Unit 2: Probability
QBA 201 – Summer 2013
Instructor: Michael Malcolm
2.1: Basic laws of probability
2.2: Mutually exclusive and independent events
2.3: The law of total probability
2.1: Basic laws of probability
Being able to understand and work with probabilities is a very important skill. There are really only a few basic laws, but understanding how they work together and how to apply them to solving problems takes practice.
The sample space 𝑆 is the set of possible outcomes of some uncertain situation. These outcomes are defined in a way that exactly one outcome will occur. The foundation of probability is an event, which is just some subset of the sample space.
For example, before a customer walks into McDonalds, he might order a small drink, a medium drink, a large drink or no drink. In this case, the sample space is {small, medium, large, none}. One possible event from this sample space is that the customer orders a small drink. Another possible event we could define is that the customer orders some kind of drink. This event is represented by the subspace {small, medium, large} from the sample space. As a second example, an individual might develop cancer or not develop cancer over the course of his life. In this case, the sample space is just {cancer, no cancer}.
The key is that there is some randomness. Before the event is observed, we don’t necessarily know whether it will happen. Rather, we assign probabilities to the events, which describe the “chance” with which the event will occur.
Legal probabilities for any event 𝐸𝑖 must lie between zero and one: 0 ≤ Pr(𝐸𝑖) ≤ 1. If the event never occurs then Pr(𝐸𝑖) = 0 and if the event always occurs then Pr(𝐸𝑖) = 1. This suggests that the probability of the sample space must be Pr(𝑆) = 1 since the sample space was defined in a way that exactly one outcome from the sample space always occurs.
Sometimes there is a natural way to assign probabilities. For example, if you flip a coin then the sample space of outcomes is {head, tail}. Assuming that the coin is fair, then the probability of the event that the coin comes up “head” is 0.5 and the probability that the coin comes up “tail” is also 0.5. When we roll a die, the sample space of outcomes is {1,2,3,4,5,6}. The probability for each number is 16. In general, if there are 𝑛 outcomes and each is equally likely, then the probability for each outcome is 1
𝑛.
Suppose that Pr(𝐴) denotes the probability that event 𝐴 will occur. Then what is the probability that event 𝐴 will not occur? This event is called the complement of 𝐴 and is written 𝐴𝐶. The probability of 𝐴𝐶 is defined as such:
Pr(𝐴𝐶) = 1 − Pr(𝐴)
This makes intuitive sense. If 𝑉 is the event that a person has blond hair and Pr(𝑉) = 0.4, then the probability that a person does not have blond hair is Pr(𝑉𝐶) = 1 − 0.4 = 0.6.
The union of two events 𝐴 and 𝐵 is the event that either 𝐴 happens, or 𝐵 happens or both happen. The union of events 𝐴 and 𝐵 is denoted 𝐴 ∪ 𝐵.
The intersection of two events 𝐴 and 𝐵 is the event that both 𝐴 happens and 𝐵 happens. The intersection of events 𝐴 and 𝐵 is denoted 𝐴 ∩ 𝐵.
For example, if 𝑉 is the event that a person is a vegetarian and 𝐹 is the event that a person is female, then the union 𝑉 ∪ 𝐹 includes all individuals who are female, or vegetarian or both, while the intersection 𝑉 ∩ 𝐹 includes only female vegetarians.
The addition law relates the probabilities of unions and intersections of events. According to this law, for any events 𝐴 and 𝐵:
Pr(𝐴 ∪ 𝐵) = Pr(𝐴) + Pr(𝐵) − Pr(𝐴 ∩ 𝐵)
The addition law is intuitive. Suppose that a community consists of 100 individuals, as follows.
50 females: 30 vegetarians and 20 non-vegetarians 50 males: 10 vegetarians and 40 non-vegetarians
For a randomly selected individual in this community, Pr(𝐹) = 0.5 and Pr(𝑉) = 0.4. Since there are 30 female vegetarians Pr(𝑉 ∩ 𝐹) = 0.3.
