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.

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.

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

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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Statement 1:
More than ½ are red, so 6 or more. If we have 6 red, the probability is 6/10 * (5/9) = 30/90 = ⅓. This means that we need 7 red marbles or more for b>⅓. Insufficient.

Statement 2:
Let x be the number of red marbles, then the probability for 1R1W or 1W1R is 2 * x/10 * (10 - x)/9 = 2*x(10-x)/90 = 7/15 (the 2x in the beginning is for different color order). Then x(10-x) = 21. Solving this gives x = 7 or x = 3 so this is insufficient.

Combined:
There must be more red so 7 red is the conclusion. Sufficient.

Answer: C
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I like spending a bit more time on the question stem with something like this, so that I don't have to do a bunch of calculations twice (once for each statement).

Let's say that the number of red marbles is R.

Then, what's the probability that both marbles are red?

The probability that the first is red is R/10, and the probability that the second is red is (R-1)/9.

So, the overall probability is R(R-1) / 90.

Therefore, the question is really asking: Is R(R-1) / 90 > 1/3?

Simplify: Is R(R-1) > 30?

When will this happen? Well, I know that 6*5 is exactly 30. So, if R is bigger than 6, the answer will be "yes". But if R is 6 or smaller, the answer will be "no".

The question really says, Is R greater than 6?

Statement 1: Insufficient - R is greater than 5. So, there could be 6 red marbles ('no') or 9 red marbles ('yes').

Statement 2: I'm going to say this is insufficient without doing any calculation. This is a safe assumption because of how probability works. Let's say that we had 3 red marbles and 7 white marbles. We'd have the same probability of getting one red and one white, as we would if we had 3 white marbles and 7 red marbles. So, just from this probability, you can't tell if there are a lot of red marbles, or only a few. Switching the number of red with the number of white will give you the same probability.

Statements 1 and 2: This seems sufficient for the same reason described in statement 2 above. Since we now know that R is greater than 5, we can no longer 'switch' the numbers of red and white marbles to end up with the same probability.

But if you want to do the calculation, here's how!

7/15 = probability of one red, one white

7/15 = probability of red first, white second + probability of white first, red second

7/15 = R/10 * (10-R)/9 + (10-R)/10 * R/9 = 2*R*(10-R)/90

Multiply both sides by 90:

42 = 2*R*(10-R)

21 = R*(10-R)

R could be 7 or 3. But, 3 is excluded by statement 1, so R can only be 7, and the answer is "yes". The statements together are sufficient.
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