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A box contains 10 light bulbs, fewer than half of which are [#permalink]
28 Aug 2010, 01:54

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

68% (02:27) correct
31% (01:44) wrong based on 195 sessions

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.

Re: Defective Bulb Probability [#permalink]
28 Aug 2010, 05:09

2

This post received KUDOS

Expert's post

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.

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.

Re: Defective Bulb Probability [#permalink]
30 Aug 2010, 07:03

2

This post received KUDOS

Expert's post

seekmba wrote:

Bunuel,

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 some 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}.
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Re: Defective Bulb Probability [#permalink]
03 Feb 2011, 21:46

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.

Re: Defective Bulb Probability [#permalink]
03 Feb 2011, 23:54

1

This post received KUDOS

Defective bulbs: n Non-defective bulbs: 10-n

and 0<=n<5.

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:

C^{n}_{2} * C^{(10-n)}_{0}

=> \frac{n!}{(n-2)!2!} * 1

\frac{n(n-1)(n-2)!}{(n-2)!2!}=\frac{n(n-1)}{2}

Number of favorable choices: \frac{n(n-1)}{2}

Number of total choices: 45

Probability(Pick both defective bulbs) = \frac{n(n-1)}{(2*45)}

Given, Probability(Pick both defective bulbs) = 1/15

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]
08 May 2012, 04:37

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]
08 Jun 2013, 01:08

2

This post received KUDOS

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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Re: Defective Bulb Probability [#permalink]
01 Apr 2014, 05:19

Bunuel wrote:

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.

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?

Re: Defective Bulb Probability [#permalink]
01 Apr 2014, 07:27

Expert's post

jlgdr wrote:

Bunuel wrote:

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.

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.
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