A dog just had 3 puppies who jump in and out of their box A

A dog just had 3 puppies who jump in and out of their box. A puppy spends an avg of 10 minutes (exponentially distributed) in the box before jumping out. Once out of the box, a puppy spends an avg of 15 minutes (exponentially distributed) before jumping back into the box.

At any given time, what is the probability that more puppies will be out of the box than will be in the box? On the average, how many puppies will be in the box?

Solution

In the long run, each puppy will on average spend 15 of 25 minutes out of the box and 10 of 25 minutes in the box (consider the Central Limit Theorem), so P(in the box) = 10/25 = 2/5. We assume the puppy jumping is independent.

Then, if X is the number of puppies in the box,
X ~ B(3, 2/5) and P(x) = C(n,x) p ^x (1-p) ^n-x = C(3,x) 2/5 ^x (3/5) ^n-x

Thus, P(0) = C(3,0) 2/5 ⁰ (3/5) ³ = 27/125 or .216

P(1) = C(3,1) 2/5¹ (3/5)² = 54/125 = .432

P(2) = C(3,2) 2/5² (3/5)¹ = 36/125 = .288

P(3) = C(3,3) 2/5³ (3/5)⁰ = 8/125 = .064

Then, the probability that more puppies will be out of the box than in the box is P(0) + P(1) =

.216 + .432 = .648.

For the binomial distribution, E(x) = np = 3(2/5) = 6/5 = 1.2