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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.

Sagot :

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

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