Probability

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crackgmat007: Legendary Member; Posts: 882; Joined: Fri Feb 20, 2009 2:57 pmFollowed by: 1 member; Thanked: 15 times; GMAT Score:690

by crackgmat007 » Fri May 08, 2009 8:49 am

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.

IMO - D. Since the probability is stated in each statement, I guessed each statement must be sufficient.

Can anyone help me compute the value of n? Tx

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Source: Beat The GMAT — Data Sufficiency | Fri May 08, 2009 8:49 am

bluementor: Master | Next Rank: 500 Posts; Posts: 418; Joined: Wed Jun 11, 2008 5:29 am; Thanked: 65 times

by bluementor » Fri May 08, 2009 11:00 am

Since less than half are defective, so n<5.

Statement 1: Prob(2 defect bulbs) = 1/15

(n/10)*((n-1)/9) = 1/15
(n)(n-1) =6

the only possible value for n is 3. Sufficient.

Statement 2: Prob(1 defective, 1 not defective) = 7/15

Prob(1 defective, 1 not defective) = 7/15
2*(n/10)*((10-n)/9) = 7/15
(n)(10-n) = 21

here, n can be 7 or 3. but since n must be less than 5, the only possible value for n is 3. Sufficient.

Choose D.

-BM-

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crackgmat007: Legendary Member; Posts: 882; Joined: Fri Feb 20, 2009 2:57 pmFollowed by: 1 member; Thanked: 15 times; GMAT Score:690

by crackgmat007 » Fri May 08, 2009 11:18 am

I have a question. In Statement 2. Why did you multiply the probability by 2? Is it because there can be two ways of getting 1 defective and other not defective? Pls clarify. Tx.

Statement 2: Prob(1 defective, 1 not defective) = 7/15

Prob(1 defective, 1 not defective) = 7/15
2*(n/10)*((10-n)/9) = 7/15
(n)(10-n) = 21

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dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members; Location: Pittsburgh, PA; Thanked: 113 times; GMAT Score:710

by dmateer25 » Fri May 08, 2009 12:34 pm

crackgmat007 wrote:I have a question. In Statement 2. Why did you multiply the probability by 2? Is it because there can be two ways of getting 1 defective and other not defective? Pls clarify. Tx.

Statement 2: Prob(1 defective, 1 not defective) = 7/15

Prob(1 defective, 1 not defective) = 7/15
2*(n/10)*((10-n)/9) = 7/15
(n)(10-n) = 21

because it can be the first one or the second one that is defective.

it can be:

(n/10) * (10-n)/9 = 7/15

and

(10-n)/10 * n/9 = 7/15

So you have to add these two together. (They give they same result so he just multiplied one of them by 2).

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