Question: 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.

1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
4. EACH statement ALONE is sufficient.
5. Statements (1) and (2) TOGETHER are not sufficient.

Correct Answer: D
Solution and Explanation:
Approach Solution 1:
Given:
Total no. of bulbs = 10
Defective bulbs, n < 5
We need to determine the value of n.
Let’s see if the information in Statement (1) alone is sufficient –
The probability that the two bulbs to be drawn will be defective = 1/15
It can be written as, n/10*[n – 1]/9 = 1/15
n[n – 1] = 1/15*90 [we can eliminate 15 & 90 because 15*6 = 90]
n[n – 1] = 6
\(n^2\)– n = 6
Solving it as Quadratic Equation,
\(n^2\) – n – 6 = 0
\(n^2\) – [3 – 2]n – 6 = 0
\(n^2\) – 3n + 2n – 6 = 0
n [n – 3] +2[n – 3] = 0
[n + 2] [n – 3] = 0
n = – 2 or n = 3
Since, n is the number of defective bulbs, it can't be negative which means n = 3 is the answer and Statement (1) alone is sufficient to answer the question.
Now, Let’s see if the information in Statement (2) alone is sufficient or not –
Probability that one bulb drawn is defective while the other isn't = 7/15
The logic that this statement uses is – the probability given to us for drawing one good bulb and the probability of drawing a defective bulb is 7/15. Now, when total number of bulbs there can be two cases where the probability of drawing a good and a defective bulb would be same due to symmetric distribution –
Case#1: 3 defective bulbs & 7 good bulbs
Case#2: 7 defective bulbs & 3 good bulbs
So, we get two values of n where n is the number of defective bulbs. The sum of these values is 10 but one of these values is greater than 5 while the other is lesser than 5.
Mathematically, 2*n/10*[10 – n]/9 = 7/15
n [10 – n] = 7/15*90*½
n[10 – n] = 21
Solving it as Quadratic Equation,
10n – \(n^2\) – 21 = 0
– \(n^2\) + 10n – 21 = 0
– \(n^2\) + [7 + 3]n – 21 = 0
– \(n^2\) + 7n + 3n – 21 = 0
– n[n – 7] + 3[n – 7] = 0
[n – 7] [3 – n] = 0
n = 7 or n = 3
Since it is given in the question itself that n < 5 which means n = 3 is the answer and Statement (2) alone is sufficient to reach the answer.

Approach Solution 2:
Given:
Total number of bulbs = 10
No. of defective bulbs [n] – 0 < /= n < 5
We need to determine the value of n with the help of additional information given in Statement (1) and (2).
Because both statements are about drawing two bulbs from the lot so let’s calculate in how many ways two bulbs can be taken out of the lot [using combination] –
C2 10 = 10!2!8! = 10*92 = 45
There are 45 ways in which 2 bulbs can be drawn from the lot.
Statement (1): it talks about the probability where both bulbs drawn are defective and the probability for the same is 1/15.

Probability [to pick both defective] = Number of Favorable choicesTotal number of choices

Number of favorable choices = 2 bulbs out of n [defective bulbs] * 0 bulbs out of 10 – n bulbs [because non-defective bulbs aren’t drawn]
Number of favorable choices = C2 n *C0 10-n
Number of favorable choices = n!(n-2)!2! *1
Number of favorable choices = n(n-1)(n-2)!(n- 2)!2!
Number of favorable choices = n(n-1)2
Total number of choices = 45 [as determined earlier it’s the total number of ways two bulbs can be drawn from the lot]
Probability [to pick both defective] = n(n-1)2*45
its given that, Probability [to pick defective] = 1/15
Hence, 1/15 = n(n-1)90
n [n – 1] = 1/15*90 [eliminating 15 & 90 by 15*6 = 90]
n [n – 1] = 6
Solving as Quadratic Equation,
\(n^2\) – n – 6 = 0
\(n^2\) – [3 – 2]n – 6 = 0
\(n^2\) – 3n + 2n – 6 = 0
n [n – 3] +2[n – 3] = 0
[n + 2] [n – 3] = 0
n = – 2 or n = 3
Since, n is the number of defective bulbs, it can't be negative which means n = 3 is the answer and Statement (1) alone is sufficient to answer the question.
Statement (2): it talks about the probability of two bulbs being drawn where one is defective and the other isn’t and the probability for the same is 7/15.

