Source : Score top Gmat Sets
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. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
The correct answer is D.I could solve with A but with B I'm lost.When you choose simultaneosly, I take them as independent events or dependent events.Can you please solve this with the steps given.
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- Shib
- StaceyKoprince
- ManhattanGMAT Staff
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- Joined: Wed Oct 19, 2005 9:05 am
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Just FYI, I am not a fan of this particular source. Many of the questions fall short of what I consider to be minimum quality and I would not study myself using questions from this source. (I don't say this because they are a "competitor" - I like some of my competitors' work. Just not the work from this one.)
"fewer than half of which are defective" means that 1, 2, 3, or 4 are defective. Those are the possible values for n. You said you understand why statement 1 works, so I'll only address statement 2.
This gives us a prob if one of the two is defective and the other is not. Note that they do not specify the order in which this must occur - so I can either first pick a defective and then a non-defective, or vice versa.
A probability of an "and" scenario (that is, one is defective AND one is not defective) is calculated by finding the individual probabilities of the two events and multiplying them together. A probability of an "or" scenario (that is, I can find defective first, then non-defective, OR I can find non-defective first, then defective) is calculated by finding the individual probabilities and adding them together.
We know the four possibilities for defective (1, 2, 3, or 4). Test them.
If 1 is defective, 9 are not. The prob. of first getting one defective and then one non-defective is 1/10*9/9 = 1/10. The prob of first getting one non-defective and then getting one defective is the same. So, 1/10+1/10 = 2/10=1/5. (Alternatively, you can just multiply 1/10 by two, which we will do from now on.)
If 2 are defective, 8 are not. The prob. of getting one defective and one non-defective is 2/10*8/9 = 8/45. 8/45 * 2 = 16/45.
If 3 are defective, 7 are not. The prob. of getting one defective and one non-defective is 3/10*7/9 = 7/30. 7/30 * 2 = 7/15. Ding ding ding.
If 4 are defective, 6 are not. The prob. of getting one defective and one non-defective is 4/10*6/9 = 4/15. 4/15 * 2 = 8/15.
Only one of the four possibilities matches the probability given in statement 2 - so that must be the only valid option. Sufficient.
"fewer than half of which are defective" means that 1, 2, 3, or 4 are defective. Those are the possible values for n. You said you understand why statement 1 works, so I'll only address statement 2.
This gives us a prob if one of the two is defective and the other is not. Note that they do not specify the order in which this must occur - so I can either first pick a defective and then a non-defective, or vice versa.
A probability of an "and" scenario (that is, one is defective AND one is not defective) is calculated by finding the individual probabilities of the two events and multiplying them together. A probability of an "or" scenario (that is, I can find defective first, then non-defective, OR I can find non-defective first, then defective) is calculated by finding the individual probabilities and adding them together.
We know the four possibilities for defective (1, 2, 3, or 4). Test them.
If 1 is defective, 9 are not. The prob. of first getting one defective and then one non-defective is 1/10*9/9 = 1/10. The prob of first getting one non-defective and then getting one defective is the same. So, 1/10+1/10 = 2/10=1/5. (Alternatively, you can just multiply 1/10 by two, which we will do from now on.)
If 2 are defective, 8 are not. The prob. of getting one defective and one non-defective is 2/10*8/9 = 8/45. 8/45 * 2 = 16/45.
If 3 are defective, 7 are not. The prob. of getting one defective and one non-defective is 3/10*7/9 = 7/30. 7/30 * 2 = 7/15. Ding ding ding.
If 4 are defective, 6 are not. The prob. of getting one defective and one non-defective is 4/10*6/9 = 4/15. 4/15 * 2 = 8/15.
Only one of the four possibilities matches the probability given in statement 2 - so that must be the only valid option. Sufficient.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
- Jeff
conditional probability
The reason you apply conditional probabilty is that this is a problem featuring "sampling without replacement". If you think about it makes no difference whether you pull out two bulbs "simultaneously" or pull out one bulb then ....pause..... pull out the second bulb. The key thing is that you do not replace the first bulb before pulling out the second.
Jeff
Jeff
- aanchoo
Re: Probability DS
Shib Wrote:Source : Score top Gmat Sets
The correct answer is D.I could solve with A but with B I'm lost.When you choose simultaneosly, I take them as independent events or dependent events.Can you please solve this with the steps given.
FYI, the correct answer cannot be D. It has to be B. Here is an explanation for A not being sufficient.
Prob of selecting two defective pieces = n (n-1)/ 10.9 = 1/15.
Solving for n , we get n =2, n=3 .
Given n< 5 both the values of n are valid , hence A is insufficient.
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