Up to this point, we have discussed probability and its relationships intuitively.
We now present general rules for understanding and calculating probabilities that are very helpful in decision making.
The General Rule of Addition
Th e general rule of addition applies whenever you want to know the probability that either of two events occurred. Statisticians express the general rule of addition symbolically as
P(A x B) 5 P(A) 1 P(B) 2 P(A y B)
Although this rule may appear complex, it is actually fairly simple. In English, the rule states that the probability of either of two events occurring [P(A x B)]
is equal to the probability that one event will occur [P(A)] plus the probability that the other event will occur [P(B)] minus the probability that both events will occur simultaneously [P(A y B)]. Remember that the probability of both events occurring at once is subtracted because this probability is contained in both the probability of the fi rst event and the probability of the second event.
Perhaps the clearest way to illustrate the union and intersection of proba- bilities is with a Venn diagram (see Figure 7.9). A Venn diagram pictures each event—for example, [P(A)]—with its own circle and can easily show intersection or overlap of diff erent events; for example, P(A y B)]. In Figure 7.9, it is clear that adding the probability of A to the probability of B counts the shaded area twice. For that reason, the shaded area [or P(A y B)] needs to be subtracted from the total.
The probability of any of three events occurring [P(A x B x C)] is a simple extension of the logic used for two events. Statisticians use the following formula:
P(A x B x C ) 5 P(A) 1 P(B) 1 P(C ) 2 P(A y B) 2 P(A y C ) 2 P(B y C)1 P(A y B y C )
In other words, the probability of any of three events occurring [P(A x B x C)]
is equal to the probability of the first event [P(A)] plus the probability of the second event [P(B)], plus the probability of the third event [P(C)], minus the probability that the fi rst and the second events both occur [P(A y B)], minus the probability that the fi rst and the third events both occur [P(A y C)], minus
Figure 7.9 Venn Diagram for P(A < B)
P(A > B)
P(A) P(B)
the probability that the second and the third events both occur [P(B y C)], plus the probability that all three events occur at once [P(A yB y C)]. The subtraction of probabilities is necessary to correct for double counting the prob- ability that pairs of events will occur.
Again, it is easier to visualize this formula using Venn diagrams (see Figure 7.10). Adding the probability of A, the probability of B, and the prob- ability of C together counts each of the lightly shaded areas twice and counts the darkly shaded area three times. Subtracting the intersection of A and B, the intersection of A and C, and the intersection of B and C removes each of the lightly shaded sections once. However, it also removes all three counts of the darkly shaded sections (A y B y C), so this probability must be added back into the formula.
An example will help here. Individuals can be placed on the England County civil service list in three ways: by passing the civil service exam, earning a college diploma, or being a member of Pi Alpha Alpha (a national public administration honorary society). Table 7.3 shows how the current list of eligibles qualifi ed.
Figure 7.10 Venn Diagram for P(A < B < C)
P(A > B)
P(A > B > C) P(B > C)
P(A > C) P(A)
P(B)
P(C)
Table 7.3 Results for England County Civil Service List Pi Alpha Alpha
Member
Not a Pi Alpha Alpha Member
Exam Result
College Degree
No
Degree Subtotal
College Degree
No
Degree Subtotal Total
Passed Exam 26 14 40 64 16 80 120
Failed Exam 4 16 20 36 74 110 130
Subtotals 30 30 60 100 90 190 250
Totals 60 190 250
What is the probability that a person is on the England County eligibles list because he or she passed the exam, had a college degree, or was a member of Pi Alpha Alpha? Th e probability of passing the exam is .48 (120 4 250). Th e probability of having a college degree is .52 (130 4 250). Th e probability of be- ing a member of Pi Alpha Alpha is .24 (60 4 250). (If you do not understand how these probabilities were derived, review the beginning of this chapter.) Th e sum of these probabilities is 1.24. From this number, we subtract the probability that someone passed the exam and had a college degree (90 4 250 5 .36), the probability that someone was a member of Pi Alpha Alpha and passed the exam (40 4 250 5 .16), and the probability that someone was a college graduate and a member of Pi Alpha Alpha (30 4 250 5 .12). Subtracting these probabilities from 1.24, we get a diff erence of .6. To this probability we add the probability that someone is a member of Pi Alpha Alpha, graduated from college, and passed the exam (26 4 250 5 .104), leaving us a probability of .704 (.6 1 .104 5 .704).
