A jar contains 8 red marbles and y white marbles. If Joan takes 2 rand

Question

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

Bunuel · Accepted Answer

# of total marbles in the jar equals to  8+y , out of which  R=8  and  W=y

P(RR)=\frac{8}{8+y}*\frac{8-1}{8+y-1}=\frac{8}{8+y}*\frac{7}{7+y} ;

P(RW)=2*\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y} , multiplying by 2 as RW can occur in two ways RW or WR;

Question: is  \frac{8}{8+y}*\frac{7}{7+y}>2\frac{8}{8+y}*\frac{y}{7+y} ? --> is  \frac{7}{2}>y ? --> is  y<3.5 ? eg 0, 1, 2, 3.

(1) y ≤ 8, not sufficient.
(2) y ≥ 4, sufficient, ( y  is not less than 3.5).

Answer: B.

gurpreetsingh · Answer

This question is from MGMAT CAT. I initially thought that it is a tough question and marked it randomly. But this question is quite easy.

Since the total number of events will be same in both the cases, the question is basically asking whether the number ways in which we can pick 2 red marble is greater than that of picking of one of each color.

=> is 8C2 > 8*y

=> is 8!/6!*2! > 8*y

is 7/2 > y

=> the question is is y < 3.5

Statement 1: y ≤ 8 not sufficient.

Statement 2: y ≥ 4 => the answer to the question is NO.. hence sufficient.

Raths · Answer

@Bunnel

will permutation be considered..?? . The question only says that 2 marbles were taken
out . It does not say that the marbles were taken out one by one . Please throw some light...me confused.

Bunuel · Answer

For probability there is no difference whether we pick some # of marbles one by one or all at once.

For example: 
Jar containing 3 red and 5 blue marbles. We pick 2 marbles one by one. What is the probability that both marbles are red? 
Jar containing 3 red and 5 blue marbles. We pick 2 marbles simultaneously. What is the probability that both marbles are red?

\frac{C^2_3}{C^2_8}=\frac{3}{28}  or  \frac{3}{8}*\frac{2}{7}=\frac{3}{28} .

As for the "permutation": what do you mean by that? Question asks to compare the probability "of 2 red marbles" with the probability of "one marble of each color". So can you pleas clarify what permutation has to do with it?

hemanthp · Answer

Bunuel, consider the case where Y = 2 . then The probability of P(RW) = 2*\frac{8}{10}*\frac{2}{9} = \frac{32}{90} P(RR) = \frac{8}{10}*\frac{7}{9} = \frac{56}{90} So even for y=2 , it is likely to pick two reds than a red and white. Isn't that contradicting the OA? As a matter of fact, as y keeps growing this likelyhood goes down and so y >!8 (y cannot be greater than 8). Or did I keep my foot in my mouth?

Bunuel · Answer

I'm not sure understand your question.

Anyway, according to the solution above when the number of white marbles (y) is 0, 1, 2, or 3 then the probability of picking two red marbles is  higher  than the probability of picking one marble of each color (and when the number of white marbles (y) is more than 3: 4, 5, ... then the probability of picking two red marbles is lower than the probability of picking one marble of each color). So clearly for  y=2 :  P(RR)>P(RW) .

Statement (1) says  y\leq{8} , not sufficient, as  y  could be for example 1 (answer YES) or 7 (answer NO).
Statement (1) says  y\geq{4}  --> as  y\geq{4}  then the probability of picking two red marbles is lower than the probability of picking one marble of each color and the answer to the question is NO. Sufficient

Answer: B.

Hope it's clear.

kannn · Answer

I used combinatorics approach. Can someone clarify whether the approach is correct? Please.

Total = (8+y)C2 
RR = 8C2 
one from each = 8C1 * yC1

Probability of RR =  \frac{8C2}{(8+y)C2} 
Proabaility of 1 from each =  \frac{8C1 * yC1}{(8+y)C2}

Question:  \frac{8C2}{(8+y)C2}  >  \frac{8C1 * yC1}{(8+y)C2} ?

Canceling denominator from both the sides : 8C2 > 8C1 * yC1 ?
 \frac{(8 * 7)}{2}  > 8 * (  \frac{y!}{(y-1)!} ) ?
 \frac{(8 * 7)}{2}  > 8 * y ?

\frac{7}{2}  > y ?

