Three bulbs are required to light a room. Out of 15 bulbs, 6 are defective. Probability that room will be lighted? Here, I can have a total of $^{15}C_3$ combination of Bulbs. Now, defective bulbs can be chosen like $^{6}C_3$ ways. Therefore, the probability of room not getting lighted = $\frac{^{6}C_3}{^{15}C_3}$ = $\frac {4}{91}$ And probability of room getting lighted = 1 - $\frac {4}{91}$ = $\frac {87}{91}$ But, why can't I do like this? Now, good bulbs can be chosen like $^{9}C_3$ ways. Therefore, probability of room getting lighted = $\frac{^{9}C_3}{^{15}C_3}$ = $\frac {12}{65}$ What am I missing out? Please help me. Thanks in advance.

First part of the answer is incorrect. Room is not getting lighted if at least one bulb is defective, not all three. Therefore, probability of room not getting lighted is:

$\dfrac{^6C_1\cdot ^9C_2+^6C_2\cdot ^9C_1+^6C_3}{^{15}C_3}=\dfrac{53}{65}$

Explanation:

Pick one defective bulb + 2 working bulbs or 2 defective bulbs + one working bulb or 3 defective bulbs. All these events lead to at least one defective bulb out of three and therefore room will not be lighted.