Question

Making high-stakes insurance decisions. The Journal of Economic Psychology (September 2008) published the results of a high-stakes experiment in which subjects were asked how much they would pay for insuring a valuable painting. The painting was threatened by fire and theft, hence, the need for insurance. To make the risk realistic, the subjects were informed that if it rained on exactly 24 days in July, the painting was considered to be stolen; if it rained on exactly 23 days in August, the painting was considered to be destroyed by fire. Although the probability of these two events, “fire” and “theft,” was ambiguous for the subjects, the researchers estimated their probabilities of occurrence at .0001. Rain frequencies for the months of July and August were shown to follow a Poisson distribution with a mean of 10 days per month.

a. Find the probability that it will rain on exactly 24 days in July.

b. Find the probability that it will rain on exactly 23 days in August.

c. Are the probabilities, parts a and b, good approximations to the probabilities of “fire” and “theft”?

Short answer

a. The probability of rain for exactly 24 days in July is, 0.00007317

b. The probability of rain for exactly 23 days in August is, 0.000175661

c. Part(a) is a good approximation and is better than part(b)

Step-by-step solution

Given information

By the Journal of Economic Psychology (September 2008),

The rain frequencies of July and August follow a Poisson distribution with a mean of 10

The estimated probability of occurrence of two events fire and theft is 0.0001

Finding the probability

a.

x is a Poisson distribution with a mean of 10,

Here, rain on exactly 24 days in July, i.e., $X=24\mathrm{and}\lambda =10$

$$\begin{gathered} P\left(x=24\right)=\frac{e^{-\lambda }{\lambda }^x}{x!} \\ =\frac{e^{-10}{10}^{24}}{24!} \\ =0.00007317 \\ P\left(x=24\right)=0.00007317 \end{gathered}$$

Therefore, the probability of rain for exactly 24 days in July is, 0.00007317

Finding the probability

b.

x is a Poisson distribution with a mean of 10,

Here, rain on exactly 23 days in August, i.e.,$\mathrm{x}=23\mathrm{and}\mathrm{\lambda }=10$

$$\begin{gathered} P\left(x=24\right)=\frac{e^{-\lambda }{\lambda }^x}{x!} \\ =\frac{e^{-10}{10}^{24}}{23!} \\ =0.00017561 \\ P\left(x=23\right)=0.00017561 \end{gathered}$$

Therefore, the probability of rain for exactly 23 days in August is, 0.000175661

Finding which probability is a good approximation

c.

In the case, of part (a),

The approximate probability of rain for exactly 24 days in July is, 0.00007317, i.e., 0.0001

And, in the case of part (b),

The probability of rain for exactly 23 days in August is 0.000175661, i.e., 0.0002

So, probability in part(a) is very close to the exact probability given in the question, but in part(b) the probability differs more from the exact probability given in the question.

So, part(a) is a good approximation and is better than part(b)