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Lightbulb Probability Problem

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by Jim@StratusPrep » Wed Jun 13, 2012 5:17 pm
What problem are you talking about?
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by Anurag@Gurome » Thu Jun 14, 2012 4:50 am
dellaboemia wrote:Can we get any of the pros to chime in here? This problem seems to have buried within it, some aspects of probability that I'm not getting.
Are you talking about this problem?

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously
from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

If yes, it's already discussed on the thread: https://www.beatthegmat.com/p-c-problem-t112709.html
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by gmattesttaker2 » Sun Jun 17, 2012 3:13 pm
Anurag@Gurome wrote:
dellaboemia wrote:Can we get any of the pros to chime in here? This problem seems to have buried within it, some aspects of probability that I'm not getting.
Are you talking about this problem?

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously
from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

If yes, it's already discussed on the thread: https://www.beatthegmat.com/p-c-problem-t112709.html

Hello,

I tried to solve this problem as follows. Can you please tell me if this approach is correct?

1) P(E) = (n(E1)/n(S1)).(n(E2)/n(S2))
= (nC1/10C1).( (n-1)C1/9C1 )
= n(n-1)/90

So, n(n-1)/90 = 1/15
=> n = 3, -2

Since n not equal to -2, n = 3

2) P(E) = ( n(E1)/n(S1) ).( n(E2)/n(S2) )
= ( nC1/10C1 ) . ( (10-n)C1 / 9C1 )
= n( 10 - n )/90

So, n(10 - n)/90 = 7/15

=> 42 = 10n - n^2

However, I am not sure if this is the correct approach since I am not sure how to proceed further from here? Can you please help? Thanks a lot.

Best Regards,
Sri
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by eagleeye » Sun Jun 17, 2012 6:44 pm
gmattesttaker2 wrote:
Anurag@Gurome wrote:
dellaboemia wrote:Can we get any of the pros to chime in here? This problem seems to have buried within it, some aspects of probability that I'm not getting.
Are you talking about this problem?

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously
from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

If yes, it's already discussed on the thread: https://www.beatthegmat.com/p-c-problem-t112709.html

Hello,

I tried to solve this problem as follows. Can you please tell me if this approach is correct?

1) P(E) = (n(E1)/n(S1)).(n(E2)/n(S2))
= (nC1/10C1).( (n-1)C1/9C1 )
= n(n-1)/90

So, n(n-1)/90 = 1/15
=> n = 3, -2

Since n not equal to -2, n = 3

2) P(E) = ( n(E1)/n(S1) ).( n(E2)/n(S2) )
= ( nC1/10C1 ) . ( (10-n)C1 / 9C1 )
= n( 10 - n )/90

So, n(10 - n)/90 = 7/15

=> 42 = 10n - n^2

However, I am not sure if this is the correct approach since I am not sure how to proceed further from here? Can you please help? Thanks a lot.

Best Regards,
Sri
Hi Sri:

1) For the first part, you have done it correctly. Well done!.

2) For the second part, you've considered only one case, where the defective one is picked first, and the "good" bulb second. There is the other case, where "good" bulb is picked first, and the defective one second. The probability should have both cases (the probability of each case is the same). So:

P(E) = 2* (n/10) * (10-n)/9 = 7/15.

Solving this we get n(10-n) = 21 => n^2-10n+21 = 0 => (n-3)(n-7)=0. => n=3 or n=7. However, we are told that n<5, so n=3. Sufficient.

Let me know if this helps :)
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by gmattesttaker2 » Sun Jun 17, 2012 10:38 pm
eagleeye wrote:
gmattesttaker2 wrote:
Anurag@Gurome wrote:
dellaboemia wrote:Can we get any of the pros to chime in here? This problem seems to have buried within it, some aspects of probability that I'm not getting.
Are you talking about this problem?

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously
from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

If yes, it's already discussed on the thread: https://www.beatthegmat.com/p-c-problem-t112709.html

Hello,

I tried to solve this problem as follows. Can you please tell me if this approach is correct?

1) P(E) = (n(E1)/n(S1)).(n(E2)/n(S2))
= (nC1/10C1).( (n-1)C1/9C1 )
= n(n-1)/90

So, n(n-1)/90 = 1/15
=> n = 3, -2

Since n not equal to -2, n = 3

2) P(E) = ( n(E1)/n(S1) ).( n(E2)/n(S2) )
= ( nC1/10C1 ) . ( (10-n)C1 / 9C1 )
= n( 10 - n )/90

So, n(10 - n)/90 = 7/15

=> 42 = 10n - n^2

However, I am not sure if this is the correct approach since I am not sure how to proceed further from here? Can you please help? Thanks a lot.

Best Regards,
Sri
Hi Sri:

1) For the first part, you have done it correctly. Well done!.

2) For the second part, you've considered only one case, where the defective one is picked first, and the "good" bulb second. There is the other case, where "good" bulb is picked first, and the defective one second. The probability should have both cases (the probability of each case is the same). So:

P(E) = 2* (n/10) * (10-n)/9 = 7/15.

Solving this we get n(10-n) = 21 => n^2-10n+21 = 0 => (n-3)(n-7)=0. => n=3 or n=7. However, we are told that n<5, so n=3. Sufficient.

Let me know if this helps :)
Hello Eagleeye,

Hope all is well and thank you very much for your prompt response and for the detailed explanation. Now it is clear to me. Thanks again.

Best Regards,
Sri
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