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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.

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  • $\begingroup$ Because 3 are required to lighten the room, you have calculated the probability of 3 defective bulbs not chosen. $\endgroup$
    – user371530
    Oct 19, 2016 at 5:59
  • $\begingroup$ 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. $\endgroup$
    – MaliMish
    Oct 19, 2016 at 6:00
  • $\begingroup$ Try en.wikipedia.org/wiki/Binomial_distribution. $\endgroup$ Oct 19, 2016 at 6:02
  • $\begingroup$ @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 ? $\endgroup$
    – lu5er
    Oct 19, 2016 at 6:11
  • $\begingroup$ @Apy Yes, all three are required. Check my answer. $\endgroup$
    – MaliMish
    Oct 19, 2016 at 6:12

2 Answers 2

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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.

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  • $\begingroup$ And just like that 1 - $\frac{53}{65}$ = $\frac{12}{65}$, probability of the room getting lighted. $\endgroup$
    – lu5er
    Oct 19, 2016 at 6:32
  • $\begingroup$ Thanks for the answer. :) Fancy, books confuse so much.. :p $\endgroup$
    – lu5er
    Oct 19, 2016 at 6:33
  • $\begingroup$ Glad I could help :) $\endgroup$
    – MaliMish
    Oct 19, 2016 at 6:34
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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.

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