Amy, Brianne, and Cedric will each choose exactly 1 container, Contain

Question

Amy, Brianne, and Cedric will each choose exactly 1 container, Container X or Container Y, and then randomly pick 1 marble, without replacement, from that container. Container X has 3 red marbles and 10 white marbles, while Container Y has 2 red marbles and 8 white marbles. If Amy and Brianne each pick 1 marble before Cedric picks 1 marble, which of these containers should Cedric choose to maximize the probability that the 1 marble he picks will be a red marble?

(1) Cedric knows that Amy picked 1 red marble from Container X.
(2) Cedric knows that Brianne picked 1 red marble from Container Y.

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Official Answer

B

stne · Accepted Answer

Initially in container  x  we have R =3  , W =10  and in  y  we have R =2  and W  =8

(1) Cedric knows that Amy picked 1 red marble from Container x.

After Amy picks, the status becomes:

x  : R =2  , W =10  and  y  : R =2,  W = 8

Now if Brian picks  1  Red from  x  the status becomes:

x : R =1 , W =10  and  y  : R =2,  W = 8

Now  p  of red  1/11< 2/10 . Therefore Cedric should choose container  y

But if Brian picks  1  red from  y  then the status becomes:

x  : R =2  , W =10  and  y  :R =1,  W = 8

Now  p  of red  2/12 > 1/9  Therefore now Cedric should choose container  x

Since we already have a YES and a NO there is no need to test more cases with white marble.

INSUFF.

(2) Cedric knows that Brianne picked 1 red marble from Container Y.

Initially in container  x  we have R =3  , W =10  and in  y  we have R =2  and W =8

After Brian picks  1  Red from  y  the status becomes:

x  we have R =3  , W =10  and in  y  we have R =1  and W =8

Red cases:

Now if Amy picks  1  Red from  x  the status becomes:

x  : R =2  , W =10  and  y  :R =1,  W = 8

p  Red  2/12 > 1/9  Therefore Cedric should choose container  x

But if Amy picks  1  red from  y  then the status becomes:

x  we have R =3  , W =10  and in  y  we have R =0  and W =8

p  Red  3/13 > 0  Therefore now also Cedric should choose container  x

White cases:

if Amy picks  1  white from  x  the status becomes:

x:   R =3  , W =9  and  y  : R =1,  W = 8

p  red  3/12 > 1/9  Hence now also Cedric needs to choose container  x.

if Amy picks  1  White from  y  the status becomes:

x  we have R =3  , W =10  and in  y  we have R =1  and W =7

p  red  3/13 >1/8,  again Cedric needs to choose container  x.

Thus we see in statement(2) after Brian picks, whatever Amy picks from whichever container, Cedric should always cointainer  x  to get higher probability of choosing red.

SUFF.

Ans B

Hope it helped.

gmatophobia · Answer

Statement 1 (1) Cedric knows that Amy picked 1 red marble from Container X. Case 1 : Amy picked 1 red marble from Container X, and Brian picked 1 red marble from Container Y When Cedric picks - Probability(Red Marble in Container X) = \frac{2}{12} = 0.167 Probability(Red Marble in Container Y) = \frac{1}{9 }= 0.111 Therefore Probability(Red Marble in Container X) > Probability(Red Marble in Container Y) Case 2 : Amy and Brian picked 1 red marble from Container X When Cedric picks - Probability(Red Marble in Container X) = \frac{2}{11} = 0.0909 Probability(Red Marble in Container Y) = \frac{2}{8 }= 0.25 Therefore Probability(Red Marble in Container Y) > Probability(Red Marble in Container X) As we are getting contradictory answers in both cases, the statement alone is not sufficient to answer the question. Eliminate A, and D. Screenshot 2024-01-28 230538.png Statement 2 (2) Cedric knows that Brianne picked 1 red marble from Container Y. Case 1 : Amy picked 1 red marble from Container X, and Brian picked 1 red marble from Container Y When Cedric picks - Probability(Red Marble in Container X) = \frac{2}{12} = 0.167 Probability(Red Marble in Container Y) = \frac{1}{9 }= 0.111 Therefore Probability(Red Marble in Container X) > Probability(Red Marble in Container Y) Case 2 : Amy and Brian picked 1 red marble from Container Y When Cedric picks - Probability(Red Marble in Container X) = \frac{3}{13} Probability(Red Marble in Container Y) = 0 Therefore, Probability(Red Marble in Container X) > Probability(Red Marble in Container Y) Screenshot 2024-01-28 230937.png Both cases lead to the same answer. Hence, the statement is sufficient to answer the question asked. Option B.

