# 12.2 - Finding Poisson Probabilities

12.2 - Finding Poisson Probabilities

## Example 12-2

Let $$X$$ equal the number of typos on a printed page with a mean of 3 typos per page. What is the probability that a randomly selected page has at least one typo on it?

We can find the requested probability directly from the p.m.f. The probability that $$X$$ is at least one is:

$$P(X\ge 1)=1-P(X=0)$$

Therefore, using the p.m.f. to find $$P(X=0)$$, we get:

$$P(X \geq 1)=1-\dfrac{e^{-3}3^0}{0!}=1-e^{-3}=1-0.0498=0.9502$$

That is, there is just over a 95% chance of finding at least one typo on a randomly selected page when the average number of typos per page is 3.

What is the probability that a randomly selected page has at most one typo on it?

The probability that $$X$$ is at most one is:

$$P(X\le 1)=P(X=0)+P(X=1)$$

Therefore, using the p.m.f., we get:

$$P(X \leq 1)=\dfrac{e^{-3}3^0}{0!}+\dfrac{e^{-3}3^1}{1!}=e^{-3}+3e^{-3}=4e^{-3}=4(0.0498)=0.1992$$

That is, there is just under a 20% chance of finding at most one typo on a randomly selected page when the average number of typos per page is 3.

Just as we used a cumulative probability table when looking for binomial probabilities, we could alternatively use a cumulative Poisson probability table, such as Table III in the back of your textbook. You should be able to use the formulas as well as the tables. If you take a look at the table, you'll see that it is three pages long. Let's just take a look at the top of the first page of the table in order to get a feel for how the table works:

In summary, to use the table in the back of your textbook, as well as that found in the back of most probability textbooks, to find cumulative Poisson probabilities, do the following:

1. Find the column headed by the relevant $$\lambda$$. Note that there are three rows containing $$\lambda$$ on the first page of the table, two rows containing $$\lambda$$ on the second page of the table, and one row containing $$\lambda$$ on the last page of the table.
2. Find the $$x$$ in the first column on the left for which you want to find $$F(x)=P(X\le x)$$.

Let's try it out on an example. If $$X$$ equals the number of typos on a printed page with a mean of 3 typos per page, what is the probability that a randomly selected page has four typos on it?

#### Solution

The probability that a randomly selected page has four typos on it can be written as $$P(X=4)$$. We can calculate $$P(X=4)$$ by subtracting $$P(X\le 3)$$ from $$P(X\le 4)$$. To find $$P(X\le 3)$$ and $$P(X\le 4)$$ using the Poisson table, we:

1. Find the column headed by $$\lambda=3$$.
2. Find the 3 in the first column on the left, since we want to find $$F(3)=P(X\le 3)$$. And, find the 4 in the first column on the left, since we want to find $$F(4)=P(X\le 4)$$.

Now, all we need to do is, first, read the probability value where the $$\lambda=3$$ column and the $$x=3$$ row intersect, and, second, read the probability value where the $$\lambda=3$$ column and the $$x=4$$ row intersect. What do you get?

$$\lambda = E(X)$$
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0 0.905 0.819 0.741 0.670 0.607 0.549 0.497 0.449 0.407 0.368
1 0.995 0.982 0.963 0.938 0.910 0.878 0.844 0.809 0.772 0.736
2 1.000 0.999 0.996 0.992 0.986 0.970 0.966 0.953 0.937 0.920
3 1.000 1.000 1.000 0.999 0.998 0.997 0.994 0.991 0.987 0.981
4 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.998 0.996
5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999
6 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

x 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
0 0.333 0.301 0.273 0.247 0.223 0.202 0.183 0.165 0.150 0.135
1 0.699 0.663 0.627 0.592 0.558 0.525 0.493 0.463 0.434 0.403
2 0.900 0.879 0.857 0.833 0.809 0.830 0.757 0.731 0.704 0.677
3 0.974 0.966 0.957 0.946 0.934 0.921 0.907 0.981 0.875 0.857
4 0.995 0.992 0.989 0.986 0.981 0.976 0.970 0.964 0.956 0.947
5 0.999 0.998 0.998 0.997 0.996 0.994 0.992 0.990 0.987 0.983
6 1.000 1.000 1.000 0.999 0.999 0.999 0.998 0.997 0.997 0.995
7 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.999
8 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

