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A box contains 10 light bulbs, fewer than half of which are [#permalink]

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28 Aug 2010, 02:54

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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.

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.

Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

I did not understand the below part in option 2. How you got '2' in the equation. I know n < 5 but why do we need to multiply by '2'

\(2 * n/10 * (n-1)/9 = 1/15\)

Thanks

The probability that one of the bulbs to be drawn will be defective and the other will not be defective is the sum of the probabilities of 2 events: the first one is defective and the second is not PLUS the first is not defective and the second one is defective = \(\frac{n}{10}*\frac{10-n}{9}+\frac{10-n}{10}*\frac{n}{9}=2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\). _________________

Yes, that it was only for two ways. However, we must see the option into the works of Bunuel. That it was the possible way to check and see the probability of the light bulb.

Two bulbs can be drawn from a lot of 10 in \(C^{10}_{2} = \frac{10!}{2!8!}=\frac{(10*9)}{2}=45\) ways.

(1) Both defective:

Probability(Pick both defective bulbs) = Number of favorable choices/Number of total choices

Number of favorable choices : If we draw both defective bulbs i.e. {draw 2 balls out of n defective balls} * {0 balls out of (10-n) balls} The number of ways that can be done is:

Probability(One defective and one non-defective) = Number of favorable choices/Number of total choices

Number of favorable choices : If we draw exactly one defective and one non-defective bulb i.e. {draw 1 balls out of n defective balls} * {1 balls out of (10-n) balls} The number of ways that can be done is:

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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08 May 2012, 05:37

Hi Bunuel , I am not very sure about the concept of symmetric probability , and when to use it ? Can you please explain with an example. Thanks _________________

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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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08 Jun 2013, 02:08

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udaymathapati wrote:

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.

As an alternative way i suggest to use plug in method.

(1) since fewer than 5 bulbs are defective we are limited to choices 4,3,2 or 1. So max 4 numbers to plug, not many. First lets take 4 defective bulbs. (4/10)*(3/9)=12/90=2/15 a little bigger than 1/15, so the number of defective bulbs must be 3, but lets check it. (3/10)*(2/9)=6/90=1/15. Yes we could find the number of defective bulbs using the statement 1 - Sufficient.

(2) we can use the same logic here as well. The only difference is that here we need to look for one defective and one not defective sequence and multiple it to 2, because the sequence could start with defective as well as with non-defective bulb. Lets take 4 as the number of defective bulbs: (4/10)*(7/9)=24/90=8/30 after multiplying to 2 we have 8/15 not the ration we are looking for. Lets take 3 as the number of defective bulbs: (3/10)*(7/9)=21/90=7/30 after multiplying it to 2 we have 7/15. This is the ratio that we were looking for, so the statement 2 is sufficient.

Since both statements are sufficient on their own the answer is D. _________________

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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.

Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

Answer: D.

Hope it's clear.

In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?

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.

Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

Answer: D.

Hope it's clear.

In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?

Thanks for clarifying Cheers J

Not sure I understand what you mean...

Anyway, mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same. _________________

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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22 Jul 2014, 09:57

D is correct

b=10 , d=n<5 , Select 2

The total possibilities to select 2 out of 10 bulbs: 10C2 = 45 . Total possibilities to select 2 out of n defective bulbs: nC2 = 1/2 * n(n-1) . (1) => (1/2 * n(n-1))/45 =1/5 => n={3;-2} => (1) sufficient . Total possibilities to either select 2 out of 10-n bulbs or n defective bulbs: nC2 + (10-n)C2 = 1/2 * n(n-1) + 1/2 * (10-n)(9-n) . (2) => the probability to either select 2 out of 10-n bulbs or n defective bulbs is 8/15 we have the equation: (1/2 * n(n-1) + 1/2 * (10-n)(9-n))/45 = 8/15 => n = {3;7} => (2) sufficient

A box contains 10 light bulbs, fewer than half of which are [#permalink]

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11 Dec 2014, 03:49

Hi Bunuel, I think I have the same question as jlgdr. (In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?)

In other words, why is the equation 2x(n/10)x((10-n)/9) and not 2x(n/10)x((10-n-1)/9). Is this because they are taken simultaneously? Or is this because the second term is (9-(n-1))/9 = (9-n+1)/9 =(10-n)/9 as there is one "n" less to deduct after pulling it out in the first draw?

Appreciate your help, Many thanks

Last edited by Parco on 11 Dec 2014, 05:17, edited 1 time in total.

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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11 Dec 2014, 04:49

as per question n ranges from 0 to 4 and is an integer statement 1 nC2/10C2 = 1/15 this gives us a quadratic equation with solutions as 3 & -2. 3 is valid. Hence sufficient statement 2 nC1*(10-n)C1/10C2 = 7/15 this also gives us a quadratic equation with 2 +ve solns of 3 & 7. But as per the range of n it cannot be 7. So n=3. Hence sufficient.

Hi Bunuel, I think I have the same question as jlgdr. (In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?)

In other words, why is the equation 2x(n/10)x((10-n)/9) and not 2x(n/10)x((10-n-1)/9). Is this because they are taken simultaneously? Or is this because the second term is (9-(n-1))/9 = (9-n+1)/9 =(10-n)/9 as there is one "n" less to deduct after pulling it out in the first draw?

Appreciate your help, Many thanks

After 1 defective bulb is drawn there are total 9 bulbs left out of which n-1 are defective and 9 - (n-1) = 10 - n non-defective. So, the probability of drawing non-defective is (non-defective)/(total) = (10 - n)/9. _________________

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