Astronomers treat the number of stars in a given volume of space as a Poisson random variable. The density in the Milky Way Galaxy in the vicinity of our solar system is 1 star per 16 cubic light years.
(a) What is the probability of 3 or more stars in 16 cubic light years?
(b) How many cubic light years of space must be studied so that the probability of 1 or more stars exceeds 0.93?
(a) The probability is [Round your answer to three decimal places (e.g. 98.765).]
(b) cubic light years of space. [Round your answer to the nearest integer.]
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To solve this problem, we can use the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k!
where λ is the average rate of occurrence (in this case, the average number of stars in a given volume of space).
(a) To find the probability of 3 or more stars in 16 cubic light years, we need to calculate P(X >= 3). First, let's calculate λ:
λ = (number of stars) / (volume of space) = 1 / 16 = 0.0625
Now, let's calculate the probability using the formula:
P(X >= 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))
P(X = 0) = (e^(-0.0625) * 0.0625^0) / 0! = 0.9392
P(X = 1) = (e^(-0.0625) * 0.0625^1) / 1! = 0.0587
P(X = 2) = (e^(-0.0625) * 0.0625^2) / 2! = 0.0018
P(X >= 3) = 1 - (0.9392 + 0.0587 + 0.0018) = 0.0003
The probability of 3 or more stars in 16 cubic light years is approximately 0.0003.
(b) To find the cubic light years of space needed so that the probability of 1 or more stars exceeds 0.93, we need to solve for the volume of space when P(X >= 1) > 0.93:
P(X >= 1) = 1 - P(X = 0) = 1 - (e^(-λ) * λ^0) / 0!
0.93 = 1 - e^(-λ)
Rearranging the equation:
e^(-λ) = 1 - 0.93 = 0.07
Taking the natural logarithm (ln) of both sides:
-λ ≈ ln(0.07) = -2.6592
Solving for λ:
λ ≈ 2.6592
Now, let's calculate the volume of space using the formula:
(volume of space) = (number of stars) / (λ)
(volume of space) = 1 / (2.6592) ≈ 0.3764 cubic light years
So, approximately 0.3764 cubic light years of space must be studied to have a probability of 1 or more stars exceeding 0.93. Rounding it to the nearest integer, we get 1 cubic light year.
Therefore:
(a) The probability is approximately 0.0003 (or 0.03%).
(b) The cubic light years of space needed is approximately 1.