x 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0
0 0.111 0.091 0.074 0.061 0.050 0.051 0.033 0.027 0.022 0.018
1 0.355 0.308 0.267 0.231 0.199 0.171 0.147 0.126 0.107 0.092
2 0.623 0.570 0.518 0.469 0.423 0.380 0.340 0.303 0.269 0.238
3 0.819 0.779 0.736 0.692 0.647 0.603 0.558 0.515 0.473 0.433
4 0.928 0.904 0.887 0.848 0.815 0.781 0.744 0.706 0.668 0.629
5 0.975 0.964 0.951 0.935 0.916 0.895 0.871 0.844 0.816 0.785
6 0.993 0.988 0.983 0.976 0.966 0.955 0.942 0.927 0.909 0.889
7 0.998 0.997 0.995 0.992 0.988 0.983 0.977 0.969 0.960 0.949
8 1.000 0.990 0.999 0.998 0.993 0.994 0.992 0.988 0.984 0.979
9 1.000 1.000 1.000 0.999 0.999 0.998 0.997 0.996 0.994 0.992
10 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.998 0.997
11 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999
12 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
$$\lambda = E(X)$$
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
0 0.905 0.819 0.741 0.670 0.607 0.549 0.497 0.449 0.407 0.368
1 0.995 0.982 0.963 0.938 0.910 0.878 0.844 0.809 0.772 0.736
2 1.000 0.999 0.996 0.992 0.986 0.970 0.966 0.953 0.937 0.920
3 1.000 1.000 1.000 0.999 0.998 0.997 0.994 0.991 0.987 0.981
4 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.998 0.996
5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999
6 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

x 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
0 0.333 0.301 0.273 0.247 0.223 0.202 0.183 0.165 0.150 0.135
1 0.699 0.663 0.627 0.592 0.558 0.525 0.493 0.463 0.434 0.403
2 0.900 0.879 0.857 0.833 0.809 0.830 0.757 0.731 0.704 0.677
3 0.974 0.966 0.957 0.946 0.934 0.921 0.907 0.981 0.875 0.857
4 0.995 0.992 0.989 0.986 0.981 0.976 0.970 0.964 0.956 0.947
5 0.999 0.998 0.998 0.997 0.996 0.994 0.992 0.990 0.987 0.983
6 1.000 1.000 1.000 0.999 0.999 0.999 0.998 0.997 0.997 0.995
7 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.999
8 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

x 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0
0 0.111 0.091 0.074 0.061 0.050 0.051 0.033 0.027 0.022 0.018
1 0.355 0.308 0.267 0.231 0.199 0.171 0.147 0.126 0.107 0.092
2 0.623 0.570 0.518 0.469 0.423 0.380 0.340 0.303 0.269 0.238
3 0.819 0.779 0.736 0.692 0.647 0.603 0.558 0.515 0.473 0.433
4 0.928 0.904 0.887 0.848 0.815 0.781 0.744 0.706 0.668 0.629
5 0.975 0.964 0.951 0.935 0.916 0.895 0.871 0.844 0.816 0.785
6 0.993 0.988 0.983 0.976 0.966 0.955 0.942 0.927 0.909 0.889
7 0.998 0.997 0.995 0.992 0.988 0.983 0.977 0.969 0.960 0.949
8 1.000 0.990 0.999 0.998 0.993 0.994 0.992 0.988 0.984 0.979
9 1.000 1.000 1.000 0.999 0.999 0.998 0.997 0.996 0.994 0.992
10 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.998 0.997
11 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999
12 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

The cumulative Poisson probability table tells us that finding $$P(X\le 4)=0.815$$ and $$P(X\le 3)=0.647$$. Therefore:

$$P(X=4)=P(X\le 4)-P(X\le 3)=0.815-0.647=0.168$$

That is, there is about a 17% chance that a randomly selected page would have four typos on it. Since it wouldn't take a lot of work in this case, you might want to verify that you'd get the same answer using the Poisson p.m.f.

What is the probability that three randomly selected pages have more than eight typos on it?

#### Solution

Solving this problem involves taking one additional step. Recall that $$X$$ denotes the number of typos on one printed page. Then, let's define a new random variable $$Y$$ that equals the number of typos on three printed pages. If the mean of $$X$$ is 3 typos per page, then the mean of $$Y$$ is:

$$\lambda_Y=3 \text{ typos per one page }\times 3\text{ pages }=9 \text{ typos per three pages}$$

Finding the desired probability then involves finding:

$$P(Y>8)=1-P(Y\le 8)$$

where $$P(Y\le 8)$$ is found by looking on the Poisson table under the column headed by $$\lambda=9.0$$ and the row headed by $$x=8$$. What do you get?

