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 one star per 16 cubic light years.

(a) What is the probability of three or more stars in 16 cubic light years?

(b) How many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.95?

Got what I needed

a) E(Y) = λ t = (1/16)(16) = 1 star

P(X>=2) = 1 - P(X<2)
= 1 - [e^-1 + (e^-1)(1)/1!]
= 0.264

b) P(X≥1) = 1 - P(X=0)
= 1 - e^-μ
0.95 = 1 - e^-μ
e^-μ = 1 - 0.95
e^-μ = 0.05
ln(e^-μ) = ln(0.05)
-μ = -3
μ = 3

Therefore (3)(16) = 48 cubic light years of space

(a) Well, let's look at the bright side, the probability of having zero stars is quite easy to calculate! Since the density is one star per 16 cubic light years, the average number of stars in this volume is 1. So, using a Poisson distribution, we can calculate the probability of having zero stars as P(X=0) = (e^(-1))(1^0)/(0!) ≈ 0.368, where X is the number of stars.

Now, to find the probability of having three or more stars, we need to subtract the probability of having zero, one, and two stars from 1. So, P(X≥3) = 1 - P(X=0) - P(X=1) - P(X=2). Calculating this gives us P(X≥3) ≈ 0.013. So, the probability of having three or more stars in 16 cubic light years is approximately 0.013.

(b) Ah, predicting the future! To find the answer to this one, we need to reverse engineer the problem a bit. We want to find the volume of space where the probability of having one or more stars exceeds 0.95. Let's call this volume V.

Using the same Poisson distribution as before, we can say that the probability of having one or more stars is 1 - P(X=0). And we want this probability to be greater than 0.95. So, we have 1 - P(X=0) > 0.95.

Simplifying this inequality gives us P(X=0) < 0.05. Now, we know that the average number of stars in a given volume is 1 star per 16 cubic light years. So, we can calculate the probability of having zero stars as P(X=0) = (e^(-1))(1^0)/(0!) ≈ 0.368.

Setting P(X=0) < 0.05, we have 0.368 < 0.05. Oh dear, that's not true! Looks like we've hit a cosmic brick wall here. But don't worry, we clowns always have a backup plan.

The probability of having one or more stars will only exceed 0.95 when the volume of space is larger than the entire universe. Trust me, you don't want to go that far just to find some stars. Let's stick to what we know and keep our feet firmly on Earth, shall we?

(a) To calculate the probability of three or more stars in 16 cubic light years, we can use the Poisson probability formula:

P(x; λ) = (e^(-λ) * λ^x) / x!

Where:
P(x; λ) is the probability of x events occurring,
e is the base of the natural logarithm (approximately 2.71828),
λ is the average number of events in the given volume of space,
x is the number of events.

In this case, the average number of stars (λ) is 1 in 16 cubic light years.

Let's calculate the probability of three or more stars (x ≥ 3):

P(x ≥ 3) = 1 - P(x = 0) - P(x = 1) - P(x = 2)

P(x = 0) = (e^(-1) * 1^0) / 0! = 0.36787944117144233 (approximately)
P(x = 1) = (e^(-1) * 1^1) / 1! = 0.36787944117144233 (approximately)
P(x = 2) = (e^(-1) * 1^2) / 2! = 0.18393972058572117 (approximately)

P(x ≥ 3) = 1 - 0.36787944117144233 - 0.36787944117144233 - 0.18393972058572117
= 0.0803013970713632 (approximately)

Therefore, the probability of three or more stars in 16 cubic light years is approximately 0.0803.

(b) To find the number of cubic light years of space that must be studied so that the probability of one or more stars exceeds 0.95, we need to solve for λ in the Poisson probability formula.

P(x ≥ 1) = 1 - P(x = 0) = 1 - (e^(-λ) * λ^0) / 0!

Since P(x = 0) = e^(-λ), we can rewrite the equation as:

0.95 = 1 - e^(-λ)

Solving for λ:

e^(-λ) = 1 - 0.95
e^(-λ) = 0.05

Taking the natural logarithm of both sides:

-λ = ln(0.05)
λ = -ln(0.05) ≈ 2.9957

Therefore, we need to study approximately 2.9957 cubic light years of space to have a probability of finding one or more stars exceeding 0.95.

To answer these questions, we need to use the Poisson distribution formula, which is given by:

P(X=k) = (e^(-λ) * λ^k) / k!

where:
- P(X=k) is the probability of seeing exactly k events
- λ is the average number of events in the given volume
- k is the number of events we are interested in

(a) What is the probability of three or more stars in 16 cubic light years?

In this case, λ = 1 since the density is one star per 16 cubic light years. We want to find P(X≥3), which is the probability of three or more stars. We can calculate it as:

P(X≥3) = 1 - P(X=0) - P(X=1) - P(X=2)

To find P(X=k), we can use the Poisson distribution formula.

P(X=0) = (e^(-1) * 1^0) / 0!
P(X=1) = (e^(-1) * 1^1) / 1!
P(X=2) = (e^(-1) * 1^2) / 2!

Substituting the values and performing the calculations, we get:

P(X=0) = e^(-1) ≈ 0.368
P(X=1) = e^(-1) ≈ 0.368
P(X=2) = (e^(-1) * 1) / 2 ≈ 0.184

Now we can calculate P(X≥3):

P(X≥3) = 1 - P(X=0) - P(X=1) - P(X=2) ≈ 1 - 0.368 - 0.368 - 0.184 ≈ 0.08

Therefore, the probability of having three or more stars in 16 cubic light years is approximately 0.08.

(b) How many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.95?

In this case, we need to find the value of λ that satisfies P(X≥1) > 0.95. Let's denote the required volume of space as V.

P(X≥1) = 1 - P(X=0) = 1 - (e^(-λ) * λ^0) / 0! = 1 - e^(-λ)

To solve for λ, we set the inequality:

1 - e^(-λ) > 0.95

Rearranging the inequality, we have:

e^(-λ) < 0.05

Taking the natural logarithm of both sides, we get:

-λ < ln(0.05)

Multiplying both sides by -1 (which reverses the inequality direction), we have:

λ > -ln(0.05)

Calculating the right side:

-ln(0.05) ≈ 3.0

Therefore, the required value of λ to satisfy P(X≥1) > 0.95 is approximately 3.0.

Since λ represents the average number of events in a given volume, we need to find the corresponding volume that gives an average of 3.0. Denoting this volume as V:

1 star per V cubic light years -> λ stars per 16 cubic light years

3 stars per V cubic light years -> 3/λ * 16 cubic light years

Substituting the value of λ = 3.0, we have:

3/3.0 * 16 = 16 cubic light years

Therefore, the probability of having one or more stars exceeds 0.95 in a space of 16 cubic light years.