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hebinomial probability distribution provides a method for estimating probability when events that concern the public or nonprofi t manager take on certain characteristics. Th e binomial allows the manager to determine the probability that an event will occur a specifi ed number of times in a certain number of trials. For example, if certain conditions are met, you can use the binomial distribution to estimate the probability that the mail will be delivered before a specifi ed time every day for a week or the probability that a certain piece of equipment will remain operational throughout a 10-day inspection period. As these examples suggest, the binomial is a discrete distribution that deals with the probability of observing a certain number of events of interest in a set number of repeated trials. In this chapter, we discuss the characteristics and use of the bino- mial probability distribution.on previous fl ips. Th e probability of a fi re occurring in a community on a given night is not aff ected by whether a fi re occurred in the community the previous night (assuming that there are no pyromaniacs living in or visiting the city).
Any process that exhibits these two characteristics is a Bernoulli process.
When a Bernoulli process exists, the probability that any number of events will occur can be determined by using the binomial probability distribution.
To determine a probability by using the binomial probability distribution, you must know three things. First, you must know the number of trials—that is, the number of times a certain event is tried. Second, you must know the number of successes—that is, the number of times you achieve or want to achieve a given event. Th ird, you must know the probability that the event in question will occur in any trial. Th is probability might be known a priori (such as rolling a 6 on a die), or you might have to calculate it (for example, from the rates of equipment failures).
With the knowledge of these three factors, the formula that follows can be used to calculate the probability of any one event for the binomial probability distribution:
Crnprqn2r where
n 5 the number of trials r 5 the number of successes
p 5 the probability that the event will be a success q 5 1 2 p
Although this formula looks complex, in actuality it is not. Th e symbol Crn refers to what some statisticians call a combination (read as “a combination of n things taken r at a time”). To illustrate a combination and its value, let us assume that we have four balls marked with the letters a, b, c, and d, respectively. We want to know how many diff erent sets of three balls we could select from the four. In statistical language, we want to know the number of combinations of four balls taken three at a time. As the following illustration shows, a combination of four things taken three at a time equals 4; that is, there are four diff erent combina- tions of four things taken three at a time.
Combination Balls Selected
1 a, b, c
2 a, b, d
3 a, c, d
4 b, c, d
Th ese four combinations of three balls represent all the possible ways that four items can be grouped into sets of three. In the space provided on page 153 use the same procedure to determine the value of a combination of four things taken two at a time.
If you found six combinations, (ab, ac, ad, bc, bd, cd ), congratulations.
Listing all possible combinations as a way to fi gure out the value of Crn often gets burdensome. For example, the following combination would take a long time to calculate by the combination-listing method:
C156
Th e number of combinations of 15 things taken six at a time will be very large.
To simplify, a shortcut method of determining the number of combinations has been found using the formula
Crn5 n!
r!1n2r2!
where n! (called n factorial) is equal to n 3 (n 2 1) 3 (n 2 2) 3 (n 2 3) 3 . . . 3 3 3 2 3 1; r ! and (n 2 r)! have the same interpretation. For example, 6 ! 5 6 3 5 3 4 3 3 3 2 3 1 5 720.
In the preceding example, we fi nd C156 5 15!
6!9!
15!
5 15314313312311310393837363534333231 1635343332312 3 1938373635343332312
6! 9!
5 5,005
Th e use of the binomial distribution can best be illustrated by working a prob- lem. Assume that we are going to fl ip an unbiased coin three times, and we want to know the probability of getting exactly three heads. In this example, the num- ber of trials (n) is equal to 3, the number of coin fl ips. Th e number of successes (r) is equal to 3, the number of heads. Th e probability of obtaining a head on one fl ip of a coin (p) is equal to .5; q, then, is equal to 1 2 p, or .5. Recall that in a Bernoulli process, only two outcomes are possible. If the probability of success is p, then the probability of failure (i.e., nonsuccess) must be 1 2 p 5 q. Substitut- ing these numbers into the binomial probability distribution formula, we get
Crnprqn2r5 a 3!
3!0!b3.533.505.125
Calculating the combination term fi rst, we fi nd that the numerator (top) of the fraction is equal to 3! (3 3 2 3 1), or 6. Th is number is divided by the denomi- nator (bottom) of the fraction, 3!, or 6, times 0!. To keep the universe orderly,
⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
⎫⎪⎪ ⎪ ⎪⎬⎪⎪⎪ ⎪⎭ ⎫⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪⎪ ⎪ ⎪ ⎪⎪ ⎪⎭
statisticians defi ne 0! as equal to 1. Th e combination term, therefore, is 6 divided by 6, or 1; that is, there is only one way in which you could fl ip a coin three times and get three heads (the sequence head-head-head). Th e probability terms are easily calculated: .53 is equal to .5 3 .5 3 .5, or .125, and .50 is equal to 1.
Note again that statisticians defi ne any number raised to the 0 power as equal to 1 to maintain an orderly universe. Th e probability of obtaining three heads in three fl ips of an unbiased coin, then, is 1 3 .125 3 1, or.125.
To test yourself, calculate the probability that in four fl ips of an unbiased coin, exactly two heads will occur. Calculate in the space provided.
You should have found a probability of .375 that two heads would occur on four fl ips of an unbiased coin. Th e problem may be defi ned as one where n 5 4, r 5 2, and p 5 .5 (the number of trials, the number of successes, and the prob- ability of success, respectively). Again, q (probability of tails) 5 1 2 p 5 .5. Th e probability can be calculated as follows:
Crnprqn2r5 a 4!
