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8 red marbles and y white marbles

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8 red marbles and y white marbles

by Makushr1 » Sat Aug 28, 2010 3:57 pm
From MGMAT CAT 3

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA B
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by Makushr1 » Sat Aug 28, 2010 4:04 pm
Makushr1 wrote:From MGMAT CAT 3

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA B
I put down E, but I'm wondering how 2 is suff, why is my method wrong?

The way I did it is to say that on 2, Y can be equal to 4.

So, 8/12 chance for a red on the first time, and then 7/11 on the 2nd time = 56/132.
Chance for pull 1 red and 1 white would be 8/12 (or 4 out of 12) x 4/11 (or 8/11) = 32/132
So there is a greater chance to pull out 2 reds. but if y were 100, there would be a greater chance to pull out 1 red and 1 white, rather than 2 reds.

The way they explain is that you have to multiply the 1 red and 1 white by 2, since it can either be w-r or r-w. Why does that matter?
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by sdotcruz » Wed Sep 01, 2010 7:04 am
Can anyone else respond? I thought E was the answer since neither statement provided a number that would definitely make y less than 7. Both statements together limit y, but still do not confirm y is less than, greater than or equal to 7. The reason i thought finding if 7>= Y would provide an answer is because the questions asks is (8 red balls first draw * 7 red balls second draw) 56/number of total items > 8 red*Y/ number of total items. The denominators will always be the same no matter what number Y is. I then looked at only the numerators and created the inequality is 56> 8*Y. Then reduced the inequality to is 7>= Y?
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by Adi_Pat » Thu Sep 02, 2010 10:37 am
Would Pick (E)

(a) y <=8 ...pick y=7 and y=8

P(2R ) = 8/15*7/14 = P ( 1R 1W ) = 8/15*7/14
P(2R) = 8/16*7/15 < P(1R 1W )= 8/16*8/15

More than one possibility. INSUFFICIENT

(b) Y >= 4

Similarly, comparing the same probabilities as above

8/12*7/11 > 8/12*4/11 ( Y =4 )
8/16*7/15 < 8/16*8/15 ( Y=8 )

INSUFFICIENT.

Combining the two statement doesnot eliminate the same possibilities as discussed in B.

hence, (E)
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by sdotcruz » Thu Sep 02, 2010 11:55 am
Actually i believe i know why the answer is B. First this is a DS questions and the question is seeking a yes/no answer. The question asks is it more likely that two red marbles are selected than one of each color. Is P(r)*P(r) > P(w)* P(r)?.

If we can determine the number of red marbles is greater than, equal to or less than the white marbles we can provide a yes or no answer. We do not need a value to answer yes or no.

B allows us to respond No the the question. For instance if there are 4 white and 4 red the probability of selecting 2 red marbles is less than selecting one white and one red. Answer to the question is No.

If there are 5 white and 3 red the probability of selecting 2 red will be less than selecting 1 red and 1 white. Answer to the question is No.

Answer A as we all know allows the number of white marbles to be 1 or 7 which does not provide a conclusive no or yes.
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by Gurpinder » Thu Sep 02, 2010 1:55 pm
I would still say (E)

8 - red marbles
y - white marbles

question: prob. of 2 red > 1 of each

(1) y<=8

lets say y = 8 and y = 3
if y =8
then the prob. of both red = .46
the prob. of 1 of each = .46

if y = 3
then the prob. of both red = .70
the prob. of 1 of each = .30

so clearly, stmt 1 is insufficient. we dont get a definite yes.

(2) Y>=4

again, lets say y = 4 and y = 8

if y =8
then the prob. of both red = .46
the prob. of 1 of each = .46

if y=4
then the prob. of both red = .63
the prob. of 1 of each = .36

this one is also insufficient! if y=4, then the probability of both red is greater than selecting 1 of each, if y increases, the probability reduces.

Using these statements together is also insufficient. The numbers I tested each statement fall within the range if we were to use both statements together (4-8).

therefore the answer should be (E).

Check the OA with MGMAT experts!

Btw....Makushr1 how did you score on the MGMAT Cat if you don't mind?
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by neerajkumar1_1 » Thu Sep 02, 2010 8:10 pm
Makushr1 wrote:
Makushr1 wrote:From MGMAT CAT 3

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA B
I put down E, but I'm wondering how 2 is suff, why is my method wrong?

The way I did it is to say that on 2, Y can be equal to 4.

So, 8/12 chance for a red on the first time, and then 7/11 on the 2nd time = 56/132.
Chance for pull 1 red and 1 white would be 8/12 (or 4 out of 12) x 4/11 (or 8/11) = 32/132
So there is a greater chance to pull out 2 reds. but if y were 100, there would be a greater chance to pull out 1 red and 1 white, rather than 2 reds.

The way they explain is that you have to multiply the 1 red and 1 white by 2, since it can either be w-r or r-w. Why does that matter?

