A worker has produced n articles. Let event A_i…

A worker has produced $n$ articles. Let event $A_i$ $(i=1,2,\dots,n)$ mean that the $i$-th article produced is defective. Write out the following events:
$1^{\circ}$ all products are non-defective;
$2^{\circ}$ at least one product is defective;
$3^{\circ}$ exactly one product is defective;
$4^{\circ}$ at most two products are defective;
$5^{\circ}$ at least two products are non-defective;
$6^{\circ}$ exactly two products are defective.

Step 1: The building blocks

We also write $\prod$ for a product of many events and $\sum$ for a sum of many events.

Step 2: $1^{\circ}$ — all products are non-defective

Article $1$ is good and article $2$ is good and … and article $n$ is good:

\[ \boxed{B_{1} = \overline{A_1}\,\overline{A_2}\cdots\overline{A_n} = \prod_{i=1}^{n}\overline{A_i}.} \]

Step 3: $2^{\circ}$ — at least one product is defective

Article $1$ is defective or article $2$ is defective or …:

\[ \boxed{B_{2} = A_1 + A_2 + \dots + A_n = \sum_{i=1}^{n} A_i = \overline{\overline{A_1}\,\overline{A_2}\cdots\overline{A_n}}.} \]

The last form follows from De Morgan's law: "at least one defective" is the opposite of "all good", so $B_2 = \overline{B_1}$.

Step 4: $3^{\circ}$ — exactly one product is defective

If article $i$ is the defective one, all others must be good: $A_i\prod_{j\ne i}\overline{A_j}$. The defective article can be any of the $n$ articles, and these $n$ cases cannot happen together:

\[ \boxed{B_{3} = \sum_{i=1}^{n} A_i\prod_{j\neq i}\overline{A_j} = A_1\overline{A_2}\cdots\overline{A_n} + \overline{A_1}A_2\overline{A_3}\cdots\overline{A_n} + \dots + \overline{A_1}\cdots\overline{A_{n-1}}A_n.} \]

Common mistake: $A_1+A_2+\dots+A_n$ is "at least one", not "exactly one" — the factors $\overline{A_j}$ are essential.

Step 5: $6^{\circ}$ — exactly two products are defective

Choose the two defective articles $i\lt j$; all others are good. There are $\binom{n}{2}$ such choices, all mutually exclusive:

\[ \boxed{B_{6} = \sum_{1\le i\lt j\le n} A_iA_j\prod_{k\neq i,j}\overline{A_k}.} \]

Step 6: $4^{\circ}$ — at most two products are defective

"At most two" means "none, or exactly one, or exactly two". Using the events above,

\[ \boxed{B_{4} = B_1 + B_3 + B_6 = \prod_{i=1}^n\overline{A_i} + \sum_{i=1}^{n} A_i\prod_{j\neq i}\overline{A_j} + \sum_{i\lt j} A_iA_j\prod_{k\neq i,j}\overline{A_k}.} \]

Alternatively, $B_4$ is the opposite of "at least three are defective": $B_4 = \overline{\sum_{i\lt j\lt k}A_iA_jA_k}$.

Step 7: $5^{\circ}$ — at least two products are non-defective

It is enough that some two articles are both good (the others may be anything):

\[ \boxed{B_{5} = \sum_{1\le i\lt j\le n}\overline{A_i}\,\overline{A_j}.} \]

Here the terms may overlap (if three articles are good, several terms occur at once) — that is fine for "at least". Equivalently, $B_5$ is the opposite of "at most one article is good".

Step 8: Written out for $n=3$

For $n=3$, "at most two defective" simply means "not all three are defective".