Now, to compute the probability that a randomly selected individual is either female, or vegetarian, or both, we would use the addition law:
Pr(𝑉 ∪ 𝐹) = Pr(𝑉) + Pr(𝐹) − Pr(𝑉 ∩ 𝐹) = 0.5 + 0.4 − 0.3 = 0.6
vegetarians are included in both Pr(𝐹) = 0.5 and Pr(𝑉) = 0.4, but obviously should only be counted once in determining that the probability that a person is either a vegetarian or a female.
Our final fundamental concept is the idea of conditional probability. Suppose that we know that the event 𝐵 occurred, then the conditional probability of the event 𝐴 given the event 𝐵 is the probability of event 𝐴 knowing that event 𝐵 has occurred. It is denoted Pr(𝐴|𝐵).
For example, suppose we know that a person is female. What is then the conditional probability that the person is vegetarian. Well, there are 50 females, out of whom 30 are vegetarian, so it is
logical that the conditional probability Pr(𝑉|𝐹) =3050= 0.6.
Formally, the conditional probability Pr(𝐴|𝐵) is given by:
Pr(𝐴|𝐵) =Pr(𝐴∩𝐵)Pr(𝐵)
The intuition for the previous example is that Pr(𝑉 ∩ 𝐹) = 0.3 is the overall probability of being a vegetarian, but this is relative to the overall probability of being female Pr(𝐹) = 0.5. So if we know that a person is female, the conditional probability of being a vegetarian after we incorporate the information that she is female is:
Pr(𝑉|𝐹) =Pr(𝑉∩𝐹)Pr(𝐹) =0.30.5 = 0.6
Finally, from the definition of conditional probability, we can derive the multiplication law. For any events 𝐴 and 𝐵:
Pr(𝐴 ∩ 𝐵) = Pr(𝐴|𝐵) Pr(𝐵)
In our previous example, we know that there is 0.5 probability that a person is female and that there is 0.6 probability that the person is vegetarian given that she is female. Thus, the probability that a randomly selected person is a vegetarian female can be calculated as such:
Pr(𝑉 ∩ 𝐹) = Pr(𝑉|𝐹) Pr(𝐹) = 0.6 ⋅ 0.5 = 0.3
EXERCISES
1. The manager of a furniture store sells between 0 and 4 couches each week. The probabilities are Pr(0) = 0.08, Pr(1) = 0.18, Pr(2) = 0.32, Pr(3) = 0.30, Pr(4) =
0.12.
a. Are these valid probability assignments?
b. Let 𝐴 be the event that fewer than 3 couches are sold. Find Pr(𝐴). c. Let 𝐵 be the event that more than 1 couch is sold. Find Pr(𝐵). d. Find Pr(𝐴 ∩ 𝐵) and describe it in words.
e. Find Pr(𝐴 ∪ 𝐵) and describe it in words.
2. A large group of people is to be checked for two common symptoms of a certain disease. 20% of the people exhibit symptom A alone, 30% exhibit symptom B alone and 10% exhibit both symptoms. The remainder have neither. For a random person in the group
a. What is the probability that the person has at least one symptom? b. What is the probability that the person has neither symptom?
c. What is the probability that the person has both symptoms, given that he has symptom B?
3. In a certain community, 36 percent of the families own a dog and 30 percent of the families own a cat. 22 percent of the families that own a dog also own a cat. What is the probability that a randomly selected family owns a dog given that it owns a cat?
4. A survey of customers in a particular community showed that 10% were dissatisfied with the plumbing jobs done in their homes. Half of the complaints dealt with Plumber A, who does 40% of the plumbing jobs.
a. Find the probability that a consumer will obtain an unsatisfactory plumbing job, given that it was plumber A.
b. Find the probability that a consumer will obtain a satisfactory plumbing job, given that it was plumber A.
5. The king comes from a family of two children. What is the probability that the other child is his sister?
2.2: Mutually exclusive and independent events
Events 𝐴 and 𝐵 are mutually exclusive when Pr(𝐴 ∩ 𝐵) = 0.
In other words, events 𝐴 and 𝐵 are mutually exclusive when they never happen together. For example, the event that a person has prostate cancer and the event that a person is female are mutually exclusive.
