# Poisson Approximation to Binomial

I have a Poisson approximation to binomial question, posted below. I'm not too sure if I'm using the proper formula:

$$P(x) = e^{-np}(np)^x/x!$$ .

Of anyone can tell me if I'm doing this right, that would be great.

Again these are just practice questions not homework.

Q: A salesperson has found that the probability of making a sale on a particular product manufactured by him or her company is .05. If the salesperson contacts 140 potential customers, what is the probability he or she will sell at least 2 of these products? Use and justify Poisson approximation to Binomial.

What I'm doing:

$$P(x) = e^{-140 (.05)}(140*.05)^2/2!$$ .

• You did not correctly apply the formula you quoted. Look carefully. – Glen_b Dec 11 '13 at 22:40
• fixed * is that the correct formula though ? – user115027 Dec 11 '13 at 22:45
• The formula is the top of the question is a correct formula, for what it represents. While you corrected an error in copying the formula, you still have not applied it correctly to the question. You might want to think about the formula and the question and consider what you might have missed. (Where does this practice question come from?) – Glen_b Dec 11 '13 at 22:47
• Its on a mock test our teacher created, its on paper cant seem to find a variation on the web. – user115027 Dec 11 '13 at 22:51
• user115027 Since you suggest it, I do imagine that you might. I would not. You might do better to read about the distinction between probability functions and cdfs before tackling this question. How could you find the probability that it's less than 2? – Glen_b Dec 11 '13 at 23:04

According to Wikipedia: Poisson approximation, the use of this approximation seems ok as n=140 and np=7

Poisson approximation The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed. Therefore the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n ≥ 20 and p ≤ 0.05, or if n ≥ 100 and np ≤ 10.[9]

From wikipedia, it is clear your P(x) is a mass function where giving the probability of a particular value of x, not a cumulative distribution function giving the probability of a value below or equal to x.

To find the probability for x (the number of sales) to be at least 2, you can calculate either:

Prob{x>=2} = Sum(P(x) for integer x>=2) which is an infinitely long sum, or

Prob{x>=2} = 1 - Prob(x<2) = 1 - P(0) - P(1)


I would imagine the second method to be easier, yielding:

Prob{x>=2} = 1 - exp(-7) - 7 exp(-7) = 1 - 8 exp(-7) = 0.9927