Awesome riddle including independence and exponential distribution

Question

The life cycles of 3 devices $A, B$ and $C$ are independent and exponentially distributed with parameters $\alpha,\beta,\gamma$. These three devices form a system that fails if not only device A fails but also device B or C fails too. Maybe $a \land (b \lor c) $ is easier to understand.

Calculate the probability that the system fails before time $t$.

This riddle is driving me insane, I have spent like 5 hours thinking about it and I just can't seem to find the answer. I am pretty sure thought that there is a simple solution, anyone creative here?

Answer 1

We first go after the complementary event, the event the system is still alive at time $t$. This event can happen in two disjoint ways: (i) $A$ is alive or (ii) $A$ is dead but $B$ and $C$ are alive.

The probability of (i) is $e^{-\alpha t}$.

The probability of (ii) is $(1-e^{-\alpha t})e^{-\beta t}e^{-\gamma t}$.

Thus the probability the system is dead is $$1-\left[e^{-\alpha t}+(1-e^{-\alpha t})e^{-\beta t}e^{-\gamma t}\right].$$

Answer 2

We see that we are trying to find the probability

$$P(A\text{ fails before time } t \wedge (B\text{ fails before time } t \vee C\text{ fails before time } t).$$

By independence, this is equal to

$$P(A\text{ fails before time } t)P(B\text{ fails before time } t \vee C\text{ fails before time } t),$$

We will label these events as $A_t,B_t,C_t$ for notational cleanliness. We see that $$P(A_t)P(B_t \vee C_t) = P(A_t)\left(P(B_t) + P(C_t) - P(B_t \wedge C_t)\right)$$

by the law of union. By independence, this is equal to

$$P(A_t)(P(B_t)+P(C_t) - P(B_t)P(C_t)).$$

Hence, this problem comes down to solving for $P(A_t),P(B_t),P(C_t).$

We see that $P(A_t)$ is the probability that the life cycle for device $A$ is less than or equal to $t$. We represent this as

$$P(A_t) = \int_0^t f_A(x) dx.$$

If we represent $\alpha$ as the non-inverse parameter (there are two ways to specify an exponential distribution, either with an inverse or non-inverse parameter), we see that

$$= \int_0^t \alpha e^{-\alpha x} dx = \alpha \int_0^t e^{-\alpha x} dx.$$

Set $u = -\alpha x \implies du = -\alpha dx.$ Thus,

$$= - \int_0^t e^{u} du = - (e^{-\alpha x} \mid_{x=0}^{x = t}) = - (e^{-\alpha t} - 1) = 1 - e^{-\alpha t}.$$

Given that they are developed with the same distribution but different parameters, solving for $P(B_t)$ and $P(C_t)$ should be similar in calculation. You will likely get

$$P(B_t) = 1-e^{-\beta t}$$

and

$$P(C_t) = 1-e^{-\gamma t}.$$