(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?