Question

In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.

a. Calculate the probability that there are at most $2$accidents occur in any given week.

b. What is the probability that there is at least two weeks between any $2$accidents?

Short answer

1. The probability that there are at most $2$accidents occur in any given week$0.4232$.
2. The probability that there is at least two weeks between any $2$accidents is$0.0025$.

Step-by-step solution

Part (a) Step 1: Given information

The number of automobile accidents occur with a Poisson distribution at an average of three per week.

Part (a) Step 2: Solution

For this probability, the random variable $X$will follow a Poisson distribution with mean $3$. Now, compute the value of needed probability that is $P(x\le 2)$ by operating the Ti-83 calculator. For this, click on $2^{\text{nd}}$, then DISTR, and then scroll down to the poissoncdf choice and enter the provided particulars. After this, click on ENTER button of the calculator to have the desired outcome. The screenshot is given as below:

Part (b) Step 1: Given information 

The number of automobile accidents occur with a Poisson distribution at an average of three per week.

Part (b) Step 2: Solution 

For this probability, the random variable X will follow an exponential distribution with decay parameter $3$. Now, the required probability can be calculated as below:

$P(x>2)=1-P(x<2)$

$=1-\left(1-e^{-3\times 2}\right)$

$=e^{-3\times 2}$

$=0.0025$