Write down the event that there is at most one defective component in a device. A device is made up of $n$ components, that are numbered $i=1,...,n$. Let $D_i$ be the event that the $i^{th}$ component is defective. Write down, in set notation, the event that at most there is $1$ defective component.
Since we want to look at the event that there is at most $1$ defective component, that means that there can be $0$ defective components or $1$ defective component. The event there are $0$ defective components is: $\bigcup_{i=1}^n D_i^c$ and the event that there is exactly $1$ defective component is: $\bigcup_{i=1}^n \left(D_i \cap \left(\bigcap_{j=1, i \neq j} D_j^c \right)\right)$.
Therefore, the event that there is at most $1$ defective component is:
$$\left(\bigcup_{i=1}^n D_i^c \right) \cup \bigcup_{i=1}^n \left(D_i \cap \left(\bigcap_{j=1, i \neq j} D_j^c \right)\right)$$
However, the answer in the book is: $\bigcup_{i=1}^n \bigcap_{j \neq i}D_j^c$ and I'm not sure why.
 A: Take for example $n=4$ (just for simplicity).
The answer in the book means this:
$D_2^cD_3^cD_4^c \cup D_1^cD_3^cD_4^c \cup D_1^cD_2^cD_4^c \cup D_1^cD_2^cD_3^c \tag{1}$
And it's logically correct because each of the 4 terms in it imposes the restriction that 3 of the components are not defective (and whether the 4th component is defective or not doesn't matter, so it's not part of the term). And then these four terms are connected with a logical OR symbol (or say set theory's union symbol).
$D_2^cD_3^cD_4^c$
$ D_1^cD_3^cD_4^c $
$ D_1^cD_2^cD_4^c  $
$ D_1^cD_2^cD_3^c$
The expression you have written would be equivalent to the answer in the book if you change one symbol in it. It should be this
$$\left(\bigcap_{i=1}^n D_i^c \right) \cup \bigcup_{i=1}^n \left(D_i \cap \left(\bigcap_{j=1, i \neq j} D_j^c \right)\right)$$
So it seems to me you meant to say this:
$D_1^cD_2^cD_3^cD_4^c \cup D_1D_2^cD_3^cD_4^c \cup D_1^cD_2D_3^cD_4^c \cup D_1^cD_2^cD_3D_4^c \cup D_1^cD_2^cD_3^cD_4 \tag{2}$
So you should have cap/intersection and not cup/union in your first pair of brackets.
Then the two expressions (1) and (2) are logically equivalent (and set theoretically equal), i.e. from set theory point of view they are identical (just like $x^2-1$ and $(x-1)(x+1)$ are two different expressions for the same thing in algebra). You can prove that they are equivalent easily if you take some small values of $n$. Say $n=4$ for example. Of course it can be proven for any $n$ too.