x 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0
0 0.002 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000
1 0.011 0.007 0.005 0.003 0.002 0.001 0.001 0.000 0.000 0.000
2 0.043 0.030 0.020 0.014 0.009 0.006 0.004 0.003 0.002 0.001
3 0.112 0.082 0.059 0.042 0.030 0.021 0.015 0.010 0.007 0.005
4 0.224 0.173 0.132 0.100 0.074 0.055 0.040 0.029 0.021 0.015
5 0.369 0.301 0.241 0.191 0.150 0.116 0.089 0.067 0.050 0.015
6 0.527 0.450 0.378 0.313 0.256 0.207 0.165 0.130 0.102 0.079
7 0.673 0.599 0.525 0.453 0.386 0.324 0.269 0.220 0.179 0.143
8 0.792 0.729 0.662 0.593 0.523 0.456 0.392 0.333 0.279 0.232
9 0.877 0.830 0.776 0.717 0.653 0.587 0.522 0.458 0.397 0.341
10 0.933 0.901 0.862 0.816 0.763 0.706 0.645 0.583 0.521 0.460
11 0.966 0.947 0.921 0.888 0.849 0.803 0.752 0.697 0.639 0.579
12 0.984 0.973 0.957 0.936 0.909 0.876 0.836 0.792 0.742 0.689
13 0.993 0.987 0.978 0.966 0.949 0.926 0.898 0.864 0.825 0.781
14 0.997 0.994 0.990 0.983 0.973 0.959 0.940 0.917 0.888 0.854
15 0.999 0.998 0.995 0.992 0.986 0.978 0.967 0.951 0.932 0.907
16 1.000 0.999 0.998 0.996 0.993 0.989 0.982 0.973 0.960 0.944
17 1.000 1.000 0.999 0.998 0.997 0.995 0.991 0.986 0.978 0.968
18 1.000 1.000 1.000 0.999 0.999 0.998 0.096 0.993 0.988 0.982
19 1.000 1.000 1.000 1.000 0.999 0.999 0.998 0.997 0.994 0.991
20 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.998 0.997 0.995
21 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.998
22 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999
23 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000
x 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0
0 0.002 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000
1 0.011 0.007 0.005 0.003 0.002 0.001 0.001 0.000 0.000 0.000
2 0.043 0.030 0.020 0.014 0.009 0.006 0.004 0.003 0.002 0.001
3 0.112 0.082 0.059 0.042 0.030 0.021 0.015 0.010 0.007 0.005
4 0.224 0.173 0.132 0.100 0.074 0.055 0.040 0.029 0.021 0.015
5 0.369 0.301 0.241 0.191 0.150 0.116 0.089 0.067 0.050 0.015
6 0.527 0.450 0.378 0.313 0.256 0.207 0.165 0.130 0.102 0.079
7 0.673 0.599 0.525 0.453 0.386 0.324 0.269 0.220 0.179 0.143
8 0.792 0.729 0.662 0.593 0.523 0.456 0.392 0.333 0.279 0.232
9 0.877 0.830 0.776 0.717 0.653 0.587 0.522 0.458 0.397 0.341
10 0.933 0.901 0.862 0.816 0.763 0.706 0.645 0.583 0.521 0.460
11 0.966 0.947 0.921 0.888 0.849 0.803 0.752 0.697 0.639 0.579
12 0.984 0.973 0.957 0.936 0.909 0.876 0.836 0.792 0.742 0.689
13 0.993 0.987 0.978 0.966 0.949 0.926 0.898 0.864 0.825 0.781
14 0.997 0.994 0.990 0.983 0.973 0.959 0.940 0.917 0.888 0.854
15 0.999 0.998 0.995 0.992 0.986 0.978 0.967 0.951 0.932 0.907
16 1.000 0.999 0.998 0.996 0.993 0.989 0.982 0.973 0.960 0.944
17 1.000 1.000 0.999 0.998 0.997 0.995 0.991 0.986 0.978 0.968
18 1.000 1.000 1.000 0.999 0.999 0.998 0.096 0.993 0.988 0.982
19 1.000 1.000 1.000 1.000 0.999 0.999 0.998 0.997 0.994 0.991
20 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.998 0.997 0.995
21 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999 0.998
22 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.999
23 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

The cumulative Poisson probability table tells us that finding $$P(X\le 8)=0.456$$. Therefore:

$$P(Y>8)=1-P(Y\le 8)=1-0.456=0.544$$

That is, there is a 54.4% chance that three randomly selected pages would have more than eight typos on it.

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