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

Answer:

$P(X\ge 1) = 0.9502$

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:

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$\lambda = density$

$\lambda = \frac{3}{10}$

$\lambda = 0.3$

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

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}$

$P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}$

$P(X=0) = (3)^0 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-3}$

$P(X=0) = e^{-3}$

$P(X=0) = 0.04979$

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