A jar has 10 marbles, a mix of red and white. Two marbles are randomly : Data Sufficiency (DS)

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A jar has 10 marbles, a mix of red and white. Two marbles are randomly

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Math Expert

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A jar has 10 marbles, a mix of red and white. Two marbles are randomly [#permalink]

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16 Jul 2015, 00:31

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Difficulty:

95% (hard)

Question Stats:

41% (02:42) correct

59% (03:02) wrong

based on 175 sessions

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Spoiler: OA

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Re: A jar has 10 marbles, a mix of red and white. Two marbles are randomly [#permalink]

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16 Jul 2015, 02:19

Bunuel wrote:

Solution -

Stmt1 - Less than 1/2 of the marbles in the jar are white. --> number of red > number of white and the numbers sum to 10.

Lets take red=7 and white=3, probability of selecting two red marbles is 7C2/10C2 = 21/45 = 7/9>1/3.
For red=6 and white=4, probability of selecting two red marbles is 6C2/10C2 = 1/3. Not sufficient.

Stmt2 - The probability that 1 white marble and 1 red marble will be chosen together is 7/15 --> WC1*RC1/10C2 = WR/45 = 7/15 -->WR=21.

Product of red and white marbles is 21. So there may be 7 red and 3 white marbles(b > 1/3) or 7 white and 3 red marbles(b < 1/3). Not Sufficient.

1+2 --> From Stmt1, there are more red than white marbles. The only combination fits this condition from stmt2 is 7 red and 3 white marbles. Sufficient.

ANS C.
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A jar has 10 marbles, a mix of red and white. Two marbles are randomly [#permalink]

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17 Jul 2015, 13:27

Bunuel wrote:

Need to Check if b > 1/3.

St 1: Less than 1/2 marbles are white
Possible Options are
W R
4 6 => Calculating the prob => 6/10 * 5/9 = 1/3(equal to 1/3).... No
3 7 => Calculating the prob => 7/10 * 6/9 = 7/15 (> 1/3)..... Yes
2 8
1 9

Hence this statement is not sufficient.

St 2:

Given the prob that 1 white and 1 red ball chosen is 7/15
i.e r/10 * w/9 * 2! = 7/15
Solving, rw = 21. we know r+w =10, solving r = 3, 7.

If r = 3, b = 3/10 * 2/9 = 1/15 (<1/3)
If r = 7, b = 7/15 (>1/3)
Hence Not sufficient.

Combining the two statements:
There is only one possibility: R = 7, W = 3 marbles. Here, b = 7/15(>1/3). Hence Sufficient.

Option C. Please let me know if there is some shortcut.

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Re: A jar has 10 marbles, a mix of red and white. Two marbles are randomly [#permalink]

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19 Jul 2015, 10:34

A jar has 10 marbles, a mix of red and white. Two marbles are randomly chosen from the jar. If b is the probability that both will be red, is b > 1/3?

(1) Less than 1/2 of the marbles in the jar are white.
(2) The probability that 1 white marble and 1 red marble will be chosen together is 7/15.

Given : Total T = 10 = Red(R) + White(W)
R + W = 10
Probability of selecting Both Red = B = (R/10) * ( R-1)/9 = R(R-1)/90

Asked ?
Is b >1/3
R(R-1)/90 > 1/3
On simplifying R^2 - R - 30 > 0
(R-6) (R + 5) > 0
Hence For above equation to be greater than Zero Either R < -5 or R > 6
Since R is positive Integer Hence Cannot be less than -5, Hence R > 6

So we need to Find if R > 6 ??

Statement 1 :
W < 5
So W can be 4 and R can be 6
or W can be 3 and R can be 7
So we are didn't get the answer for question R > 6 (it can be greater, also it can be equal to 6) - Insufficient

Statement 2 :
The probability that 1 white marble and 1 red marble will be chosen together is 7/15.

(R/10)(W/9) * 2 = 7 /15 (Left hand side is multiplied by 2 since, First i can draw red and then White, Or first white and then red)
RW = 21
Also R + W = 10
Hence R (10-R) = 21
Solving we get R = 7 or R = 3

So it R can be greater than 6 or less than 6 -- Hence Not sufficient.

Combining statement 1 and 2 :
Statement 1 says : W < 5 hence R >5,
Statement 2 says : R = 3 or R = 7, Combining we get R = 7
Hence R > 6 --- Answer the question

Hence Option C is sufficient to answer the question

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Re: A jar has 10 marbles, a mix of red and white. Two marbles are randomly [#permalink]

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19 Jul 2015, 13:06

Bunuel wrote:

800score Official Solution:

This is a dependent probability problem. To find the probability of choosing 2 red marbles, you need to figure out the probability that the first marble will be red and that the second marble will be red. In this case, the question wants to know if that probability is larger than 1/3.

Statement (1) says that less than half the marbles are white, which means that more than half the marbles are red. The best way to approach this is to systematically (but quickly) figure out what the probability of drawing two red marbles is for each scenario. Make a table where (number of red) > (number of white) and the numbers sum to 10.

When less than half the marbles are white, the probability of choosing 2 red marbles can be greater than or equal to 1/3. Statement (1) is not sufficient.

Statement (2) says the probability of choosing one red marble and one white marble is 7/15. This is a trap. Since the probability given is exact, it may seem that only one scenario of red marbles and white marbles will work.

If you make a table of all the scenarios, you will see that when there are 7 red marbles and 3 white marbles, the probability of choosing one of each is 7/15. However, it is also true in reverse: If there were 7 white marbles and 3 red marbles, the probability would also be 7/15. Therefore, Statement (2) is not sufficient.

Combining Statements (1) and (2) does give enough information. From Statement (2) you know that there must be 7 of one color and 3 of the other. From Statement (1) you know that there must be more red than white. The only combination that fits this is 7 red marbles and 3 white marbles.

The correct answer is choice (C).

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Re: A jar has 10 marbles, a mix of red and white. Two marbles are randomly [#permalink]

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