You're responsible for maintaining four ATMs (E,W,N, and S). The time between failures for each ATM is exponentially distributed with mean time between failures 6 hours, 5 hours, 8 hours, and 8 hours, respectively. The ATMs can be serviced between 8 A.M. and 8 P.M.
(a) All ATMs are working at 8 A.M. What's the probability that the first failure will occur at or before 8:30 A.M. ?
(b) What is the probability that the first ATM to fail will be W ?
(c) ATM E just stopped working. The time to service it is exponentially distributed with mean 3 hours. Find the probability that another ATM will stop working before you're able to repair ATM E. Then, find the probability that ATMs W, N and S will all stop working before you're able to repair ATM E.
Here is my solution, so far :
Let $W$, $X$, $Y$, and $Z$ be the exponential random variable representing the time (in hours) between failures for ATMs E, W, N and S, respectively. Then, their probability density functions are given by
$f(w) = \frac{1}{6}e^{\frac{-1}{6}w}$ for $w \geq 0$ and $f(w) = 0$ otherwise
$f(x) = \frac{1}{5}e^{\frac{-1}{5}x}$ for $x \geq 0$ and $f(x) = 0$ otherwise
$f(y) = \frac{1}{8}e^{\frac{-1}{8}y}$ for $y \geq 0$ and $f(y) = 0$ otherwise
$f(z) = \frac{1}{8}e^{\frac{-1}{8}z}$ for $z \geq 0$ and $f(z) = 0$ otherwise
(a) $P(W < 0.5$ or $X < 0.5$ or $Y < 0.5$ or $Z < 0.5)$
$= P(W < 0.5) + P(X < 0.5) + P(Y < 0.5) + P(Z < 0.5) - P(W < 0.5, X < 0.5, Y < 0.5, Z < 0.5)$
$ = \int_{0}^{0.5} f(w)dw + \int_{0}^{0.5} f(x)dx + \int_{0}^{0.5} f(y)dy + \int_{0}^{0.5} f(z)dz - \int_{0}^{0.5} f(w)dw\int_{0}^{0.5} f(x)dx\int_{0}^{0.5} f(y)dy\int_{0}^{0.5} f(z)dz$ (where we've used that $W,X,Y,Z$ are independent for the last term)
$= 0.29625$.
Is this correct for part (a) ?
(b) I believe that the relevant probability would be $P(W < X, W < Y, W < Z)$. I also know have that $W, X, Y , Z$ are independent. How do I go about finding this probability, then ?
(c) Let $T$ be the exponential random variable representing the time to service ATM E. Then, it has probability density function given by $f(t) = \frac{1}{3}e^{\frac{-1}{3}t}$ for $t \geq 0$ and $f(t) = 0$ otherwise. I believe the probability that another ATM will stop working before you're able to repair ATM E is $P(X < T$ or $Y < T$ or $Z < T)$. Also, I believe that the probability that ATMs W, N and S will all stop working before you're able to repair ATM E is $P(X < T, Y < T, Z < T)$. Would this be correct ?
Thank you!
Part (a) is not correct because you are adding probabilities for outcomes that are not necessarily mutually exclusive. For example, if you toss three fair coins, the probability that you get at least one heads obviously is not $1/2 + 1/2 + 1/2 = 3/2$. That is what you are doing in your calculation.
Instead, the correct calculation should consider the complementary event, namely that the "opposite" outcome of at least one machine failing within the first half hour, is the outcome in which all machines remain operational after the first half hour. That is to say, $$\Pr[(W \le 0.5) \cup (X \le 0.5) \cup (Y \le 0.5) \cup (Z \le 0.5)] \\ = 1 - \Pr[(W > 0.5) \cap (X > 0.5) \cap (Y > 0.5) \cap (Z > 0.5)].$$ And now since the lifetimes are independent, the joint probability of each machine remaining operational after $0.5$ hours is the product of the individual probabilities that each machine lasts that long. This again is analogous to the coin example, in which the correct computation is $$1 - (1 - 1/2)(1 - 1/2)(1 - 1/2) = 7/8.$$
For (b), there are a number of ways to do the computation. One way is to integrate over the joint density of all four variables, over the region for which the desired outcome occurs; e.g., $$\Pr[W < X, Y, Z] = \int_{x=0}^\infty \int_{y=0}^\infty \int_{z=0}^\infty \int_{w = 0}^{\min(x,y,z)} f(x,y,z,w) \, dw \, dz \, dy \, dx.$$ But there's a better way. You can compute the first order statistic of machines $X$, $Y$, and $Z$. In other words, figure out what the probability density function is for the random variable $M = \min(X, Y, Z)$ by observing $$\Pr[M > m] = \Pr[\min(X,Y,Z) > m] = \Pr[(X > m) \cap (Y > m) \cap (Z > m)] \overset{\text{ind}}{=} \Pr[X > m]\Pr[Y > m]\Pr[Z > m].$$ Then compute the resulting density for $M$, and use that to compute $$\Pr[W < X, Y, Z] = \Pr[W < M].$$
For (c), exploit the memorylessness property of the exponential distribution. You want to compute the density of the first order statistic of the other three machines, just like you did for part (b). Then given that Machine E has failed, the first time to failure of the other three machines is the first order statistic, and the calculation proceeds in a similar fashion as (b). For the second part of (c), you calculate the maximum or last order statistic, $$L = \max(X, Y, Z)$$ and repeat the computation as above.