**The Poisson Binomial Approximation Formula**

The Poisson formula can be used to approximate the probability of T successes in n binomial trials in situations where n is large and the probability of success p is small. The rule of thumb followed by most statisticians is that n is large enough and p is small enough if n is 20 or more and pis 0.05 or less; that is, n 2: 20 and p :5 0.05. In general, the larger n is and the smaller pis, the better will be the approximation. The conditions of large n and small p are often encountered in applied work. For example, examination of 100 bank balances would bring to light very few, if any, erroneous balances, because the probability that a particular balance is in error is’very small. Again, a sample of a large number of homeowners would contain few owners whose home had a major fire last year because the probability of such a fire is very small. The Poisson formula is easier to use than the binomial formula. Another important advantage is that Poisson probabilities can be tabulated more efficiently than binomial probabilities. The Poisson formula for approximating P(r), the number of successes in n binomial trials, is

where u = np e = 2.71828 The only new quantity in the Poisson formula is the constant number e, which, rounded to six digits, is 2.71828. To evaluate the Poisson formula, we need values of e to negative powers. Values for some of these powers can be found in Table II at the end of the book. There you can verify that

e-2 = 0.1353 e-3.2 = 0.0408 e-O.8 = 0.4493 Some hand calculators have a key which provides values of e to a power Of course these powers can also be found on any calculator which has a power key For example,

rounded to four digits. Hand calculators are easier to use than a table, and provide more accurate values for many more powers than a table. **EXAMPLE** Two hundred passengers have made reservations for an airplane flight. If the probability that a passenger who has a reservation will-not show up is 0.01, what is the probability that exactly three will not show up? **SOLUTION** Here a “no show” is a success. We assume that the 200 reservations constitute a sample from a large population (size unspecified) of reservations. Moreover, for the population, p = 0.01 is the probability of success.

Note that n = 200 is large enough and p = 0.01 is small enough to satisfy the rule of thumb that n should be at least 20 and p should not be greater than 0.05. We first find

For comparison, let’s compute P(3) for the example by the binomial formula

We have n = 200, P = 0.01, q = 1 – P = 1 – 0.01 = 0.99. Then

Observe that the Poisson approximation) 0.1804, is nearly the same as the binomial value, 0.1814. It is clear that the Poisson calculation is simpler than the binomial calculation. However, a greater advantage of the Poisson formula, if it is applicable, is that it has only one parameter, p., whereas the binomial distribution has two parameters, n and p; consequently, Poisson probabilities can be tabulated more compactly than binomial probabilities. For example, the Poisson probability P(3) is the same for n = 200, P = 0.01 as it is for n = 100, P = 0.02, and for any other pair of nand p values whose product is p. = n p = 2. Therefore, all Poisson distributions with the same value of p.can be tabulated in a single column of a table. An illustrative tabulation of the probabilities of exactly r successes for various values of p. is provided in Table I l I A at the end of the book. In this table, the entry for the airplane reservation example, where p. = 2 and r = 3, is found to be the Poisson value, 0:1804, computed in the example. As stated earlier, the Poisson formula provides good approximations of . binomial distributions when n ~ 20 and p s 0.05. Table 7.4 and Figure 7.7 (at the top of page 210) compare the binomial-Poisson probabilities for n = 20 and p = 0.05. As noted in Table 7.4, with n = 20, more than six successes are possible, but the probabilities for more than six successes are very small. Moreover, it should be noted that with n = 20, it is impossible to have

more than 20 successes. However, the Poisson formula, if applied with r larger than 20, will give a very small nonzero probability-so small that for practical purposes it is zero. In fact, the Poisson formula will give a nonzero probability for any value of r, no matter how large; but, again, only 0, 1, 2, 3, and a small additional number of r values have probabilities large enough to be of practical significance. **EXERCISE** (a) Five percent of the items in a large lot are defective. A sample of 60 is inspected. Write the Poisson expression for the probability that the sample contains exactly four defectives: and calculate this probability by the Poisson formula. (b) Verify your answer to (a) by referring to Table IliA. (c) Calculate the probability of exactly four defectives if 4 percent of the items in the lot are defective. **ANSWER** (a) P(4) = [e-~(3)·]/4! = 0.1680; (b) 0.1680; (c) 0.125.

Table IIIB at the end of the book contains cumulative Poisson probabilities,

E P(r)

for selected values of p.,to be used in problems. **EXAMPLE** The manager of Music, Inc., is planning to stock and advertise a very expensive stereo system. The manager expects to draw 250 interested customers and assesses the probability of selling a system to a customer at 0.02. Suppose the manager decides to stock six systems. What then would be the probability of (a) being able to satisfy customer demand? (b) not being able to satisfy customer demand? **SOLUTION** Here we have n = 250, P = 0.02; so u= n p = 250(0.02) = 5 is the mean or expected value of the number demanded

a. The probability of being able to satisfy demand is the inability that demand’ ‘;ll be from 0 through 6 systems. From Table IlIA, with u 5; this is

E Per) = 0.7622

b. The probability of not being able to satisfy demand is the probability that demand is more than 6, that is, 7 or 8 or 9 and so on. Demands from 0 through 6, and demands of more than 6, include all possible demands; so p(demand 0 through 6) + p(demand more than 6) •• 1 P(demand more than 6) = 1 – P(demand 0 through 6)

= 1 – E Per) = 1 – 0.7622 = 0.2378

**EXERCISE** In the last example, there is a rather high probability, 0.2378, of not being able to satisfy demand. What would be the probability of not being able to satisfy demand if nine stereo systems were in stock? **ANSWER** 0.0318