Data Sufficiency

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.


OA given is Op D, see in for both the Op'ns i made separate eq'ns and for Op A i got the solution but m not able to get the solution for Op B, so if someone can solve Op B that will be really helpful.

Total number of bulbs = 10
Total number of defective bulbs = n
Total number of bulbs not defective = 10 - n

(1)The probability that the two bulbs to be drawn will be defective is 1/15.

When first bulb is drawn, probability that it is defective = n/10
When second bulb is drawn, probability that it is defective = (n-1)/(10-1) = (n-1)/9

probability that the two bulbs to be drawn will be defective = (n/10) * (n-1)/9 which is equal to 1/15.
Solving the equation n(n-1) = 6
So n = 3


(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15

When first bulb is drawn, probability that it is defective = n/10
When second bulb is drawn, probability that it is not defective = (10 - (n-1))/(10-1) = (11-n)/9

probability that one of the bulbs to be drawn will be defective and the other will not be defective = (n/10) * (11-n)/9 which is equal to 7/15.
Solving the equation n(11-n) = 42
Considering factors of 42, 3(11-3) = 42. So n = 3

So answer is D

Experts, please suggest if the above approach is correct.

Total number of bulbs = 10
Total number of defective bulbs = n
Total number of bulbs not defective = 10 - n

(1)The probability that the two bulbs to be drawn will be defective is 1/15.

When first bulb is drawn, probability that it is defective = n/10
When second bulb is drawn, probability that it is defective = (n-1)/(10-1) = (n-1)/9

probability that the two bulbs to be drawn will be defective = (n/10) * (n-1)/9 which is equal to 1/15.
Solving the equation n(n-1) = 6
So n = 3


(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15

When first bulb is drawn, probability that it is defective = n/10
When second bulb is drawn, probability that it is not defective = (10 - (n-1))/(10-1) = (11-n)/9

probability that one of the bulbs to be drawn will be defective and the other will not be defective = (n/10) * (11-n)/9 which is equal to 7/15.
Solving the equation n(11-n) = 42
Considering factors of 42, 3(11-3) = 42. So n = 3

So answer is D

Experts, please suggest if the above approach is correct.

GAMATO wrote:Total number of bulbs = 10
Total number of defective bulbs = n
Total number of bulbs not defective = 10 - n

(1)The probability that the two bulbs to be drawn will be defective is 1/15.

When first bulb is drawn, probability that it is defective = n/10
When second bulb is drawn, probability that it is defective = (n-1)/(10-1) = (n-1)/9

probability that the two bulbs to be drawn will be defective = (n/10) * (n-1)/9 which is equal to 1/15.
Solving the equation n(n-1) = 6
So n = 3


(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15

When first bulb is drawn, probability that it is defective = n/10
When second bulb is drawn, probability that it is not defective = (10 - (n-1))/(10-1) = (11-n)/9

probability that one of the bulbs to be drawn will be defective and the other will not be defective = (n/10) * (11-n)/9 which is equal to 7/15.
Solving the equation n(11-n) = 42
Considering factors of 42, 3(11-3) = 42. So n = 3

So answer is D

Experts, please suggest if the above approach is correct.

n(11-n) = 42
Considering factors of 42, 3(11-3) = 42. So n = 3

here u are saying n = 3, how???? if u put n =3 in n(11-n), u get, 3*8 = 24

Remember that we need to consider two equally probable cases for (2) : first is defective and second is not defective, and vice versa.

Thus 2(n/10)(10-n)/9 = 7/15

n (10 - n) = 21

n= 3 or n= 7

Since n < 5 from the question, n=3 SUFF

gmat25 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.


OA given is Op D, see in for both the Op'ns i made separate eq'ns and for Op A i got the solution but m not able to get the solution for Op B, so if someone can solve Op B that will be really helpful.

Statement 1 gives something away: since P(both bulbs are defective) > 0, there must be at least 2 defective bulbs.
Since the question stems indicates that n<5, we know that the only possible values are n=2, n=3, or n=4.

Statement 1: P(both are defective) = 1/15.
If n=2, P(both are defective) = 2/10 * 1/9 = 1/45. Doesn't work.
If n=3, P(both are defective) = 3/10 * 2/9 = 1/15. This works.
Clearly, n=4 can't work, since it will increase all the numerators.
Thus, n=3.
Sufficient.

Statement 2: P(exactly 1 is defective) = 7/15.
Since only n=3 worked in statement 1, we should start with n=3.

