Answer:
[tex]P(X\ge 1) = 0.9502[/tex]
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:
[tex]P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}[/tex]
[tex]\lambda = density[/tex]
[tex]\lambda = \frac{3}{10}[/tex]
[tex]\lambda = 0.3[/tex]
T = the light years
[tex]T = 10[/tex]
Calculating [tex]P(X \ge 1)[/tex]
In probability:
[tex]P(X \ge 1) = 1 - P(X = 0)[/tex]
Calculating P(X=0)
Substitute 0 for k and the values for [tex]\lambda[/tex] and T in
[tex]P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}[/tex]
[tex]P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}[/tex]
[tex]P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}[/tex]
[tex]P(X=0) = (3)^0 * e^{-0.3 * 10}[/tex]
[tex]P(X=0) = 1 * e^{-0.3 * 10}[/tex]
[tex]P(X=0) = 1 * e^{-3}[/tex]
[tex]P(X=0) = e^{-3}[/tex]
[tex]P(X=0) = 0.04979[/tex]
Substitute 0.04979 for P(X=0) in [tex]P(X \ge 1) = 1 - P(X = 0)[/tex]
[tex]P(X\ge 1) = 1 - 0.04979[/tex]
[tex]P(X\ge 1) = 0.95021[/tex]
[tex]P(X\ge 1) = 0.9502[/tex] --- approximated
Hence, the required probability is 0.9502