2!2!b 3.523.52 5 a4333231
2313231b 3.253.25 563.253.25
5.375
Suppose you want to know the probability of obtaining two or more heads on four fl ips of an unbiased coin. In this situation, you would calculate the probabil- ity of obtaining four heads (.0625), the probability of obtaining three heads (.25), and the probability of obtaining two heads (.375). Because these three events are mutually exclusive, the probability of two, three, or four heads occurring is equal to the probability of two heads plus the probability of three heads plus the prob- ability of four heads, or .6875. Th e calculations for two heads appear earlier; the calculations for three and four heads follow.
Three heads: C443.533.515 a4!
3!1!b3.1253.5 5 a4333231
3323131b3.0625 543.0625
5.25
Four heads: C443.543.505 a 4!
4!0!b 3.062531 5 a4333231
433323131b 3.0625 513.0625
5.0625
An example of how the binomial probability distribution can be used in a public management setting is in order. Suppose that the Stermerville Public Works Depart- ment has been charged with racial discrimination in hiring practices. Last year, 40%
of the persons who passed the department’s civil service exam and were eligible to be hired were minorities. From this group, the Public Works Department hired 10 individuals; 2 were minorities. What is the probability that if the Stermerville Public Works Department did not discriminate, it would have hired two or fewer minori- ties? (Note that we assume, as a personnel analyst would have to in this situation, that anyone who passed the exam was capable of successful job performance. In other words, we assume that every individual had the same chance to be hired.)
Th e easiest way to solve a problem of this nature is to identify n, r, p, and q fi rst and then perform the necessary calculations. In this case p, the probability that a minority is hired, is equal to the proportion of minorities in the job pool, or .4. And q, of course, is equal to 1 2 p 5 1 2 .4 5 .6. Th e number of trials (n) in this case is equal to the number of persons hired, or 10. Th e number of successes (r) is equal to the number of minorities selected in this group, or 2.We are interested in the prob- ability of two or fewer minorities being selected, so we must calculate the binomial probability distribution for two minorities, one minority, and zero minorities.
Th e probability calculations for two minorities are as follows:
Crnprqn2r5 a10!
2!8!b 3.423.685453.423.685453.163.68 57.23.6857.23.01685.120
Th e probability calculations for one minority are as follows:
C101 3.413.695 a10!
1!9!b3.43.695103.43.69543.69543.0105.040 Th e probability calculations for zero minorities are as follows:
C100 3.403.6105a 10!
0!10!b3.403.610513.403.61051313.6105.006 Statistically, a person can conclude that the probability that the Stermer- ville Public Works Department could select two or fewer minorities out of 10 individuals hired from a pool that consists of 40% minorities is .166 (.120 1 .040 1 .006 5 .166) if the department shows no preference in regard to hiring minorities. In other words, a pattern such as this could happen about 1 time in 6.
Th is is a statistical statement. As a manager, you have to arrive at managerial conclusions. You need to know whether the department is engaged in discrimi- nation. Th e statistical fi nding is only one piece of evidence that a manager uses in arriving at a conclusion.
After receiving statistical information, the manager should always ask whether anything else could have produced a result similar to this, other than discrimination (or whatever other question is involved). For example, perhaps more minorities were off ered jobs, but they turned them down for some reason.
Sometimes it is helpful to examine past behavior. If last year the department hired 42% minorities, you might not be too concerned. If it hired only 20% minorities last year, you might become more concerned. Th e manager’s task at this point is to examine all potential reasons why the statistical pattern occurred and eliminate them. Can you think of other causes for these statistical results?
If no other reason (other than discrimination) can be found to account for the statistical result (p 5 .166), then the manager must decide how sure he or she wants to be. Because this pattern could have occurred by random chance 16.6%
of the time, if the manager concludes that discrimination has occurred, there is a 1 in 6 chance that he or she is wrong. Th ere is no magic number for how sure a manager should be. Social scientists often use a probability of less than 5% for making decisions. Th is is generally not a good idea for a manager. In some cases, a 5% risk is far too high. For example, would you authorize the launch of the space shuttle if there were a 5% chance of an explosion? Such a decision would mean that the manager would fi nd it acceptable if 1 of every 20 launches resulted in an explosion. Other decisions might not require levels as low as .05. A probability of .25, or even .35, might be acceptable in deciding between two alternative meth- ods of collecting trash. In short, the manager must take the responsibility for establishing a level of risk based on how important the decision is. Th e manager should never let his or her statistical analyst set the probability level. Th is decision requires managerial judgment based on an individual’s experience and willingness to take risks. Managers would do well to remember that statistical analysts are rarely fi red when the manager they report to makes a bad decision. In Chapter 12 we revisit the issue of how sure the manager should be.
Although the binomial probability distribution provides a good method for estimating the probability that a given number of events will occur, often the task of calculating the binomial probability distribution becomes fairly complex mathematically. For example, suppose that instead of hiring 2 minorities out of 10 persons, the Stermerville Public Works Department hired 43 minorities out of 150 persons. Th ese fi gures would require the following probability to be estimated:
C15043 3.4433.6107
Th e estimation of this term for the probability of hiring exactly 43 minorities is complex. In addition, because the binomial is a discrete probability distribution, we would need to estimate the probability of hiring 42, 41, 40. . . 0 minorities out
of 150. Clearly the diffi culty in estimating this probability exceeds the patience of most individuals. In situations like this, however, the binomial probability dis- tribution can be estimated fairly well by using the continuous normal probability distribution discussed in Chapter 8.