Three devices are being tested. Let A be the event that…

Three devices are being tested. Let $A$ be the event that exactly one of the three devices is defective, and let $B$ be the event that the devices are functional. What do the events $A+B$ and $AB$ represent?

Step 1: The sample space

Each device is either functional (F) or defective (D). Writing the states of devices 1, 2, 3 in order, there are $2^3 = 8$ outcomes:

\[ \Omega = \{\text{FFF},\ \text{DFF},\ \text{FDF},\ \text{FFD},\ \text{DDF},\ \text{DFD},\ \text{FDD},\ \text{DDD}\}. \]

Let $k$ be the number of defective devices, $k\in\{0,1,2,3\}$.

Step 2: The events $A$ and $B$

$A$ = "exactly one device is defective" ($k=1$):

\[ A = \{\text{DFF},\ \text{FDF},\ \text{FFD}\}. \]

$B$ = "the (tested) devices are functional", i.e. all three are functional ($k=0$):

\[ B = \{\text{FFF}\}. \]

Step 3: The event $AB$

$AB$ requires both events at the same time: exactly one device is defective and all devices are functional. These two statements contradict each other — the column for $AB$ contains no "yes". Therefore

\[ \boxed{AB = \varnothing \ \text{(the impossible event)}}. \]

Events whose product is impossible are called mutually exclusive (incompatible): $A$ and $B$ can never occur together.

Step 4: The event $A+B$

$A+B$ occurs if at least one of the events occurs: either exactly one device is defective, or none is. Joining the outcomes,

\[ A+B = \{\text{FFF},\ \text{DFF},\ \text{FDF},\ \text{FFD}\} = \{k \le 1\}. \]

\[ \boxed{A+B = \text{"at most one device is defective"} = \text{"at least two devices are functional"}.} \]

It does not occur exactly when two or three devices are defective.

Step 5: Remark — the well-known textbook version

In many problem collections this exercise is stated with $A$ = "at least one of the three devices is defective" and $B$ = "all devices are functional". Then $A = \{k\ge 1\}$ and $B=\{k=0\}$, so $B$ is exactly the opposite event of $A$: $B = \overline{A}$. Consequently

\[ A+B = A+\overline{A} = \Omega \ \ \text{(the certain event)}, \qquad AB = A\overline{A} = \varnothing \ \ \text{(the impossible event)}. \]

In both versions $AB$ is impossible; the sum $A+B$ is "at most one defective" for "exactly one", and certain for "at least one".

Step 6: Remark — another possible reading of $B$

If $B$ were read as "at least one device is functional" ($k\le 2$), then every outcome of $A$ would also be in $B$ (with one defective device, two are functional), i.e. $A\subseteq B$. Then $AB = A$ ("exactly one device is defective") and $A+B = B$ ("at least one device is functional"). Always check exactly what an event says before combining it with others.