
One thing that is a bit curious is the fact that conditional probabilities are often useful in computing unconditional probabilities. The way in which this comes about is that we can turn around the definition of conditional probability and look at it in the form
P(A ∩ B) = P(A) P(B|A).
In some situations the two probabilities on the right are easier to obtain than the one on the left.
Example 5.28. If 2 cards are dealt from a standard 52-card deck, what is the probability that both are hearts?
Solution: We have done this problem previously using some of the basic counting techniques. This time we are going to look at it from the standpoint of conditional probability.
It’s useful for us to think of the cards in terms of a first card and a second card. It really doesn’t matter whether they were actually dealt one at a time or not. We can think of them that way if we like. It really just amounts to labeling one of the cards with the label “first” and the other with the label “second.” We also need to name the events that we are going to be interested in:
A = event that the first card is a heart B = event that the second card is a heart A ∩ B = event that both cards are hearts
The idea is to compute P(A ∩ B) by using the fact that P(A ∩ B) = P(A) P(B|A). The term P(A) on the right is easy, because when we draw the first card it is coming from a deck of 52 cards including 13 hearts. Therefore P(A) = 13/52. Now what about P(B|A)? This is the conditional probability that the second card is a heart based on the information that the first card is a heart. Knowing that the first card is a heart means that when the second is drawn it is coming from a deck of 51 cards in which there are 12 hearts
Section 5.4: Conditional Probability 187 remaining. Therefore P(B|A) = 12/51. And so P(A ∩ B) is given by
P(A ∩ B) = P(A) P(B|A) = 1352 × 1251 = 17 . 1
This of course agrees with the answer we got earlier for this problem by using the basic counting techniques.
This idea can be extended to more than two events. For example, with three events the equation becomes
P(A ∩ B ∩ C) = P(A) P(B|A) P(C | A ∩ B).
The last term on the right here means the conditional probability of C given that both A and B occur.
If three cards are dealt, we could use this idea to determine the probability that all three are hearts. The computation would be
P(A ∩ B ∩ C) = P(A) P(B|A) P(C | A ∩ B) = 52 13 × 1251 × 1150 ≈ .0129.
The third factor 11/50 on the right is the conditional probability that the third card is a heart given that the first two cards are hearts. The explanation is that if we know that the first two are hearts, then there are 11 hearts left in the 50 remaining cards when the third card is to be drawn.
In the next section we will see how to apply conditional probabilities to a wide variety of problems by placing them appropriately in tree diagrams.
Problems
1. If P(A) = .6, P(B) = .4, and P(A ∩ B) = .3, (a) find P(B | A) (b) find P(A | B) 2. If P(A) = .7, P(B) = .6 and P(A ∩ B) = .4,
(a) find P(A | Bc) (b) find P(Bc | A) (c) find P(Bc | Ac)
3. If a card is drawn from a deck of 52 cards, what is the probability the card is a heart if you know it is not a club?
4. If two dice are rolled (one red and one green),
(a) what is the probability of getting a pair of sixes?
(b) what is the conditional probability of getting a pair of sixes if you know that the red die shows a six?
(c) what is the conditional probability of getting a pair of sixes if you know that the green die shows a six?
(d) what is the conditional probability of getting a pair of sixes if you know that at least one of the two dice shows a six?
5. A committee has 3 members from California, 4 from Florida, 2 from Colorado, and 2 from North Carolina.
(a) If the committee president is picked randomly, what is the probability the president will be from California?
(b) What is the conditional probability the president will be from California if the 2 members from Colorado remove themselves from the selection process?
6. If a coin is tossed 3 times,
(a) what is the probability that all three tosses come up heads given that at least two of the tosses come up heads?
(b) what is the probability that all three tosses come up heads given that the first two tosses come up heads?
[Hint: List the sample space for this experiment as well as the events related to each of these questions.]
7. A political party has decided to hold political rallies in two of the following states: Missouri, Montana, Mississippi, Michigan, Alabama, Connecticut, and Pennsylvania. Assuming that the states have equal probabilities of being chosen, (a) what is the probability that both of the states chosen begin with the letter “M”? (b) what is the probability that both start with “M” given that at least one does? (c) what is the probability that both start with “M” given that one of the states is
Michigan?
8. Two men (Jim and Bob) and three women (Ann, Beth, and Carol) are on a committee. Two of the five are to be chosen to serve as officers. If the officers are chosen randomly,
(a) what is the probability that both officers will be women? (b) what is the probability at least one officer will be a woman?
(c) what is the probability that both officers will be women given that at least one is a woman?
(d) what is the probability both officers are women if you know that Beth is an officer?
9. Evaluate the following probabilities.
(a) P(A | Ac) (b) P(A |A ∩ B) (c) P(A | A) (d) P(A ∪ B | A)
10. Six married couples are attending a party where two door prizes are awarded to two different people.
(a) If John’s wife Mary wins a prize, what is the probability that John will also win a prize?
