In a jar, 9 balls are white and the rest are red. If two balls are to

Question

In a jar, 9 balls are white and the rest are red. If two balls are to be chosen at random from the jar without replacement, the probability that the balls chosen will both be white is 6/11. What is the number of balls in the jar?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 15

(A) 10
(B) 11
(C) 12
(D) 13
(E) 15

BrentGMATPrepNow · Accepted Answer

P(both white) = P(1st chosen ball is white   AND   2nd chosen ball is white)
= P(1st chosen ball is white)   x   P(2nd chosen ball is white)

Let R = the number of red balls in the jar
So, at the beginning of the experiment, the total number of balls = R + 9

Continuing, we get....
P(both white) =  9/(R + 9)    x    8/(R + 8)

Aside: Before the first draw, there are R+9 balls, and 9 of them are white. This is why the 1st probability is 9/(R + 9)
After we draw a white ball on the first turn, there are R+8 balls remaining, and 8 of them are white. This is why the 2nd probability is 8(R+8)

We are told that P(both white) = 6/11
So, we can write:  9/(R + 9)    x    8/(R + 8)  = 6/11
Simplify: 72/(R+9)(R+8) = 6/11
Divide both sides by 6 to get: 12/(R+9)(R+8) = 1/11
Cross multiply: (R+9)(R+8)(1) = (12)(11)
Simplify: R² + 17R + 72 = 132
Subtract 132 from both sides to get: R² + 17R - 60 = 0
Factor to get: (R + 20)(R - 3) = 0
Solve: R = -20 or R = 3
Since our must be a positive number, it must be the case that R = 3 (in other words there are 3 red balls)

What is the number of balls in the jar?  
Now that we know there are 3 red balls, we can add this to the original 9 white balls to get a total of 12 balls

Answer: C

Cheers,
Brent

harsha3699 · Answer

The jar has 9 white balls, and the rest are red balls.
So, let total balls = t
The probability of picking 2 white balls without replacement is = 6/11
So, (9/t)x(8/t-1) = 6/11
on simplifying the above equation we would get a quadratic equation of the form,
t^2-t-132=0
Further simplifying we would get,
(t-12)(t+11)=0
Implies, t = 12, or t = -11
Since the no. of balls is supposed to be a positive integer, so t =12
So, the correct answer is C.

CrackverbalGMAT · Answer

In a jar, 9 balls are white and the rest are red. If two balls are to be chosen at random from the jar without replacement, the probability that the balls chosen will both be white is 6/11. What is the number of balls in the jar?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 15

Let say there are n balls, out of which 9 are white and the rest are red.

Total no of ways to select 2 balls out of n = nC2

P( both balls selected are white ) = 9C2/nC2 = 6/11

Since denominator is a multiple of 11, the only possible answer choices are either option B or C. Rest of the options you can eliminate.

It will not be option B as 11C2= 11*10/2 = 11*5 and there is no factor of 5 in 9C2 . Therefore , 9C2/11C2 will never give 6/11 as the answer. Option B is also eliminated.

So,  Option C is the correct answer .

Thanks,
Clifin J Francis

RastogiSarthak99 · Answer

My 2 cents on this:

Trying to keep it simple. Let total = x; We are told that When both balls are white, probability is 6/11. Therefore the equation then is

9/x - 8/x-1 = 6/11

Solve for x and you'll get x = 12

Hence, C

ThatDudeKnows · Answer

I have no objection to the solutions already provided...just offering another way to think about things.

How about PITA (Plugging In The Answers)? I like trying B and D. 
Let's start with B. There are 11 balls. 9 are white and 2 are red. (9/11)*(8/10) = 72/110 = 7.2/11. That's too high. B is wrong.
Let's try D. There are 13 balls. 9 are white and 4 are red. (9/13)*(8/12) = 72/156. That's less than 1/2, so it's too low. D is wrong.
We need something between B and D. The only option is C.

Answer choice C.

This question is a great example of why I like trying B and D. Once we tried B and knew we could eliminate it, how confident were you that we needed something bigger? If you were able to reason your way to that conclusion, that's awesome. But the great thing about BD on PITA is that you only need to test those two answer choices even if you had no idea whether you needed something bigger than B or smaller than B. If you'd tried D and it got you farther away from the right answer, you'd have known that you went the wrong way and needed to go back to A (you don't have to actually test A since it's the only thing the opposite direction). If you'd tried D and it got you closer but not all the way to the right answer, you'd have known that you needed to keep going and could go with E (again without actually testing it).

ThatDudeKnowsPITA

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