Thanks

subhashghosh · Answer

(1) Prob 2 red marbles = 8/16 * 7/15 = 7/30

Prob of 2 marbles of diff color = 8/16 * 8/15 + 8/16 * 8/15 = 2 * 8/16 * 8/15 = 8/15 = 16/30
 
 if y = 0, then we have a different result
 
 So (1) is not sufficient

(2)
 
 Prob 2 red marbles = 8/12 * 7/11 = 14/33
 
 Prob of 2 marbles of diff color = 8/12 * 8/11 + 8/12 * 8/11 = 2 * 8/12 * 8/11 = 32/33
 
 Anything greater than 4 also gives the same result.
 
 So (2) is sufficient
 
Answer - B

gmat1220 · Answer

Actually you don't have to do any calculation in 1). Consider the worst case.

y can be zero. In that P(two red) > P(either color) Hence Statement 1) y <= 8 fails to give one answer. On the other hand P(two red) < P (either color) for every y >=4 . Here you can pick y =4 and y = 5 to convince yourself. Probability of picking two of the same color will always be less than than picking two of different colors

Two red Vs 2 different colors
y = 4 P(two red) = 8C2 / 12C2 P(either color) = 8 * 4 / 12C2
y = 5 P(two red) = 8C2 / 13C2 P(either color) = 8 * 5 / 12C2

The first probability decreases with the increase in y. The second probability increases with the increase in y

amit2k9 · Answer

56/(8+y) * (7+y) > 16/(8+y) * (7+y)

56 > 16y => y > 3.

Clear B.

imhimanshu · Answer

Hey Bunuel,
Could you please clarify the below doubt. 
I know that we can calculate the probability by using Combination approach or by using Probability Approach.

The only doubt that I have is why do we need to multiply by 2 when I use the Probability approach.
i.e 
 P(RW)= 2 *\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y} 
Is it because we are considering cases over here 
i.e RW can occur in 2 ways - (R and then W) or (W and then R). 
If yes, then when we calculate the probability(RW) by using combination approach.. we don't multiply the expression by 2. Why. 
P(RW) =  /

Just to make it a bit clearer-

Question #1: A bag has 6 red marbles and 4 marbles. What are the chances of pulling out a red and blue marble.

Answer: (Chance of picking red * chance of picking blue) + Chance of picking blue*chance of picking red)

6/10*4/9 + 4/10*6/9 = 48/90

***Here, it appears that order matters, because you have to find the probability of getting red-blue, and add it to the probability of getting blue-red.

Question #2: 
A bag has 4 red marbles, 3 yellow, and 2 green, what is the probability of getting 2 Red, 2 Green, and 1 yellow, if the marbles aren't replaced:

# of ways to get 2R, 2G, 1Y 2: 4C2*2C2*3C1 =18 
Total # of ways to pick 5 balls: 9C5= 126

Answer: 18/126

Now here, it appears order DOES NOT matter, since we're using combinations (instead of permutations).

How come in Q#1, blue-red, red-blue are distinct, which means you have to add the probabilities of each together, but in Q#2, it uses combinations, which means order doesn't matter.

WHY? Completely lost

Please throw some light on this.
Thanks
-H

KarishmaB · Answer

Responding to a pm:
Order matters. That is the reason we multiply by 2 in the probability approach. We can pick R and then W or W and then R. So R and W can be picked in two different sequences with the same end result.

As for the combination approach - it works and you can use it but I wouldn't like to in this case. When we say 8C1, we are assuming that the balls are distinct and we can select one ball out of 8 Red balls in 8 ways. But it is kind of implied in the question that the balls are identical. Still, you can assume all the balls to be distinct and select 1 out of 8 Red balls in 8 ways and 1 out of y white balls in y ways.
Number of ways of selecting R and W = 8*y
Number of ways of selecting W and R = y*8
Total number of ways of selecting 2 balls = (8+y)(7+y) (first ball can be selected in 8+y ways and second in 8+y-1 ways)
Required probability = 16y/(8+y)(7+y) (which is the same as obtained in the probability method)
It works because in the numerator we assume all balls to be distinct and we do the same in the denominator too. Hence, the probability stays the same.

You can also calculate the probability assuming that order doesn't matter (for both numerator and denominator). Since it is probability, you will get the same probability since the 2 of the numerator will get canceled with the 2 of the denominator.

Number of ways of picking R and W = 8C1*yC1 (order doesn't matter so we don't multiply by 2)
Number of ways of picking 2 balls = (8+y)C2 (order doesn't matter when we use nCr)

Probability = 8C1*yC1/(8+y)C2 = 8y/(8+y)(7+y)/2 = 16y/(8+y)(7+y)
Obviously, the result is the same.

imhimanshu · Answer

Thanks Karishma for the explanation. I understand both the approaches now. However, just needed a little bit of more explanation on the concept of identical vs distinct as you mentioned in the below excerpt.

