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605-655 (Medium)|   Probability|         
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dave13
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A box contains 10 light bulbs, fewer than half of which are defective. 11 Mar 2022, 06:52
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Bunuel
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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.

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.

Answer: D.

Hope it's clear.
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KarishmaB why in the second statement both defective and non defective are given the same unknown

Shouldnt it be \(2*\frac{d}{10}*\frac{n}{9}\) when "d" is defective and "n" is non defective .

in this case st 2 is not sufficient cause there are two unknowns
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A box contains 10 light bulbs, fewer than half of which are defective. 11 Mar 2022, 23:14
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Bunuel
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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.

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.

Answer: D.

Hope it's clear.

KarishmaB why in the second statement both defective and non defective are given the same unknown

Shouldnt it be \(2*\frac{d}{10}*\frac{n}{9}\) when "d" is defective and "n" is non defective .

in this case st 2 is not sufficient cause there are two unknowns
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Note that we know that there are total 10 light bulbs. This means that if d are defective, (10 - d) are not defective because each bulb must be either defective or not defective.
So we have only one variable in statement 2 too.
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A box contains 10 light bulbs, fewer than half of which are defective. 14 Sep 2022, 23:47
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Bunuel
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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.

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.

Answer: D.

Hope it's clear.
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Hi Beunel,

Can you please tell me in evaluating option B, Why did you multiply the equation by 2?
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A box contains 10 light bulbs, fewer than half of which are defective. 15 Sep 2022, 02:26
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Bunuel
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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.

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.

Answer: D.

Hope it's clear.

Hi Beunel,

Can you please tell me in evaluating option B, Why did you multiply the equation by 2?
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I tried explaining this in this post on page 1. Let me know if still unclear.
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A box contains 10 light bulbs, fewer than half of which are defective. 16 Jan 2024, 02:27
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Bunuel

Hi Beunel,

Can you please tell me in evaluating option B, Why did you multiply the equation by 2?

I tried explaining this [ur=https://gmatclub.com/forum/a-box-contains-10-light-bulbs-fewer-than-half-of-which-are-defective-99940.html#p1072120l]in this post on page 1[/url]. Let me know if still unclear.
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Hi Bunuel,

Can I ask, that we can always multiply by 2 in situations like these, irrespective of the value of n? in other words, would the outcomes in such cases always be symmetric, or are there situations where they may not be symmetric?
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A box contains 10 light bulbs, fewer than half of which are defective. 04 Feb 2024, 06:10
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Hi.
Maybe my method helps !
I didnt let the equation turn into a quadratic one.
Once I had massaged the question well, i knew that n<=4(defective bulbs) and the non-defective bulbs were (10-n).

Moving to statements now:-
Statement 1- Boiled down to (n)(n-1)=6.
I already know that the values of n could be 1,2, 3 and 4.
But only one value of n satisfies this equation i.e. 3.

Statement 2 – boiled down to (n)(10-n)=21
Again, only for n=3, the statement could hold true.

Side note- In DS questions Statements will never contradict each other. Once I have figured, from statement 1, that the value of n is 3, I can use this as a "hint" while i am trying to figure out value of n in statement 2.
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A box contains 10 light bulbs, fewer than half of which are defective. 27 May 2024, 20:41
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­Try it this way:

­
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A box contains 10 light bulbs, fewer than half of which are defective. 19 May 2026, 23:32
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Learning is : had the question not given n<5, answer would have been E.

refer to bunuel solution for it, well explained
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A box contains 10 light bulbs, fewer than half of which are defective. 08 Aug 2026, 03:36
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Important learning is in statement 2 where both the values are positive but given one of the value is more than 5 it got rejected. In Statement (2), both roots are positive, but one is greater than 5, and the stem explicitly tells us that n must be less than 5. Therefore, only n = 3 is possible.

Statement (1): Probability that both bulbs are defective = 1/15.

If n bulbs are defective, the number of ways to choose 2 defective bulbs is C(n,2). Total ways to choose any 2 bulbs is C(10,2) = 45.

Therefore, C(n,2)/45 = 1/15, so C(n,2) = 3.

Thus, n(n-1)/2 = 3, giving n(n-1) = 6. Recognize n(n-1) as the product of two consecutive integers. The consecutive integers whose product is 6 are 2 and 3, so n = 3. The other mathematical root is n = -2, which is impossible because n represents the number of defective bulbs. Therefore, Statement (1) alone is sufficient.

---

Statement (2): Probability that one bulb is defective and the other is not defective = 7/15.

There are n ways to choose the defective bulb and 10-n ways to choose the non-defective bulb. Therefore, the number of mixed pairs is n(10-n). Total possible pairs are C(10,2) = 45.

Therefore, n(10-n)/45 = 7/15, so n(10-n) = 21.

This gives 10n - n^2 = 21, or n^2 - 10n + 21 = 0. Factoring gives (n-3)(n-7) = 0, so n = 3 or n = 7.

Important learning: In Statement (2), both possible values are positive. Unlike Statement (1), where the other root was negative and could immediately be rejected, here both values are mathematically valid positive values. However, the stem gives additional information: fewer than half of the 10 bulbs are defective, so n < 5. Therefore, n = 7 must be rejected, leaving n = 3.

So Statement (2) alone is also sufficient.

---

Answer: D - Each statement alone is sufficient.
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