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A box contains 10 light bulbs, fewer than half of which are [#permalink]

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12 Jan 2008, 02:54

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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 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.

D

a fast 30-sec way without calculations:

1. the probability depends on n. for any given n there is only one value for probability. SUFF.

2. the probability is symmetric for defective and good bulbs. Therefore, there are two solutions (for example, (6,4) and (4,6)). But there is the restriction in the question: "fewer than half of which are defective". Therefore, SUFF.

a usual 2-and-more-min way:

1. p=n/10*(n-1)/9=1/15 ==> n²-n=6 ==> n=3,-2 ==> only positive n ==> n=3. SUFF.

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.

Soln: n bulbs are defective and n < 5 total number of bulbs is 10.

Considering statement 1 alone (1) The probability that the two bulbs to be drawn will be defective is 1/15. The probability that the two bulbs to be drawn will be defective = nC2/10C2 thus we have = nC2/10C2 = 1/15 solving we get n = 3 Statement 1 alone is 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. The probability that one of the bulbs to be drawn will be defective and the other will not be defective is also written as => nC1 * (10-n)C1/10C2 = 7/15 => n * (10-n) = (7/15) * 10C2 => n * (10-n) = 7 * 3 n = 3 or 7 Since n < 5, Thus n = 3 alone holds good. Statement 2 alone is sufficient

I also think this is D, but do not see the point in all the explanation, it is clear that after some calculation you will get exact number, but what is the point. Why to lose time, then it is obious that th both statement are suff.

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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07 Jan 2012, 15:51

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Let me throw a couple of rules to crack this problem:

1. First follow the AD/BCE stragegy. That is start with the easiest statement and then use POE. 2. GMAT always prefers that both statements provide same solution, numerically. If you didn't, you effed up.

Rephrase: The only possibilities of defective:non-defective ratios are: 4:6, 3:9, 2:8, 1:9

1. Probability that both defective is 1/15. Notice that denominator is 15, out of all defective numbers in the ratios above, 3 is a multiple. Let's start with 3. Prob = 3/10 x 2/9 = 2/30 = 1/15. So defective = 3. Suff 2. NOTE: We should try to get the same solution as 1 here and meet the S2 requirements as well. Lets start with 3 again. Probability = 3/10 x 7/9 = 7/30. But, there are two ways this can happen, so 2 x 7/30 = 7/15. Suff.

D _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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17 Jan 2012, 23:16

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.

t = 10 n < 5

n (n-1) / t (t-1)= 1/15 15 n (n-1) = t (t-1) = 10 * 9 n (n-1) = 6 n = 3

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.

Hi Bunel,, Can you please explain the second case as to why we use 2 *....( why do we take two cases)

May be with a simpler example might make things clearer!

Thanks

Statement (2) says: the probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15. Now, this event can occur in two ways: 1. The first bulb is defective and the second one is not: \(\frac{n}{10}*\frac{10-n}{9}\); 2. The first bulb is NOT defective and the second one is: \(\frac{10-n}{10}*\frac{n}{9}\);

So, \(P=\frac{n}{10}*\frac{10-n}{9}+\frac{10-n}{10}*\frac{n}{9}=2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\).

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]

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01 May 2013, 08:27

marcodonzelli 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.

here the bulbs are drawn simultaneously so the better approach would be : statement 1: nc2/10c2 =1/15 ;solve for n and choose a positive value. statement 2: (nc1*(10-n)c1)/10c2 =7/15. solve for n and choose a value less than 5 (because fewer than half are defective)

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