Question

The number of field mice found in a wheat field follows a Poission process. If the average number of field mice in a 5-acre wheat field is estimated to be 12. Find the probability that (a) fewer than 3 field mice are found on a given acre. (b) in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.

Answer 1

Answer:

a) 56.97% probability that fewer than 3 field mice are found on a given acre.

b) 41.90% probability that in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.

Step-by-step explanation:

To solve this question, we use the Poisson probability distribution and the binomial probability distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

If the average number of field mice in a 5-acre wheat field is estimated to be 12.

Find the probability that

(a) fewer than 3 field mice are found on a given acre.

5 acres have a mean of 12. For one acre

$\mu = \frac{12}{5} = 2.4$

This probability is

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2.4}*(2.4)^{0}}{(0)!} = 0.0907$

$P(X = 1) = \frac{e^{-2.4}*(2.4)^{1}}{(1)!} = 0.2177$

$P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.2613$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0907 + 0.2177 + 0.2613 = 0.5697$

56.97% probability that fewer than 3 field mice are found on a given acre.

(b) in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.

Here we use the binomial probability distribution.

56.97% probability that fewer than 3 field mice are found on a given acre, which means that $p = 0.5697$

This probability is P(X = 2) when n = 3.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{3,2}.(0.5697)^{2}.(0.4303)^{1} = 0.4190$

41.90% probability that in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.