In Exercises 9–16, use the Poisson distribution to find the indicated probabilities.

Disease Cluster Neuroblastoma, a rare form of cancer, occurs in 11 children in a million, so its probability is 0.000011. Four cases of neuroblastoma occurred in Oak Park, Illinois, which had 12,429 children.

a. Assuming that neuroblastoma occurs as usual, find the mean number of cases in groups of 12,429 children.

b. Using the unrounded mean from part (a), find the probability that the number of neuroblastoma cases in a group of 12,429 children is 0 or 1.

c. What is the probability of more than one case of neuroblastoma?

d. Does the cluster of four cases appear to be attributable to random chance? Why or why not?

It is given that 4 cases of neuroblastoma have occurred in the population of 12429 children in Oak Park, Illinois.

a.

The probability ofoccurrence of neuroblastoma cases is equal to p=0.000011.

The number of children in the group is equal to 12429.

The mean number of cases in Illinois:

\(\begin{aligned}{c}\mu = np\\ = 12429\left( {0.000011} \right)\\ = 0.137\end{aligned}\)

Thus, the mean number of cases of neuroblastoma is equal to 0.137.

b.

Let X be the number of cases in a group of 12429 children.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 0.137\).

The probability that the number of neuroblastoma cases in a group of 12429 children is equal to 0 or 1 is computed below:

\(P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\)

\(\begin{aligned}{c}P\left( {0\;{\rm{or}}\;1} \right) = P\left( 0 \right) + P\left( 1 \right)\\ = \frac{{{{\left( {0.137} \right)}^0}{{\left( {2.71828} \right)}^{ - 0.137}}}}{{0!}} + \frac{{{{\left( {0.137} \right)}^1}{{\left( {2.71828} \right)}^{ - 0.137}}}}{{1!}}\\ = 0.87197 + 0.11946\\ = 0.991430\end{aligned}\)

\( \approx 0.991\)

Therefore, the probability that the number of neuroblastoma cases in a group of 12429 children is equal to 0 or 1 is equal to 0.991.

c.

The probability of more than 1 case is computed below:

\(\begin{aligned}{c}P\left( {X > 1} \right) = 1 - P\left( {X \le 1} \right)\\ = 1 - \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right)} \right)\\ = 1 - 0.991\\ = 0.009\end{aligned}\)

Therefore, the probability that the number of neuroblastoma cases in a group of 12429 children is more than 1 is equal to 0.009.

d.

It can be seen that the probability of more than 1 case is extremely small.

Thus, the probability of 4 cases will be even smaller.

Therefore, it can be concluded that the occurrence of 4 cases cannot be considered a random event because the probability of the event happening is negligible.