There is an important implication for mutually exclusive events. Recall that the addition law of probability is Pr(𝐴 ∪ 𝐵) = Pr(𝐴) + Pr(𝐵) − Pr(𝐴 ∩ 𝐵). But, for mutually exclusive events,
Pr(𝐴 ∩ 𝐵) = 0 and thus for mutually exclusive events only:
Pr(𝐴 ∪ 𝐵) = Pr(𝐴) + Pr(𝐵)
When events never happen together, we can find the probability that any one of them happens just by adding up the respective probabilities of each event. There is no need to worry about double counting in this case.
Events 𝐴 and 𝐵 are independent when Pr(𝐴|𝐵) = Pr(𝐴).
In other words, events 𝐴 and 𝐵 are independent when knowing 𝐵 gives you no information about
𝐴. The conditional probability that 𝐴 will occur, knowing that 𝐵 occurred Pr(𝐴|𝐵) is just the same as the simple unconditional probability Pr(𝐴), since knowing 𝐵 tells you nothing about 𝐴. A simple example is to flip a coin twice. If 𝐴 is getting heads on the first toss and 𝐵 is getting heads on the second toss, then 𝐴 and 𝐵 are independent since the outcome of the first toss gives you no information about what will happen on the second toss.
There is an important implication for independent events. Recall that the multiplication law of probability is Pr(𝐴 ∩ 𝐵) = Pr(𝐴|𝐵) Pr(𝐵). But, for independent events, Pr(𝐴|𝐵) = Pr(𝐴), and thus for independent events only:
Pr(𝐴 ∩ 𝐵) = Pr(𝐴) Pr(𝐵)
For independent events, the probability that two events happen jointly is just the product of the probabilities of each event separately. There is no need to worry about conditional probabilities.
A critical point is that mutually exclusive and independent events are not the same thing. For mutually exclusive events, Pr(𝐴 ∩ 𝐵) = 0 while for independent events Pr(𝐴 ∩ 𝐵) =
happen together, but knowing whether the first happened doesn’t help you guess whether or not the second will happen.
There are two important extensions of these concepts. First, the simplified addition and multiplication laws hold for any number of events, not just two.
When the events {𝐴1, 𝐴2, ⋯ , 𝐴𝑛} are all mutually exclusive of each other, then the probability that one of them happens is:
Pr(𝐴1∪ 𝐴2∪ ⋯ ∪ 𝐴𝑛) = Pr(𝐴1) + Pr(𝐴2) + ⋯ + Pr(𝐴𝑛)
To emphasize again, this rule only holds for events that are mutually exclusive. The informal comment that “or” statements are added together in figuring out probabilities only holds when the events are mutually exclusive. If the events happen together sometimes, then the answer is not this simple.
In addition, when the events {𝐴1, 𝐴2, ⋯ , 𝐴𝑛} are all mutually independent of each other, then the joint probability that all happen is:
Pr(𝐴1∩ 𝐴2∩ ⋯ ∩ 𝐴𝑛) = Pr(𝐴1) Pr(𝐴2) ⋯ Pr(𝐴𝑛)
To emphasize again, this rule only holds for events that are independent. The informal comment that “and” statements are multiplied together in figuring out probabilities only holds when the events are independent. If the events depend on each other, then the answer is not simple and you have to worry about conditional probabilities.
EXERCISES
1. A general contractor has submitted two bids for projects (A and B). The probability of getting project A is 0.65. The probability of getting project B is 0.77. The probability of getting at least one of the projects is 0.9.
a. What is the probability that he will get both projects? b. Are the events of getting the projects mutually exclusive? c. Are the events of getting the projects independent?
2. A purchasing agent has rushed orders for a particular machine part with two different suppliers, A and B. If neither order arrives on time, then the production process must be shut down. The probability that A will deliver the part on time is 0.55. The probability that B will deliver the material on time is 0.35. The two suppliers operate independently of each other.
a. What is the probability that both suppliers will deliver the material? b. What is the probability that at least one supplier will deliver the material? c. What is the probability that the production process will be shut down?
