I'm having trouble understanding the intuition behind the answer to a practice question.
An internet company has three redundant servers for its web site. Thus, the site functions properly as long as any of the servers function. Suppose that all three servers are placed into service at the same time, and a server is not replaced when it fails. Also,assume that the times X1; X2; X3(in months) that the servers function properly are independent random variables with cumulative distribution function (cdf) $1-e^{-(x^2)}$ when $x\ge0$ and $0$ otherwise}
If W is the time that at least one server is functioning, find the pdf of W.
The answer our professor provided is the derivative of $1-(e^{-(x^2)})^3$ or $3(1-e^{-(x^2)})\times 2xe^{-(x^2)}$. The professor also notes that $W=Max(X1, X2, X3)$.
It seems like the cdfs for X1, X2, and X3 are each a "success" when $x\ge0$ (or they run more than a month), but with the professor's answer, wouldn't one failure (running less than a month => cdf=0) obviate all of the other successes?
It seems that the probability that at least one server is running is $P(X1=x1)+P(X2=x2)+P(X3=x3)$ since they are independent servers- is that what the professor's answer represents?
Let's be clear about the meaning of the problem.
First, each server's behavior is independent of the others. If one fails, that does not affect the state of the others.
Second, the random variables $X_1, X_2, X_3$ are failure times. Specifically, if you went into the room where the servers were located, and labeled each one with the numbers $1, 2, 3$, then $X_1, X_2, X_3$ represent the random time from initial activation until the respective numbered server fails.
So for example, if we observe $X_1 = 3$, then the first numbered server failed at exactly 3 months from the time it was activated. It does not mean that the first server to fail took 3 months to fail.
Next, we are given that for each $X_1, X_2, X_3$, the failure time is distributed as $$\Pr[X_i \le x] = 1 - e^{-x^2}, \quad x \ge 0, \quad i = 1, 2, 3.$$ This tells us, for example, that the probability that Server #1 fails within 3 months is $$\Pr[X_1 \le 3] = 1 - e^{-3^2} = 1 - e^{-9} \approx 0.999877.$$
Your stated probability $$\Pr[X_1 = x_1] + \Pr[X_2 = x_2] + \Pr[X_3 = x_3]$$ makes no sense whatsoever, because (a) you haven't defined the meaning of $x_1, x_2, x_3$; (b) the instantaneous probability that any of the failure times is exactly some given value is zero, because the failure time is a continuous random variable; and (c) the sum of probabilities of independent but not mutually exclusive events could exceed $1$. That you seem to think that such an expression means something or is related to the correct answer is strongly indicative of a lack of understanding of the prerequisite knowledge of probability theory to solve this question or understand its solution.
That said, suppose we observed the outcome $$(X_1, X_2, X_3) = (0.5, 1.1, 0.8);$$ that is to say, Server #1 failed in $0.5$ months; Server #2 failed in $1.1$ months, and Server #3 failed in $0.8$ months. Then the site was up for a total time of $W = \max(X_1, X_2, X_3) = 1.1$ months, because the site remains up as long as any of the three servers is functioning, and the last one to fail was Server #2 at $1.1$ months. This much should be clear.
Now, suppose we are interested in computing the probability $$\Pr[W \le w] = \Pr[\max(X_1, X_2, X_3) \le w],$$ for some fixed time $w > 0$; that is to say, we want to calculate the probability that the site goes down within $w$ months. The key idea is to note $$\Pr[\max(X_1, X_2, X_3) \le w] = \Pr[(X_1 \le w) \cap (X_2 \le w) \cap (X_3 \le w)],$$ because if the longest surviving server fails before time $w$, then all of the other servers must also have failed before time $w$; and conversely, if each of the three servers failed before time $w$, then the last surviving server also failed before time $w$. Therefore, the events enclosed within each probability expression on each side of the equation, are equivalent. Next, because the failure times are independent, we can write $$\Pr[(X_1 \le w) \cap (X_2 \le w) \cap (X_3 \le w)] = \Pr[X_1 \le w]\Pr[X_2 \le w]\Pr[X_3 \le w],$$ since for independent events $A$ and $B$, $$\Pr[A \cap B] = \Pr[A]\Pr[B].$$ Then the rest is algebra: $$F_W(w) = \Pr[W \le w] = \left(1 - e^{-w^2}\right)^3, \quad w \ge 0,$$ hence $$f_W(w) = F'_W(w) = 3\left(1 - e^{-w^2}\right)^2 \cdot \frac{d}{dw}\left[1 - e^{-w^2}\right] = 6w e^{-w^2} \left(1 - e^{-w^2}\right)^2, \quad w \ge 0.$$