Solution. The number of equally probable ways to plug in $A_{7}^{3}=7 \cdot 6 \cdot 5=210$. The lamp can be powered through 0, 1, or 2 extension cords.

0) The lamp is plugged into the socket, the other 2 plugs are plugged in randomly. The number of such possibilities $A_{6}^{2}=30$.
1) The lamp is plugged into one of the 6 sockets of an extension cord, the connected extension cord is plugged into the socket, the other is plugged in randomly. The number of possibilities $6 \cdot 5=30$.
2) The lamp is plugged into one of the 6 sockets of an extension cord, the connected extension cord is plugged into one of the 3 sockets of another extension cord, and that one is plugged into the socket. The number of possibilities $6 \cdot 3=18$.

Answer: $\mathbf{P}=\frac{30+30+18}{210}=13 / 35$
2. There is one power outlet connected to the network, two extension cords with three outlets each, and one table lamp included. Nosy Nick randomly plugged all three plugs into 3 out of 7 outlets. What is the probability that the lamp will light up? (16 points)