Solution:
Solution
A room has $$3$$ electric lamps.Out of $$10$$ bulbs present, $$6$$ bulbs are good.
$$\Rightarrow$$ $$4$$ bulbs are defective.
The probability that selected $$3$$ bulbs are good $$=1-$$ (Probability that selected $$3$$ bulbs are defective)
$$=1-\displaystyle\dfrac{{}^{4}C_{3}}{{}^{10}C_{3}}$$
$$=1-\displaystyle\dfrac{(4\times 3\times 2)}{(10\times 9\times 8)}=1-\dfrac{1}{30}$$
$$=\displaystyle\frac{29}{30}$$
