# Poisson or Binomial distribution?

Read the following questions and decide whether the Poisson or the Binomial distribution should be used to answer it. Calculate the required probabilities. As a guide,

• If a mean or average probability of an event happening per unit time/per page/per mile cycled etc., is given, and you are asked to calculate a probability of n events happening in a given time/number of pages/number of miles cycled, then the Poisson Distribution is used.
• If, on the other hand, an exact probability of an event happening is given, or implied, in the question, and you are asked to caclulate the probability of this event happening k times out of n, then the Binomial Distribution must be used.
1. A typist makes on average 2 mistakes per page. What is the probability of a particular page having no errors on it?
2. A computer crashes once every 2 days on average. What is the probability of there being 2 crashes in one week?
3. Components are packed in boxes of 20. The probability of a component being defective is 0.1. What is the probability of a box containing 2 defective components?
4. ICs are packaged in boxes of 10. The probability of an ic being faulty is 2%. What is the probability of a box containing 2 faulty ics?
5. The mean number of faults in a new house is 8. What is the probability of buying a new house with exactly 1 fault?
6. A box contains a large number of washers; there are twice as many steel washers as brass ones. Four washers are selected at random from the box. What is the probability that 0, 1, 2, 3, 4 are brass?

1. A typist makes on average 2 mistakes per page. What is the probability of a particular page having no errors on it?

```We have an average rate here: lambda = 2 errors per page.
We don't have an exact probability (e.g. something like "there is a
probability of 1/2 that a page contains errors").
Hence, Poisson distribution.
(lambda t) = (2 errors per page * 1 page) = 2.
Hence P0 = 2^0/0! * exp(-2) = 0.135.
```

2. A computer crashes once every 2 days on average. What is the probability of there being 2 crashes in one week?

```Again, average rate given: lambda = 0.5 crashes/day. Hence, Poisson.
(lambda t) = (0.5 per day * 7 days) = 3.5/week and n = 2.
P2 = (3.5)^2/2! * exp (-3.5) = 0.185.
```

3. Components are packed in boxes of 20. The probability of a component being defective is 0.1. What is the probability of a box containing 2 defective components?

```Here we are given a definite probability, in this case, of
defective components, p = 0.1 and hence q = 0.9 = Prob. not
defective. Hence, Binomial, with n = 20.  Expand (q + p)^20
to get

q^20 + 20 q^19 p + 20(20-1)/2! q^18 p^2 + ...

0      1              2                   No. faulty

So P(2) = 20(20-1)/2! q^18 p^2 = 0.285.
```

4. ICs are packaged in boxes of 10. The probability of an ic being faulty is 2%. What is the probability of a box containing 2 faulty ics?

```We have a probability of something being true and the
same thing not being true; in this case, an ic being
faulty. Hence, Binomial distribution.
p = Prob. faulty = 0.02, q = Prob. not faulty = 0.98.
n = 10. Expand (q + p)^10 to get

q^10 + 10 q^9 p + 10(10-1)/2! q^8 p^2 + ...

0      1              2        No. of faulty ics.

So, Prob of a box containing 2 faulty ics
P2 = 10(10-1)/2! q^8 p^2 = 0.015.
```

5. The mean number of faults in a new house is 8. What is the probability of buying a new house with exactly 1 fault?

```Here we have an average rate of faults occurring: 8
per house.  Hence, Poisson, with
(lambda t) = (8 faults/house * 1 house) = 8.
[1 house because we're only buying one new house.]
n = 1 too, so P1 = 8^1/1! * exp(-8) = 0.0027.
```

6. A box contains a large number of washers; there are twice as many steel washers as brass ones. Four washers are selected at random from the box. What is the probability that 0, 1, 2, 3, 4 are brass?

```Here too we have a probability of brass (1/3) and of not brass
--- i.e. steel --- which is 2/3. Hence, use the Binomial
distribution with p = 1/3, q = 2/3 and n = 4 to get

(p + q)^4 = p^4 + 4 p^3 q + 6 p^2 q^2 + 4 p q^3 + q^4

4         3        2        1          0   No. of brass

so P(0) = (2/3)^4 = 0.197, P(1) = 4 (1/3)(2/3)^3 = 0.395,
P(2) = 6 (1/3)^2 (2/3)^2 = 0.296, P(3) = 4 (1/3)^3 (2/3) = 0.099
and P(4) = 0.012.

```

J.Deane@surrey.ac.uk
January 2004