3. A very short quiz has one true/false question and one multiple choice question with five possible choices. You have no idea what the correct answers are, but you take the quiz anyway by guessing randomly.
a. What is the probability that you give the correct answer to both questions? b. What is the probability that neither answer is correct?
c. What is the probability that only the multiple choice question is correct? d. What is the probability that one of the two answers is correct?
4. A football team has probability 0.75 of winning any particular game. Assume that the games are independent.
a. What is the probability that the team wins all four games?
b. What is the probability that the team wins three out of four games?
6. Consider two events A and B such that Pr(𝐴) = 0.4 and Pr(𝐵) = 0.3. Compute Pr(𝐴|𝐵) for each of the following cases
a. Pr(𝐴 ∩ 𝐵) = 0.1 b. Pr(𝐴 ∪ 𝐵) = 0.5
c. B always happens when A happens (i.e. B is a subset of A) d. A and B are independent
e. A and B are mutually exclusive
7. Consider the following events corresponding to a single toss of a six-sided die:
A: Number shown is odd B: Number shown is even C: Number shown is 1 or 2
a. Are A and B independent events? b. Are A and C independent events?
2.3: The law of total probability
When we say that the events {𝐵1, 𝐵2, … , 𝐵𝑘} form a partition of the sample space, what we mean is that the events are mutually exclusive and that they cover the entire sample space. Informally, what this means is that every member of the population belongs to exactly one of the categories. For example {male, female} forms a partition of a population of people.
Now, the basic idea behind the law of total probability is that, when the event 𝐴 occurs, it always happens together with exactly one element of the set {𝐵1, 𝐵2, … , 𝐵𝑘}. This holds by definition because these events form a partition of the sample space. Then the following must hold:
Pr(𝐴) = Pr(𝐴 ∩ 𝐵1) + Pr(𝐴 ∩ 𝐵2) + ⋯ + Pr(𝐴 ∩ 𝐵𝑘)
For example, if a person is a vegetarian, then he is either a vegetarian male or a vegetarian female since {male, female} forms a mutually exclusive partition of the sample space. Thus, the probability that a randomly selected person is a vegetarian is equal to the probability that the person is a vegetarian male plus the probability that the person is a vegetarian female.
But recall that, according to the multiplication rule, Pr(𝐴 ∩ 𝐵) = Pr(𝐴|𝐵) Pr(𝐵). Applying this to each term in the sum above gives the law of total probability:
Pr(𝐴) = Pr(𝐴|𝐵1) Pr(𝐵1) + Pr(𝐴|𝐵2) Pr(𝐵2) + ⋯ + Pr(𝐴|𝐵𝑘) Pr(𝐵𝑘)
This looks complicated, but it actually makes good intuitive sense. Recall our example from earlier. With 100 individuals in the community, we had
50 females: 30 vegetarians and 20 non-vegetarians 50 males: 10 vegetarians and 40 non-vegetarians
Here, {male, female} forms a mutually exclusive partition of the population with Pr(𝑀) = 0.5 and Pr(𝐹) = 0.5. Of males, 20% are vegetarians and, of females, 60% are vegetarians. Stated precisely, Pr(𝑉|𝑀) = 0.2 and Pr(𝑉|𝐹) = 0.6.
Choose a person at random. What is the probability that the person is a vegetarian. Using the law of total probability, since 𝑀 and 𝐹 form a mutually exclusive partition of the population:
Pr(𝑉) = Pr(𝑉|𝑀) Pr(𝑀) + Pr(𝑉|𝐹) Pr(𝐹) = 0.2 ⋅ 0.5 + 0.6 ⋅ 0.5 = 0.4
Here is another example. Suppose that a particular community is made up of 50% whites (W), 30% Hispanics (H) and 20% blacks (B). We assume that this is a mutually exclusive partition in the sense that each member of the community belongs to exactly one group.
In this community, 45% of whites are Democrats (D), 55% of Hispanics are Democrats and 90% of blacks are Democrats. We can use this information to calculate the probability that a randomly selected individual from the community is a Democrat. Applying the law of total probability:
Pr(𝐷) = Pr(𝐷|𝑊) Pr(𝑊) + Pr(𝐷|𝐻) Pr(𝐻) + Pr(𝐷|𝐵) Pr(𝐵)
Substituting in the values given in the problem:
Pr(𝐷) = 0.45 ⋅ 0.5 + 0.55 ⋅ 0.3 + 0.9 ⋅ 0.2 = 0.57
Again, the idea is that a Democrat is either a white Democrat, an Hispanic Democrat or a black Democrat, so we just tally up each category.
