Answer:
The three survival functions are e-t, e-2/3t and e-1/3t
Then, the rate of the second death equals (multiplying all combinations of 1 survival + 1 death * the rate of death of the third device)
(e-t*(1-e-2/3t)+(1-e-t)e-2/3t)1/3e-1/3t +
(e-t*(1-e-1/3t)+(1-e-t)e-1/3t)2/3e-2/3t +
(e-2/3t*(1-e-1/3t)+(1-e-2/3t)e-1/3t)e-t =
1/3e-4/3t + 1/3e-t -2/3e -2t +
2/3e -5/3t + 2/3e-t -4/3e -2t +
e -4/3 + e -5/3 -2e -2t =
e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t
Note that this integrates to 1, thus satisfying one of the rules of a density. We also can see that this function is greater than 0, as e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t =
e-t + 4/3e-4/3t + 5/3e-5/3t - e-2t - 4/3e-2t - 5/3e-2t =
e-t - e-2t + 4/3e-4/3t - 4/3e-2t + 5/3e-5/3t- 5/3e-2t =
e-t (1 - e-t) + 4/3e-4/3t(1 - e-2/3t) + 5/3e-5/3t(1 - e-1/3t) > 0 for t > 0
Then, to find the average length that the ship is at sea, we must integrate
t(e-t + 4/3e-4/3t + 5/3e-5/3t - 4e-2t) from 0 to infinity.
However, as a shortcut, note that we can easily identify the integral of the first three functions,
te-t, 4/3te-4/3t, and 5/3te-5/3t, as calculating the expected value of the exponential distribution with parameters 1, 4/3, and 5/3, and the expected value is 1, 3/4, and 3/5
For the final function, 4te-2t = 2*2e-2t , this is twice the expected value of the exponential distribution with parameter 2 = 2 * 1/2 = 1
1 + 3/4 + 3/5 - 1 = 27/20
Step-by-step explanation:
