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Sladkaya [172]
3 years ago
13

Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v

icinity 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.
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

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

<em>Hence, the required probability is 0.9502</em>

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