Three electric lamps are fitted in a room. 3 bulbs are chosen at random from 10 bulbs having 6 good bulbs. The probability that the room is lighted is

Given that there are 10 bulbs in which 6 are good

To lighten the room we need to select at least one bulb from the 6 good bulbs

Let $S$ be the sample and let $E$ be the event of lightining the room. 

$n\left(S\right)={10}_{C_3}$ and $n\left(\overline{E}\right)$ is the number of ways not lighting the room

it means $n\left(\overline{E}\right)=4_{C_3}$

Hence,

       $\begin{array}{rcl}P\left(\overline{E}\right)& =& \frac{n\left(\overline{E}\right)}{n\left(S\right)}\\ & =& \frac{4_{C_3}}{{10}_{C_3}}\\ & =& \frac{4·3·2}{10·9·8}\\ & =& \frac{1}{30}\end{array}$

Therefore, the probability of the event of " lightining the room" is

        $\begin{array}{rcl}P\left(E\right)& =& 1-P\left(\overline{E}\right)\\ & =& 1-\frac{1}{30}\\ & =& \boxed{\frac{29}{30}}\end{array}$