There is an important special case of the law of total probability. For any event 𝐵, it is the case that 𝐵 and 𝐵𝐶 form a mutually exclusive partition of the sample space. By definition, event 𝐵 either happens or it does not, so all members of the population are in either 𝐵 or 𝐵𝐶. Thus, an important special case of the law of total probability is:
Pr(𝐴) = Pr(𝐴|𝐵) Pr(𝐵) + Pr(𝐴|𝐵𝐶) Pr(𝐵𝐶)
Here is an example. Suppose in a certain population that 1% of people are infected with a disease (D). There is a test for the disease that is quite reliable: 97% of infected people get a positive test result (P) and only 5% of non-infected people get a positive test result. What is the probability of getting a positive test result?
Since 𝐷 and 𝐷𝐶 form a mutually exclusive partition of the population, we can apply the law of total probability.
Pr(𝑃) = Pr(𝑃|𝐷) Pr(𝐷) + Pr(𝑃|𝐷𝐶) Pr(𝐷𝐶)
Substituting in the values given in the example:
To emphasize one final time, the law of total probability can only be applied for events
{𝐵1, 𝐵2, … , 𝐵𝑘} that form a mutually exclusive partition of the population. If there is overlap or if
EXERCISES
1. 60% of statistics students attended class on Fridays. Of students who attend class on Friday, 98% pass the class. Of students who do not attend class on Friday, only 20% pass the class. What is the probability that a randomly chosen student will pass the class?
2. Many universities use a “no-pass, no-play” rule for athletes. Under this system, a student who fails a course is disqualified from participating in extracurricular activities during the next semester. Suppose that the probability is 0.15 that an athlete who has not previously been disqualified will be disqualified in the next semester. For athletes who have been previously disqualified, the probability is 0.5 that the athlete will be disqualified in the next semester. 30% of athletes have been disqualified in previous semesters. What is the probability that a randomly selected athlete will be disqualified next semester?
2.4: Bayes’ Rule
Bayes’ Rule is just combination of the law of total probability together with the definition of conditional probability. But Bayes’ Rule turns out to be tremendously important for business and economics applications.
Recall that when {𝐵1, 𝐵2, … , 𝐵𝑘} forms a mutually exclusive partition of the population, then the probability of any event 𝐴 could be calculated using the law of total probability as:
Pr(𝐴) = Pr(𝐴|𝐵1) Pr(𝐵1) + Pr(𝐴|𝐵2) Pr(𝐵2) + ⋯ + Pr(𝐴|𝐵𝑘) Pr(𝐵𝑘)
Now suppose we turn this around. We know that 𝐴 happened. What is the probability that it happened together with 𝐵𝑖? Using the definition of conditional probability:
Pr(𝐵𝑖|𝐴) =Pr(𝐴∩𝐵𝑖)
Pr(𝐴)
If we use the multiplication rule for the numerator and use the law of total probability for the denominator, we obtain Bayes’ Rule. When {𝐵1, 𝐵2, … , 𝐵𝑘} forms a mutually exclusive partition of the a population, then for any element 𝐵𝑖 and any event 𝐴:
Pr(𝐵𝑖|𝐴) = Pr(𝐴|𝐵𝑖) Pr(𝐵𝑖)
Pr(𝐴|𝐵1) Pr(𝐵1)+Pr(𝐴|𝐵2) Pr(𝐵2)+⋯+Pr(𝐴|𝐵𝑘) Pr(𝐵𝑘)
The basic idea is that the denominator is all of the ways that the event 𝐴 can happen, and the numerator is when 𝐴 happens together with 𝐵𝑖. Here, Pr(𝐵𝑖) is called the prior probability or the ex ante probability of event 𝐵𝑖, which is the probability of 𝐵𝑖 before learning anything about
𝐴. By contrast, Pr(𝐵𝑖|𝐴) is called the posterior probability or the ex post probability of event
𝐵𝑖, which is the probability of 𝐵𝑖 after we learn that 𝐴 occurred.
