# 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 = $\frac{^{9}C_3}{^{15}C_3}$ = $\frac {12}{65}$

• Because 3 are required to lighten the room, you have calculated the probability of 3 defective bulbs not chosen. Oct 19, 2016 at 5:59
• There is an error in your reasoning i think. You don't have to pick all three defective bulbs to get the probability of room not getting lighted, just one is enough. Oct 19, 2016 at 6:00
• Oct 19, 2016 at 6:02
• @MaliMish, thanks for the comment. :) Yes in that case, the 1st method given tallies with the solution given in book. Because, I'm choosing all possible cases where all 3 bulbs will be defective and in the second case I'm not considering 1 or 2 bulbs being defective. I get it. But just having a final doubt, "3 bulbs are required" - is'nt this saying that all 3 are necessary ? Oct 19, 2016 at 6:11
• @Apy Yes, all three are required. Check my answer. Oct 19, 2016 at 6:12

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.

• And just like that 1 - $\frac{53}{65}$ = $\frac{12}{65}$, probability of the room getting lighted. Oct 19, 2016 at 6:32
• Thanks for the answer. :) Fancy, books confuse so much.. :p Oct 19, 2016 at 6:33
• Glad I could help :) Oct 19, 2016 at 6:34

Therefore, I probability of room not getting lighted = $\frac{^{6}C_3}{^{15}C_3}$ = $\frac {4}{91}$

No, this is the probability of choosing $0$ functional light bulbs.

The room will not get lighted with $1$ or $2$ functional light bulbs either.

You need to calculate the probability of choosing $3$ functional light bulbs.