sayan640 · Answer

gmatophobia  , Why did you not consider the "white" cases ? Amy and Brianne could have picked up white balls also..

Is it because picking up 'red' balls will lead to the worst possible cases ? Please explain and help me understand..

gmatophobia · Answer

sayan640

We have to assume the worst case possible here. If Amy or Brianne picks while marble, the action is anyways in favor of Cedric. They now get more red marbles to choose from.

user1937 · Answer

What is the optimal timing for this question? Is it a 2 min question?

Ankit__7182 · Answer

approx 1 minute. This is a sub 600 level question.

KarishmaB · Answer

X: Probability of picking a red marble = 3/13

Y: Probability of picking a red marble = 2/10

Cedric should pick from the container from which his probability of picking a red marble is higher. Currently, probability of red in container X is higher. But after Amy and Brianne pick, we do not know which container has higher probability.

(1) Cedric knows that Amy picked 1 red marble from Container X.

X: Probability of picking a red marble = 2/12 now. But we do not know what Brienne picked. If she picked a red marble from X, Cedric should select Y. If she picked a red marble from Y, Cedric should select X. 
Not sufficient.

(2) Cedric knows that Brianne picked 1 red marble from Container Y.

Y: Probability of picking a red marble = 1/9 now. 
We do not know what Amy picked, but if she picked a red marble from X, X will still have higher probability than Y because it will still have 2/12 probability of a red marble which is higher than 1/9. 
If Amy picked from Y, then anyway, Cedric should select X again.
So in either case here, Cedric should select X.

Sufficient alone.

Answer (B)

Note that this is an easy (C) trap question.

bumbleBee3445 · Answer

Q: which container has max probability of Cedric choosing 1 red marble?

Given: 
    
 x 
 y 
 
  red 
 3 
 2 
 
  white 
 10 
 8

Option 1: Amy picked 1 red marble from cont. x. now X has 2 red and 10 white. 
now, Brianne can pick from any container.

case a:  B picks from X: now x has 1 red and 10 white. 
hence P(C choosing red from x)=  \frac{1}{11} 
Similarly, P(C choosing red from y)=  \frac{2}{10} =  \frac{1}{5} 
 P(C choosing red from y) is greater for this case.

case b:  B picks from y: now x has 2 red and 10 white, y has 2 red and 8 white
P(C choosing red from x)=  \frac{2}{12} =  \frac{1}{6}  
P(C choosing red from y)=  \frac{2}{10} =  \frac{1}{5}  
 P(C choosing red from x) is greater for this case.

Hence not conclusive and insufficient.

Option 2: Brainne picked 1 red marble from cont. y. now X has 3 red and 10 white and y has 1 red and 8 white 
 now, Ann can pick from any container.  
 
 case a: A picks from X: now x has 2 red and 10 white.  
hence P(C choosing red from x)=  \frac{1}{5} 
Similarly, P(C choosing red from y)=  \frac{1}{9} 
 P(C choosing red from x) is greater for this case.

case b: A picks from y: now x has 3 red and 10 white, y has 0 red and 8 white 
P(C choosing red from x)=  \frac{3}{12} =  \frac{1}{4}  
P(C choosing red from y)=  \frac{0}{8} = 0 
 P(C choosing red from x) is greater for this case.

Hence P(C choosing red from x) is higher, sufficient to answer

Hence Ans is B.

PeachSnapple1 · Answer

No it's not. And this kind of questions take average 2 minutes, and up to 2'30".

Amy, Brianne, and Cedric will each choose exactly 1 container, Contain

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