Probability [of picking one good & one defective] = Number of Favorable choicesTotal number of choices

Total number of choices = 45 [as determined earlier it’s the total number of ways two bulbs can be drawn from the lot]
Number of favorable choices = 1 bulb out of n [defective bulbs] * 1 bulbs out of 10 – n bulbs
Number of favorable choices = C1 n *C1 10-n
Number of favorable choices = n!(n-1)!1! * (10-n)!(10-n-1)!1!
Number of favorable choices = n(n-1)!(n- 1)!1! * 10-n!10-n-1!(10-n-1)!1!
Number of favorable choices = n [10 – n]
Probability [of picking one good & one defective] = Number of Favorable choicesTotal number of choices
Probability [of picking one good & one defective] = 7/15 [its given in the statement]
7/15 = n [10 – n] / 45
n [10 – n] = 7/15*45 [eliminating 15 & 45 through 15*3 = 45]
n [10 – n] = 21
Solving as Quadratic Equation,
10n – \(n^2\) – 21 = 0
– \(n^2\) + 10n – 21 = 0
– \(n^2\) + [7 + 3]n – 21 = 0
– \(n^2\)+ 7n + 3n – 21 = 0
– n[n – 7] + 3[n – 7] = 0
[n – 7] [3 – n] = 0
n = 7 or n = 3
Since it is given in the question itself that n < 5 which means n = 3 is the answer and Statement (2) alone is sufficient to reach the answer.

Approach Solution 3:
Given:
There are 10 bulbs in total
Let's assume, no. of defective bulbs = n
Now, Let’s dissect Statement (1) –
Theoretically, it's given that the probability of drawing two defective bulbs is 1/15 which depends upon the value of n. for any given value of n there is only one value of probability as well. This information is sufficient to determine the value of n because 1/15 is one value of probability which will be attached to a very specific value of n and if n is more or less than that the probability will vary. Hence, Statement (1) alone is sufficient.
Mathematically,
P = n/10*[n – 1]/9
1/15 = n[n – 1]*90
n[n – 1] = 1/15*90
n[n – 1] = 6
Solving as Quadratic Equation,
\(n^2\) – n – 6 = 0
\(n^2\) – [3 – 2]n – 6 = 0
\(n^2\) – 3n + 2n – 6 = 0
n [n – 3] +2[n – 3] = 0
[n + 2] [n – 3] = 0
n = – 2 or n = 3
Since, n is the number of defective bulbs, it can't be negative which means n = 3 is the answer and Statement (1) alone is sufficient to answer the question.
Now, Let’s dissect Statement (2) –
Theoretically, it's given that the probability of drawing one good and one defective bulb is 7/15 – the probability here is symmetric for defective and good bulbs. Therefore, there can be two solutions wherein the sum of both values of defective/good bulbs will be ten. The question adds a restriction to this solution which is n < 5. So, two values that we get the one that is greater than 5 can be discarded. Hence, Statement (2) alone is sufficient.
Mathematically,
P = 2*n/10*[10 – n]/9 = 7/15
7/15 = 2*n/10*[10 – n]/9
n [10 – n] = 7/15*90*½
n[10 – n] = 21
10n – \(n^2\) = 21
Solving as Quadratic Equation,
10n – \(n^2\) – 21 = 0
– \(n^2\) + 10n – 21 = 0
– \(n^2\) + [7 + 3]n – 21 = 0
– \(n^2\) + 7n + 3n – 21 = 0
– n[n – 7] + 3[n – 7] = 0
[n – 7] [3 – n] = 0
n = 7 or n = 3
Since it is given in the question itself that n < 5 which means n = 3 is the answer and Statement (2) alone is sufficient to reach the answer.

“A box contains 10 light bulbs, fewer than half of which are defective.”- is a topic of the GMAT Quantitative reasoning section of GMAT. GMAT Quant section consists of a total of 31 questions. GMAT Data Sufficiency questions consist of a problem statement followed by two factual statements. GMAT data sufficiency comprises 15 questions which are two-fifths of the total 31 GMAT quant questions.

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