Examining the table directly, we see that 176 persons were on the eligibles list (these three ways are the only ways to get on the list); the probability of being on the eligibles list, therefore, is .704 (176 4 250).
You can calculate the probability that any of four or more events occurred by using the same logic that was used here. In practice, however, this calcula- tion becomes fairly complex. If you are ever faced with such a problem, consult a statistics text or a statistician (or delegate the task to an exceptionally qualifi ed and motivated intern!).
Whenever two events are mutually exclusive, a special rule of addition is used.
Two events are mutually exclusive if the occurrence of A means that B will not occur. In other words, two events are mutually exclusive if only one of the events can occur at a time. As statisticians state it, if the probability of the intersection of A and B is zero, then A and B are mutually exclusive events. When two events are mutually exclusive, the probability of either A or B occurring is equal to the prob- ability of A plus the probability of B (because there is no overlap to subtract).
The special rule of addition for mutually exclusive events is expressed symbolically as
P(A x B) 5 P(A) 1 P(B) Th e rule is illustrated in the Venn diagram in Figure 7.11.
Figure 7.11 Venn Diagram for P(A : B) 5 P(A) 1 P(B)
P(A) P(B)
The General Rule of Multiplication
Th e general rule of multiplication is applied when you want to know the joint probability of two events (the probability that both events will occur). Th e general rule of multiplication is denoted symbolically as follows:
P(A y B) 5 P(A) 3 P(B | A) 5 P(B) 3 P(A | B)
In other words, the joint probability of two events is equal to the probability of one event multiplied by the conditional probability of the other event, given that the fi rst event occurs.
Th e Venn diagram in Figure 7.12 shows the intersection of A and B, which is equivalent to the probability of A times the probability of B given A (Figure 7.13), or the probability of B times the probability of A given B (Figure 7.14).
To illustrate the general rule of multiplication, let’s use information supplied by the Bessmer (Michigan) Hospital, a nonprofi t health care facility; the data are shown in Table 7.4. Th e Bessmer Hospital offi cials collected these data because they perceived that their workload was much higher in the summer months (refl ecting the fact that babies are conceived in the winter months when recreation activities in Bessmer are fairly limited).
Figure 7.12
P(A > B)
P(A) P(B)
Figure 7.13
P(BlA) P(A)
What is the probability that a woman is at least 3 months pregnant and it is winter? From the table, we can see the probability is .42 (216 4 511). Th e general rule of multiplication says we can also fi nd this probability by multi- plying the probability that a woman is at least 3 months pregnant (.53, or 269 4 511) by the probability that it is winter, given that a woman is at least 3 months pregnant (.80, or 216 4 269). (Not every example can make sense!) Th e product of these probabilities is .42.
When two events are independent, a special rule of multiplication applies.
Two events are independent if the probability of one event is not aff ected by whether or not the other event occurs. For example, when a coin is fl ipped twice, the probability of obtaining a head on the second fl ip is independent of the result of the fi rst fl ip. Th e probability remains .5 regardless of what happened on the fi rst fl ip. Statisticians say two events A and B are independent if P(A|B) 5 P(A).
If the events are independent, it also follows that P(B|A) 5 P(B).
When two events are independent, the probability of two events occurring is equal to the probability that one event will occur times the probability that the other event will occur. Th at is,
P(A y B) 5 P(A) 3 P(B)
If you think about it, the formula for the probability of two independent events occurring is an extension of the general rule of multiplication explained earlier.
Th e multiplication rule states P(A y B) 5 P(A) 3 P(B|A). But when events are independent, P(B) 5 P(B|A), so in this situation (only), the formula becomes easier: P(A y B) 5 P(A) 3 P(B).