This seems at first like a pedantic exercise in conditional probabilities, but actually what underlies Bayes’ Rule is a very deep and important idea. It represents how we process new information and change our guesses depending on new things that we learn. Let us start with a simple motivating example.
Returning to our earlier problem, 50% of the population are males and 50% are females. 20% of men are vegetarians and 60% of females are vegetarians.
Suppose we learn that a person is vegetarian – what then is the probability that the person is female? Let’s use intuitive reasoning, and then do it formally through Bayes’ Rule. Overall, 0.2 ∙
population is composed of female vegetarians. Out of the 0.4 of the population composed of vegetarians, 0.1 are male vegetarians, so intuitively, once we learn that the person is a vegetarian
then the probability that the person is also female is 0.30.4 = 0.75.
Using Bayes’ Rule formally:
Pr(𝐹|𝑉) =Pr(𝑉|𝐹) Pr(𝐹)+Pr(𝑉|𝑀) Pr(𝑀)Pr(𝑉|𝐹) Pr(𝐹) =0.6∙0.5+0.2∙0.50.6∙0.5 = 0.75
The idea is that the denominator represents all vegetarians, while the numerator represents the case of female vegetarians, which is what we are interested in.
Using Bayesian language, before we have any information about whether a person is vegetarian, our best guess is probability 0.5 that a randomly selected person is female. This is the prior probability. However, once we incorporate the information that the person is a vegetarian, then our posterior probability that the person is female is 0.75. This process is the essence of Bayesian statistics: starting with a prior probability, then incorporating new information to update our prior probabilities to a posterior probability.
Recall the earlier example of race and voter registration. A community is made up of 50% whites (W), 30% Hispanics (H) and 20% blacks (B). We assume that this forms a mutually exclusive partition. In this community, 45% of whites are Democrats (D), 55% of Hispanics are Democrats and 90% of blacks are Democrats.
Suppose we learn that a person is a Democrat. What now is the probability that he is black? Using Bayes’ Rule:
Pr(𝐵|𝐷) =Pr(𝐷|𝑊) Pr(𝑊)+Pr(𝐷|𝐻) Pr(𝐻)+Pr(𝐷|𝐵) Pr(𝐵)Pr(𝐷|𝐵) Pr(𝐵) =0.45∙0.5+0.55∙0.3+0.9∙0.20.9∙0.2 = 0.3158
If we choose a randomly selected person, the prior probability that he is black is Pr(𝐵) = 0.2, which is based on the overall sample space. But once we update with the information that the person is a Democrat, we can update our prior guess to probability Pr(𝐵|𝐷) = 0.3158 that he is black.
An important special case of Bayes’ Rule is the case where the partition of the sample space is made up of {𝐵, 𝐵𝐶}. This always forms a partition because, by definition, each element belongs to exactly one of the two. Conditional on knowing the event 𝐴, Bayes’ Rule gives:
As a final example, let us consider the medical test from the previous section. 1% of people are infected with a disease (D). There is a test for the disease that is quite reliable: 97% of infected people get a positive test result (P) and only 5% of non-infected people get a positive test result. Suppose you get a positive test result. What is the probability that you are infected with the disease? Applying Bayes’ Rule:
Pr(𝐷|𝑃) =Pr(𝑃|𝐷) Pr(𝐷)+Pr(𝑃|𝐷Pr(𝑃|𝐷) Pr(𝐷)𝐶) Pr(𝐷𝐶)= 0.97∙0.01+0.05∙0.990.97∙0.01 = 0.1639
This seems quite low to many people. The test is pretty reliable, and only returns false positives to non-infected people 5% of the time. How, then, can a positive test result imply only a 16% chance of having the disease. The answer is that the disease is infrequent overall. So, even with the low rate of false positives, it’s still more likely that you are an uninfected person with a positive result (probability 0.05 ∙ 0.99) than an infected person with a positive test result (probability 0.97 ∙ 0.01).