If n=3, P(1st is defective and the 2nd is not) = 3/10 * 7/9 = 7/30.
Since P(1st is not defective and the 2nd is defective) will yield the same probability, the result above must be multiplied by 2:
P(exactly 1 is defective) = 2 * 7/30 = 7/15. This works.

Clearly n=4 can't work, since it will increase all the numerators.
Using similar logic, n=2 can't work, since it will decrease all the numerators.
Thus, n=3.
Sufficient.

The correct answer is D.

+1 for D.

Stat 1 : n/10 * n/10 = 1/15

Hence n can be gound

Stat 2 : n/10 * 1-n/10 = 7/5

sufficient

GMATGuruNY wrote:

gmat25 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.


OA given is Op D, see in for both the Op'ns i made separate eq'ns and for Op A i got the solution but m not able to get the solution for Op B, so if someone can solve Op B that will be really helpful.

Statement 1 gives something away: since P(both bulbs are defective) > 0, there must be at least 2 defective bulbs.
Since the question stems indicates that n<5, we know that the only possible values are n=2, n=3, or n=4.

Statement 1: P(both are defective) = 1/15.
If n=2, P(both are defective) = 2/10 * 1/9 = 1/45. Doesn't work.
If n=3, P(both are defective) = 3/10 * 2/9 = 1/15. This works.
Clearly, n=4 can't work, since it will increase all the numerators.
Thus, n=3.
Sufficient.

Statement 2: P(exactly 1 is defective) = 7/15.
Since only n=3 worked in statement 1, we should start with n=3.

If n=3, P(1st is defective and the 2nd is not) = 3/10 * 7/9 = 7/30.
Since P(1st is not defective and the 2nd is defective) will yield the same probability, the result above must be multiplied by 2:
P(exactly 1 is defective) = 2 * 7/30 = 7/15. This works.

Clearly n=4 can't work, since it will increase all the numerators.
Using similar logic, n=2 can't work, since it will decrease all the numerators.
Thus, n=3.
Sufficient.

The correct answer is D.

Thanks a lot Mitch, in Op B, i didn't multiplied with 2 and that's where i was going wrong. Thanks a lot!!!

Statement 2: P(exactly 1 is defective) = 7/15.
Since only n=3 worked in statement 1, we should start with n=3.

If n=3, P(1st is defective and the 2nd is not) = 3/10 * 7/9 = 7/30.
Since P(1st is not defective and the 2nd is defective) will yield the same probability, the result above must be multiplied by 2:
P(exactly 1 is defective) = 2 * 7/30 = 7/15. This works.

Clearly n=4 can't work, since it will increase all the numerators.
Using similar logic, n=2 can't work, since it will decrease all the numerators.
Thus, n=3.
Sufficient.

Hi Mitch, I don't get this. It's a selection. why are we bothered about which one is picked first? I don't know what logic I'm missing here.. I thought 3/10 * 7/9 = 7/30 is the probability for picking a defective and a good bulb.

thanks.

nailGmat2012 wrote:

Statement 2: P(exactly 1 is defective) = 7/15.
Since only n=3 worked in statement 1, we should start with n=3.

If n=3, P(1st is defective and the 2nd is not) = 3/10 * 7/9 = 7/30.
Since P(1st is not defective and the 2nd is defective) will yield the same probability, the result above must be multiplied by 2:
P(exactly 1 is defective) = 2 * 7/30 = 7/15. This works.

Clearly n=4 can't work, since it will increase all the numerators.
Using similar logic, n=2 can't work, since it will decrease all the numerators.
Thus, n=3.
Sufficient.

Hi Mitch, I don't get this. It's a selection. why are we bothered about which one is picked first? I don't know what logic I'm missing here.. I thought 3/10 * 7/9 = 7/30 is the probability for picking a defective and a good bulb.

thanks.

We have to account for ALL of the different ways to get a favorable outcome.
If 3 of the bulbs are defective and 7 are not, there are TWO WAYS to select EXACTLY ONE defective bulb:

Case 1: the first bulb is defective, the second bulb is not
P(the first bulb is defective) * P(the second bulb is not defective) = 3/10 * 7/9 = 7/30.

Case 2: the first bulb is not defective, the second bulb is defective
P(the first bulb is not defective) * P(the second bulb is defective) = 7/10 * 3/9 = 7/30.

Since either Case 1 OR Case 2 will yield a favorable outcome, we ADD the probabilities:
7/30 + 7/30 = 14/30 = 7/15.

The key word here is EXACTLY: whenever we have to determine the probability that something happens EXACTLY n times, we must account for ALL of the different ways to get a favorable outcome.

To find similar problems, type "probability", "exactly" and "gmatguruny" into the search bar.