Figure 7.14
P(AlB)
P(B)
Table 7.4 Case Data on Bessmer Hospital Births Time of Year
October–March April–September Total
Women pregnant 3 months or more 216 53 269
Women pregnant less than 3 months 47 195 242
Total 263 248 511
Th e probability of independent events can be illustrated by the probability of obtaining two heads on consecutive fl ips of an unbiased coin. Th e probability of obtaining two heads is equal to the probability of obtaining a head on the fi rst fl ip times the probability of obtaining a head on the second (.5 3 .5 5 .25). Note that the probability of a head remains the same on the second fl ip [P(B|A) 5 P(B) 5 .5], regardless of what happened on the fi rst fl ip. Note, too, that these probabilities cor- respond to the ones you calculated earlier in the chapter for coin fl ips (see Table 7.1 and Figure 7.2).
Chapter Summary
Th is chapter provided a basic introduction to probability. Probability provides a numerical estimate of how likely it is that particular events will occur. Concepts introduced include the basic law of probability, probability distribution, a priori probability, posterior probability, the intersection of probabilities, the union of probabilities, joint probability, conditional probability, and the rules of addition and multiplication.
The basic law of probability (for equally likely events) states that the probability of an outcome is equal to the ratio of the number of ways that the outcome could occur to the total number of ways all possible outcomes could occur. A probability that can be determined before an event occurs is an a priori probability. For example, the probability of rolling a 6 on a die can be determined logically before the event. By contrast, posterior probabilities are those gener- ated by numerous trials. Th ese are empirical probabilities, which require that the public or nonprofi t manager gather the necessary data and calculate the relative frequency of occurrence. For example, the probability of earning a promotion within 2 years in local nonprofit agencies would have to be determined by obtaining the data and performing the calculations described in the chapter.
Th e union of two probabilities is the probability that one event or the other will occur. The probability of two events occurring simultaneously is a joint probability. A conditional probability is the probability that one event will occur, given that another event has already happened.
The general rule of addition for two events A and B is as follows: The probability of either A or B occurring is equal to the probability of A plus the probability of B minus the joint probability of A and B (the subtraction is neces- sary to eliminate double counting). Th is rule can be extended to more than two events. Th e general rule of multiplication for two events A and B states that the joint probability of A and B is equal to the probability of one event times the conditional probability of the other event, given that the fi rst event has occurred.
Th is rule can also be extended to more than two events.
Two events are independent if the probability of one event is not aff ected by whether or not the other event occurs. In that situation, the probability of both events occurring is found by multiplying the probability of one event times the probability of the other.
Problems
7.1 Answer the following questions by referring to the accompanying table, which lists the highway patrol’s reasons for stopping cars and the numbers of tickets issued.
Reason for Stopping Car
Issued Ticket
Did Not Issue Ticket Total
Speeding 40 170 210
No taillights 10 35 45
Failure to use signals 5 25 30
Careless driving 45 70 115
Total 100 300 400
Assume that a car was stopped.
(a) What is the probability that a ticket was issued?
(b) What is the probability that the car did not have taillights?
(c) What is the probability that a driver will get a ticket, given that he or she was stopped for careless driving?
(d) What is the probability that the person was stopped for speeding or for failure to use signals?
(e) Given that a ticket was not issued, what is the probability that the person was stopped for speeding?
7.2 David Morgan, the city manager of Yukon, Oklahoma, must negotiate new contracts with both the fi refi ghters and the police offi cers. He plans to off er both groups a 7% wage increase and hold fi rm. Mr. Morgan feels that there is one chance in three that the fi refi ghters will strike and one chance in seven that the police will strike. Assume that the events are independent.
(a) What is the probability that both will strike?
(b) What is the probability that neither the police nor the firefighters will strike?
(c) What is the probability that the police will strike and the firefighters will not?
(d) What is the probability that the firefighters will strike and the police will not?