After the first test, you go back and get a second positive test. What probability do you assign now to having the disease?
The essence of Bayesian statistics is to continue updating our prior with new information. In this case, before the first test, the prior probability that you had the disease was only Pr(𝐷) = 0.01. But after you have already gotten that first positive test result, your new prior of having the disease is Pr(𝐷) = 0.1639. This is the prior you should use for assessing the probability of a positive result following the second test. Using Bayes’ Rule again:
Pr(𝐷|𝑃) =Pr(𝑃|𝐷) Pr(𝐷)+Pr(𝑃|𝐷Pr(𝑃|𝐷) Pr(𝐷)𝐶) Pr(𝐷𝐶)= 0.97∙0.1639+0.05∙0.83610.97∙0.1639 = 0.7918
EXERCISES
1. Males and females are observed to react differently to a given set of circumstances: 70% of males react positively but only 40% of females react positively. A group of 20 people, 15 female and 5 male, were subjected to the circumstance and asked to describe their reactions. A response picked at random from among the 20 responses was positive. What is the probability that the response is from a male?
2. A student answers a multiple choice question that offers four possible answers. The probability that the student knows the answer is 0.8 and the probability that he guesses is 0.2. If the student guesses, then the probability of selecting the correct answer is 0.25. If the student answers the question correctly, what is the probability that he actually knew the correct answer?
3. Of all travelers arriving at a small airport, 60% fly on major airlines, 10% fly on corporate jets and 30% fly on private planes. Of those who fly on major airlines, 50% are traveling for business reasons. 90% of those flying on corporate jets are traveling for business reasons and 60% of those flying on private planes.
a. What is the probability that a randomly selected person is traveling for business? b. What is the probability that a randomly selected person is traveling for business
on a privately owned plane?
c. What is the probability that a randomly selected person is traveling for business, given that he arrived on a private plane?
2.5: Counting rules
In the previous sections, we took information on some probabilities that were given and used probability rules to derive related probabilities. But, in some cases, you can systematically determine probabilities by enumerating possible outcomes. Specifically, many situations with randomness involve selecting 𝑛 objects out of a possible set of 𝑁 objects.
If order does not matter, then the various ways to select 𝑛 objects out of a larger set of 𝑁 objects are called combinations. We can systematically count the number of combinations as such:
𝐶𝑛𝑁 = (𝑁
𝑛) =
𝑁! 𝑛!(𝑁−𝑛)!
Recall that the ! notation indicates a factorial. For example, 5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120. The expression (𝑁
𝑛) is read in English as “𝑁 choose 𝑛.”
For example, suppose that a quality control inspector has a box of five parts, and he intends to select two of these parts for inspection. If the five parts are labeled A, B, C, D and E, then the possible combinations are AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. There are evidently ten possible combinations. We could have determined this systematically using the rule for combinations:
(5 2) =
5! 2!(5−2)!=
5! 2!3!= 10
In the example above, order did not matter in the sense that AB and BA were the same thing since the same two parts were chosen. If order makes a difference, then the various ways to select 𝑛 objects out of a larger set of 𝑁 objects are called permutations. We can systematically count the number of permutations as such:
𝑃𝑛𝑁= 𝑛! (𝑁𝑛) =(𝑁−𝑛)!𝑁!
For a given 𝑁 and 𝑛 with 𝑁 > 𝑛, the number of permutations will always be larger than the number of combinations. If we take the example of our inspector above selecting two parts out of five, there are actually 20 possible permutations: AB, AC, AD, AE, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EA, EB, EC and ED. But rather than enumerating all of them by hand, we could have just used the permutation rule to figure out how many there are:
𝑁! (𝑁−𝑛)!=
5! (5−2)! =
EXERCISES
1. Of eight cars produced at a particular factory, four are defective. Nevertheless, the eight cars were shipped out: four of them to the Buyer A and four of them to buyer B. What is the probability that Buyer A ends up with all four of the defective cars?
2. A church is conducting a raffle where 50 tickets are distributed, out of which there are three winning tickets. There are four organizers of the raffle, each of whom bought one ticket. What is the probability that all three of the winning tickets will go to the organizers?
