6. A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times are exponential with means 1 year, 1.5 years, and 3 years. What is the average length of time the boat can remain at sea

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