I have noticed that you have given emphasis on the distinct vs identical objects. I want to understand that as to how does the word identical vs distinct makes a difference when using combinations. I have understood that for solving probability questions,it won't make much of impact since we assume the same in both numerator and denominator. Please clear this doubt also, but wanted to understand this concept from Combinations point of view

Thanks
H

KarishmaB · Answer

You have four distinct balls. In how many ways can you select a ball? In 4 ways. In how many ways can you select 2 balls? In 4C2 ways. etc... You have four identical balls. In how many ways can you select a ball? In 1 way. No matter which ball you pick, it is no different from the other 3 balls so every case is the same. In how many ways can you select 2 balls? Again, only 1 way. You can select any two balls - they are all the same. Check out the following posts for more on identical and distinct objects. https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... 93-part-1/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... s-part-ii/

cjliu49 · Answer

Hi Bunuel, I have a question. I got B, and below is how I conceptualized it. On second thought, however, I realized that the logic behind my reasoning is weak. But my answer was still right. Was my intuition correct or was I just lucky?

Basically, I figured that if you wanted twice the chance to pull RR as RW, you would want twice as many R as W - so you would have 8 R and 4 W. In this case the chance of pulling R is 2x the chance of pulling W. Then, in order for it to be more likely to pull RR, W would have to be <4.

My basic intuition might be in the right direction, but I was just lucky that the #s worked out in my favor, right? Because I don't think my reasoning is mathematically logically sound, as it only accounts for one pull from the jar. Thank you!

EDIT: Just for anyone else thinking this way, I realized that I was just lucky to get the right answer. If instead you say there were 9 Red Marbles, there is obviously no way to have 4.5 White Marbles, and also, using Bunuel's method, the correct answer would be that the # of White Marbles must <4, not 4.5

gmatonline · Answer

'm not sure understand your question.

Anyway, according to the solution above when the number of white marbles (y) is 0, 1, 2, or 3 then the probability of picking two red marbles is higher than the probability of picking one marble of each color (and when the number of white marbles (y) is more than 3: 4, 5, ... then the probability of picking two red marbles is lower than the probability of picking one marble of each color). So clearly for y=2: P(RR)>P(RW).

Statement (1) says y\leq{8}, not sufficient, as y could be for example 1 (answer YES) or 7 (answer NO).
Statement (1) says y\geq{4} --> as y\geq{4} then the probability of picking two red marbles is lower than the probability of picking one marble of each color and the answer to the question is NO. Sufficient

Answer: B.

tnx dear

dheeraj24 · Answer

is P(rr) > p(rw)+p(wr) (rephrased question)

Stmt1 : (1) y ≤ 8
Let us assume y=8 ;
case1 : 8 red , 8 white

p(rr) = (8/16)*(7/15) = 7/30;
p(rw) = 2[(8/16) * (8/15)] = 16/30;

p(rr) < p(rw) ;
let assume y=2;
8 red , 2 white

p(rr) = (8/10)*(7/9) = 28/45
p(rw) = 2 * (8/10)* (2/9) = 16/45;

p(rr)> p(rw); Hence Insufficient;

Stmt2 : y ≥ 4

Assume y=4 then 8 red , 4 white ;

P(rr) = (8/12)* (7/11) =14/33
p(rw) = 2* (8/12)* (4/11) = 16/33;

P(rr) < P(rw)

Assume y =8 [from the case 1 we know that P(rr) < P(rw)]

Hence Sufficient; B

iPen · Answer

I did it a tiny bit differently...

Combinations....

Given: the question is  2R > 1R1Y ? Or, is  \frac{8!}{(2!*6!)} > \frac{8!}{(1!*7!)} * \frac{Y!}{(1!*(Y-1)!} ? Note that we don't even care about the total possible combinations, since both sides will be divided by the same amount and will cancel each other out.

They're simply asking, is  28 > 8*Y ? Or, is  \frac{7}{2} > Y  (is 3.5 > Y)?

(1) y ≤ 8

Y = 1: 3.5 > 1? Yes.
Y = 8: 3.5 > 8? No.

Insufficient.

(2) y ≥ 4

Y = 4: 3.5 > 4? No. (3.5 will never be greater than any integer 4 and greater.)

Sufficient.

Answer B.

GMATinsight · Answer

The solution is as mentioned in the image attached.a

Answer option B

A jar contains 8 red marbles and y white marbles. If Joan takes 2 rand

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A jar contains 8 red marbles and y white marbles. If Joan takes 2 rand

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