Probability

[crackgmat007]

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

[bluementor]

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-

[crackgmat007]

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

[dmateer25]

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).