Question

Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the vicinity of our solar system is 3 stars per 10 cubic light-years. What is the probability of one or more stars in 10 cubic light-years? Round your answer to 4 decimal places.

Answer 1

`P(X\ge 1) = 0.9502`

Step-by-step explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

`P(X=k) = f(x) = (\lambda T)^k( e^(-\lambda T))/(k!)`

`\lambda = density`

`\lambda = (3)/(10)`

`\lambda = 0.3`

T = the light years

`T = 10`

Calculating
`P(X \ge 1)`

In probability:

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

Calculating P(X=0)

Substitute 0 for k and the values for
`\lambda` and T in

`P(X=k) = f(x) = (\lambda T)^k( e^(-\lambda T))/(k!)`

`P(X=0) = (0.3* 10)^0 * ( e^(-0.3 * 10))/(0!)`

`P(X=0) = (3)^0 * ( e^(-0.3 * 10))/(1)`

`P(X=0) = (3)^0 * e^(-0.3 * 10)`

`P(X=0) = 1 * e^(-0.3 * 10)`

`P(X=0) = 1 * e^(-3)`

`P(X=0) = e^(-3)`

`P(X=0) = 0.04979`

Substitute 0.04979 for P(X=0) in
`P(X \ge 1) = 1 - P(X = 0)`

`P(X\ge 1) = 1 - 0.04979`

`P(X\ge 1) = 0.95021`

`P(X\ge 1) = 0.9502` --- approximated

Hence, the required probability is 0.9502