Probability of exponential events: Suppose that we have two groups of mice....

[vanheer]

Suppose that we have two groups of mice. The first group placed in a puzzle where they lived before and they can find the exit with rate theta _1.
The second group placed in a new puzzle, still, they can find the exit with rate theta _2.
Here we assume theta_1 > theta_2.
1) Now for simplicity, suppose there are six mice in the first group and four mice in the second group, after 3 hours we observe that:
- After 1 hour 3 mice found the exit.
- After 2 hours 8 mice found the exit.
- After 3 hours all 10 mice found the exit.
What is the likelihood of the above happening, given values of theta_1 and theta_2?
What is the probability that they all find the exit in one hour? What is the probability that half of them finds the exit in one hour?
2) If we look at it from a different angle:
-Given two mice from group 1, what is the probability that they both find the exit within one hour? What is the probability that one of them finds the exit in one hour and the other one doesn’t?
-Given that one mouse is placed in group 1 and one in group 2, what is the probability that they both find the exit within one hour? What is the probability that the one in group 1 finds the exit and the other doesn’t?

 

[stapel]

Suppose that we have two groups of mice. The first group placed in a puzzle where they lived before and they can find the exit with rate theta _1.
The second group placed in a new puzzle, still, they can find the exit with rate theta _2.
Here we assume theta_1 > theta_2.
1) Now for simplicity, suppose there are six mice in the first group and four mice in the second group, after 3 hours we observe that:
- After 1 hour 3 mice found the exit.
- After 2 hours 8 mice found the exit.
- After 3 hours all 10 mice found the exit.
What is the likelihood of the above happening, given values of theta_1 and theta_2?
What is the probability that they all find the exit in one hour? What is the probability that half of them finds the exit in one hour?
2) If we look at it from a different angle:
-Given two mice from group 1, what is the probability that they both find the exit within one hour? What is the probability that one of them finds the exit in one hour and the other one doesn’t?
-Given that one mouse is placed in group 1 and one in group 2, what is the probability that they both find the exit within one hour? What is the probability that the one in group 1 finds the exit and the other doesn’t?

Please reply with a clear listing of your thoughts and efforts so far, so we can see where you're getting stuck. Thank you!

 

[vanheer]

Please reply with a clear listing of your thoughts and efforts so far, so we can see where you're getting stuck. Thank you! :wink:

The time that a mouse finds the exit is exponentially distributed with parameter theta_i, where i=1,2 (group 1 and 2).
So the time till a mouse from group i finds the exit is F(t)=1-exp(-theta_i * t).
For the first year: F(1)=1-exp(- theta_i * 1)
For the second year: F(2)=1-exp(- theta_i * 2)
For the third year: F(3)=1-exp(- theta_i * 3)
The probability that no mouse from group i found the exit at time t= exp(-theta_i * t)

The probability that at least one finds the exit at time t is the min{X}, X is the set of all mice from a specific group, which is= exp(-n theta_i * t), n: number of mice in group i.
let x1,x2,...,x6 denotes the time until the ith mouse found the exit then the probability that 3 from group 1 exit is:
The probability that 3 items occur and 3 are not at time t so it should be something includes combinations of three mice among the six mice.

but here it becomes so hard to continue!