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To explain my doubt, here is the question which I am stuck thinking on:

Calculate the reliability of the system in the figure if each component is operable with probability $0.92$ independently of the other components. enter image description here

  1. According to the text book, the upper link $A$-$B$ works if both $A$ and $B$ work (taking in to account the two are connected sequentially $$P\{A\cap B\}=(0.92)^2=0.8464$$ This is straightforward and easily understood, the confusion lies in the second step.
  2. Components $D$ and $E$, connected in parallel, are operable with probability $$P\{D\cup E\}=1-(1-0.92)^2=0.9936$$

I do not understand why the author has calculated the probability this way; I figured that he took the complement twice in order to calculate the intersection between the events rather than the union. $$P\{\overline{\overline{D}\cap \overline{E}}\}=P\{D\cup E\}=0.9936$$

Although that solves it, I know that I can also calculate the probability of independent events easily in this scenario as well. The only issue is figuring out whether the events are mutually exclusive or not. I found that $$P\{D\cup E\}=P\{D\}+P\{E\}-P\{D\cap E\}=0.9936$$ but my reasoning is because $P\{D\cup E\}=P\{D\}+P\{E\} > 1$. But what if each component is operable with probability $0.30$? That way I would not be sure whether my answer is correct or not since in that case $P\{D\cup E\}=P\{D\}+P\{E\} < 1$. What evidence is there that can tell me that the two events are mutually exclusive other than seeing that their sum is greater than $1$?

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    $\begingroup$ This is not clear. In general, the probability of the union of two events $X,Y$ is $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$. The subtraction is required because the intersection occurs in both $X$ and $Y$ hence is counted twice when you add $P(X)$ and $P(Y)$. $\endgroup$
    – lulu
    Mar 8, 2022 at 13:52
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    $\begingroup$ Knowing $\Pr(D)+\Pr(E)>1$ directly implies that $D$ and $E$ are not mutually exclusive. Knowing $\Pr(D)+\Pr(E)<1$ does not tell us anything about whether or not $D$ and $E$ are mutually exclusive. There are examples who are and there are examples who are not. Knowing that $D,E$ are independent and nonzero directly implies $D$ and $E$ are not mutually exclusive. Knowing that $D,E$ are not independent does not tell us anything about mutual exclusivity of $D$ and $E$. Some are, some are not. $\endgroup$
    – JMoravitz
    Mar 8, 2022 at 14:03

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What evidence is there that can tell me that the two events are mutually exclusive other than seeing that their sum is greater than 1?

In the question, it is stated that whether or not each component is broken is independent of each of the other components. Intuitively, this means that knowing whether $D$ broke tells you nothing about whether $E$ broke. This should make it clear why the events $D$ and $E$ are not disjoint/mutually exclusive: that would mean that if $D$ is broken, $E$ cannot be broken, and vice versa. That's not independent at all!

If you are looking for a more strict, formal approach, $D$ and $E$ are independent (by definition) if and only if $P(D \cap E) = P(D)P(E)$. And, as you point out, $$P(D \cup E) = P(D) + P(E) - P(D \cap E) = P(D) + P(E) - P(D)P(E).$$

Since $P(D)$ and $P(E)$ are both non-zero, it follows that $P(D \cup E) \neq P(D) + P(E)$, and the events are not mutually exclusive.

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