7.3 What do we mean when we say that two events are independent?
7.4 In your own words, explain what mutually exclusive events are.
7.5 Th e probability that a smallpox case will be found in Metro City in any given week is .0034. In the week of July 17, four unrelated cases of smallpox are reported. If these are independent events, what is the probability that this would occur? Are these likely to be independent events?
7.6 Th e National Crop Reporting Service receives the accompanying information from the Central Intelligence Agency (CIA) on the Russian wheat harvest.
Russian demand for wheat is approximately 210 million metric tons.
(a) What is the probability that the harvest will be adequate?
(b) If U.S. exports make up the diff erence, what is the probability that the United States will have to sell more than 20 million metric tons to the Russians?
Millions of Metric Tons Probability
Less than 170 .05
170–180 .15
180–190 .23
190–200 .31
200–210 .15
210–220 .07
More than 220 .04
7.7 A successful launch of an intercontinental ballistic missile (ICBM) requires three steps: the signal to fi re must reach the missile team, the ignition mechanism must work properly, and the missile must not explode prematurely. Th e Department of Defense estimates that the probability that the signal to fi re an ICBM will reach a missile team is .96. For any given missile, the probability that the ignition mechanism will work is .89. For missiles with operative ignition systems, 15 of every 100 will explode in the silo or before reaching orbit. Given the command to fi re, what is the probability of a successful launch for any given missile?
7.8 From a survey of the households in Grayson, Missouri, the crime fi gures shown in the accompanying table were estimated.
Crime Number Reported Number Not Reported Total
Murder 12 12 24
Robbery 145 105 250
Assault 85 177 262
Rape 12 60 72
Auto theft 314 62 376
Total 568 416 984
(a) What is the probability that a crime is reported?
(b) What is the probability that a crime is reported, given that an assault occurs?
(c) What is the probability that a nonreported crime is either a robbery or an assault?
(d) What is the probability that a crime is a rape and that it is reported?
(e) What is the probability that a crime is reported or that the crime is a robbery?
7.9 Th e city water department estimates that for any day in January the probability of a water main freezing and breaking is .2. For six consecutive days, no water mains break.
(a) What is the probability that this happens?
(b) Are the events independent?
7.10 Jane Watson is running a youth recreation program to reduce the juvenile crime rate. If Jane can get a Law Enforcement Assistance Administration (LEAA) grant to expand the program, she feels that there is a .9 probability that the program will work. If she fails to get the grant, the probability of success falls to .3. If the probability of getting the LEAA grant is .6, what is the probability that Jane’s program will be successful?
7.11 Given P(A) 5 .45, P(B) 5 .31, and P(A y B) 5 .26, calculate:
(a) P(A x B) (b) P(A|B) (c) P(B|A)
7.12 Given P(A) 5 .21, P(B|A) 5 .75, and P(A|B) 5 .41, calculate:
(a) P(A y B) (b) P(B)
7.13 Given P(A) 5 .3, P(B) 5 .61, and P(A|B) 5 .3, calculate:
(a) P(A y B) (b) P(A x B)
7.14 Sheriff Joe Bob Stewart thinks the odds of any law enforcement grant being funded are .25. He believes that the decision to fund one grant is independent of the decision to fund any other grant.
(a) Use a probability tree to tell Sheriff Stewart what the probability is that he can submit three grants and get none funded.
(b) What is the probability of getting exactly one funded?
(c) Exactly two?
(d) Two or more?
8
T
he most common probability distribution is the normal distribution.The normal distribution describes a great many phenomena: people’s heights and weights, scores on numerous tests of physical dexterity and mental c apability, psychological attributes, and so forth. Given this great range of phenomena, the normal probability distribution is very useful to managers in public and nonprofi t organizations. Th is chapter elaborates the characteris- tics of the normal distribution and the calculation and use of standard normal scores or “z scores.” It explains how to use the normal distribution table at the end of this book (Table 1) to determine and interpret probabilities and discusses the application of the normal distribution to problems of public and nonprofi t administration.