Hey even i would have gone for E.. but as u have mentioned about multiplying by 2... then it struck to me that indeed B is sufficient all by itself...

consider y=4

when u take out 2 red marbles... we know the prob = 8/12 x 7/11 = 14 /33

next one of each color .. we know the prob = 8/12 x 4/11 = 8/33 but in this case we have taken only one arrangement i.e first red and then white...
to take all arrangements into consideration... we will have the same prob for first white and then red...
hence total number of arrangements = 2
therefore prob of one of each color = 8/33 * 2 = 16/33

prob of 2 reds is less

now as y increases you will notice that the prob of 2 reds always remains less compared to one of each colour...
for e.g
take y= 100

prob of 2 reds = 8/108 * 7/107 = 14/(107 x 27)

prob of one of each colour = 8/108 x 100/107 x 2 = 400/(107 x 27)


Hope this helps....
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by blaster » Fri Sep 03, 2010 1:32 am
one more vote for E
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by singhsa » Tue Sep 07, 2010 12:44 pm
Makushr1 wrote:From MGMAT CAT 3

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA B
Statement 1 is clearly not sufficient as y=1 or y=8, which will give conflicting results.

from statement 2, we know that y=4or5or6...so the least no of white marbles in this case would be when y=4.
So, say y=4
Now, we have 8 red marbles and 4 white marbles.
Prob of picking 2 red marbles = 8c2/12c2=28/66
Prob of picking 1 red and 1 white marble = 8c1*4c1/12c2=32/66.
Here the prob of picking 2 red is less than prob of picking marbles of different color.
It is but obvious that any value of y more than 4 will increase further the prob of picking marbles of different color.
Thus the answer as per statement 2 is a definite no. Therefore, statement 2 is sufficient
IMO - [/spoiler]
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by Adi_Pat » Wed Sep 08, 2010 12:36 pm
Gurpinder wrote:I would still say (E)

8 - red marbles
y - white marbles

question: prob. of 2 red > 1 of each

(1) y<=8

lets say y = 8 and y = 3
if y =8
then the prob. of both red = .46
the prob. of 1 of each = .46

if y = 3
then the prob. of both red = .70
the prob. of 1 of each = .30

so clearly, stmt 1 is insufficient. we dont get a definite yes.

(2) Y>=4

again, lets say y = 4 and y = 8

if y =8
then the prob. of both red = .46
the prob. of 1 of each = .46

if y=4
then the prob. of both red = .63
the prob. of 1 of each = .36

this one is also insufficient! if y=4, then the probability of both red is greater than selecting 1 of each, if y increases, the probability reduces.

Using these statements together is also insufficient. The numbers I tested each statement fall within the range if we were to use both statements together (4-8).

therefore the answer should be (E).

Check the OA with MGMAT experts!

Btw....Makushr1 how did you score on the MGMAT Cat if you don't mind?

you're not considering the or part in the probability... the 1R and 1W or 1W and 1R

P( 2 Reds ) = 8/12 * 7/11=56/12*11
P ( 1 Red and 1 White ) or P(1w and 1r ) = 8/12*4/11 + 4/12*8/11 = 64/12*11
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by debmalya_dutta » Wed Sep 08, 2010 1:28 pm
Makushr1 wrote:From MGMAT CAT 3

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

OA B
the question rephrased
P(R,R) > P(R,W) - hope I am thinking it the correct way

P(R,R) = 8C2/ (8+y)C2 = 28 / (8+y)C2 - EQN (1)
P(R,W) = 8* y/ (8+y)C2 = 8* y/ (8+y)C2 EQN (2)

The statement 1 states y ≤ 8
when Y = 2,3,4,5 , 6, 7 , 8 EQN (2) maybe > or lesser than EQN (1)

Hence St 1 is insufficient

Now take statement (2)
It says y ≥ 4
Put the values of y starting from 4 and then go on ...5,6,,7,8,9,10 etc
For each of the values of Y, 8* y > 28 and hence EQN (2) always > EQN (1)

Hence Statement 2 is sufficient

Answer is B
@Deb
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by ishkaran » Thu Sep 23, 2010 11:27 am
I found a simpler approach
As we have to choose from 8 Red and Y whites marbles

I rephrased the questions as:-
Is Possibility of (2 Reds) > (1Red, 1 White)
Is 8C2 > 8C1 * YC1
Solving, I get 3.5>Y

Therefore rephrasing the questions: Is Y<3.5 ?


A) Statement A says that Y<=8

Taking different values of Y= 8 (Answer is No)
Taking Y=2 (Answer is Yes)

Therefore we get multiple answer using statement A, hence it can be eliminated.


B) Statement B clearly says that Y>=4

I get a distinct answer



So using statement B only I can solve the question hence B
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by this_time_i_will » Thu Sep 23, 2010 6:27 pm
This is a real good question.
Above posts have given beautiful mathematical explanation.
Also, i was trying to solve logically and understand why E is not the right answer. It is to be noted that the event of picking two red balls togethar (8c2) is different from event of picking two red balls one-by-one (8c1*7c1). And people who selected E, solved it as per the latter event.
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