niketdoshi123 wrote: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

Plug in the THRESHOLDS and try EXTREMES.
Here, the THRESHOLDS are y=8 and y=4.

Statement 1: y≤8
Case 1: 8 white marbles, 8 red marbles
P(RR) = 8/16 * 7/15 = 7/30.

P(one of each color):
P(RW) = 8/16 * 8/15 = 8/30.
Since RW can be reversed to WR, we multiply by 2:
2(8/30) = 16/30.

In this case, P(RR) < P(one of each color).

Case 2: 0 yellow marbles, 8 red marbles
Here, P(RR) = 1 and P(one of each color) = 0.
In this case, P(RR) > P(one of each color).
INSUFFICIENT.

Statement 2: y≥4
Case 1: 4 white marbles, 8 red marbles
P(RR) = 8/12 * 7/11 = 14/33.

P(one of each color):
P(RW) = 8/12 * 4/11 = 8/33.
Since RW can be reversed to WR, we multiply by 2:
2(8/33) = 16/33.

In this case, P(RR) < P(one of each color).

When y=4, P(RR) < P(one of each color).
When y=8, P(RR) < P(one of each color).
Increasing the number of yellow marbles beyond y=8 will only DECREASE the likelihood of getting RR.
Thus, P(RR) < P(one of each color).
SUFFICIENT.

The correct answer is B.

niketdoshi123 wrote: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

We need to find if the probability of 2 reds is greater than one red, one white. Let's do that:

Probability of both reds = p = 8/(8+y) * 7/(8+y-1)
Probaility of one red, one white = q = 2! * 8/(8+y) * y/(8+y-1)
we need p>q
8*7 > 2* 8*y
=> y < 7/2. since y is an integer, the question rephrases to Is y <= 3? Now its an easy question.

(1) y ≤ 8
y may be less than 3 or greater than 3. Insufficient.

(2) y ≥ 4
y is definitely greater than 3. Sufficient.

B is correct.

GMATGuruNY wrote: Since RW can be reversed to WR, we multiply by 2

eagleeye wrote: Probability of one red, one white = q = 2! * 8/(8+y) * y/(8+y-1)

Here the order in which we take 2 marbles is not important, so why do we have to multiply by 2??

niketdoshi123 wrote:

GMATGuruNY wrote: Since RW can be reversed to WR, we multiply by 2

eagleeye wrote: Probability of one red, one white = q = 2! * 8/(8+y) * y/(8+y-1)

Here the order in which we take 2 marbles is not important, so why do we have to multiply by 2??

When you do a probability calculation, it doesn't matter whether the order is considered.
Now you can go two ways about it.

1. Use the permutations approach using different ordered cases, like Mitch and I did here. Then you need to consider what's picked first. Whenever you are using FCP, (Fundamental Counting Principle aka the slot method) you are inherently considering order in the options. Hence we multiply by 2!.

The other way, of course is the basic selection probability approach. Then:
2. Use the combinatorics approach without the consideration of order. Then the probability is

P (one red, one white) = 8C1*yC1/( (8+y)C2) = 8*y / ((8+y)(8+y-1)/2!) = 2! *8*y /((8+y)(8+y-1)) = 2* 8/(8+y) * y/(8+y-1) which is the same as with the FCP approach.

Let me know if the solution/explanation helped

eagleeye wrote:

niketdoshi123 wrote:

GMATGuruNY wrote: Since RW can be reversed to WR, we multiply by 2

eagleeye wrote: Probability of one red, one white = q = 2! * 8/(8+y) * y/(8+y-1)

Here the order in which we take 2 marbles is not important, so why do we have to multiply by 2??

When you do a probability calculation, it doesn't matter whether the order is considered.
Now you can go two ways about it.

1. Use the permutations approach using different ordered cases, like Mitch and I did here. Then you need to consider what's picked first. Whenever you are using FCP, (Fundamental Counting Principle aka the slot method) you are inherently considering order in the options. Hence we multiply by 2!.

The other way, of course is the basic selection probability approach. Then:
2. Use the combinatorics approach without the consideration of order. Then the probability is

P (one red, one white) = 8C1*yC1/( (8+y)C2) = 8*y / ((8+y)(8+y-1)/2!) = 2! *8*y /((8+y)(8+y-1)) = 2* 8/(8+y) * y/(8+y-1) which is the same as with the FCP approach.

Let me know if the solution/explanation helped

It is still not 100% clear to me..
Maybe I am unable to state my doubt precisely.

Here's the gist of what I did while solving the question

P(RR) = 8/(8+y) * 7/(7+y)
P(RW) = 8/(8+y) * y/(7+y) Here I thought that P(RW) = P(WR), so considering any one of them will solve the purpose. This is where I got wrong

Now , what I understand from yours and Mitch's posts is that

P(one of each color) = P(WR) + P(RW) or 2!*P(WR) (2! -- since there are 2 colors to chose from)

If there were 3 different colors -- red, white and green, then
would P(one of each color) be 3!*P(WRG) ?

niketdoshi123 wrote: If there were 3 different colors -- red, white and green, then
would P(one of each color) be 3!*P(WRG) ?

Yes. You are right. It will be multiplied by 3!.

For example, let's make up a question where we have 5 White, 4 Red, and 3 Green marbles. We are asked the probability that if a player picks three marbles at random:
(i) No two marbles are of the same color.
(ii) All of the marbles picked are white.

(i) P (no two are the same color) = 3! * 5/12 * 4/11 * 3/10.

(ii) P (all three are white) = 3!* 5/12 * 4/11 * 3/10 * 1/(3!). (Here we divide by 3! because all 3 colors are identical so the two factorials cancel out)

Works?

niketdoshi123 wrote:

eagleeye wrote:

niketdoshi123 wrote:

GMATGuruNY wrote: Since RW can be reversed to WR, we multiply by 2

eagleeye wrote: Probability of one red, one white = q = 2! * 8/(8+y) * y/(8+y-1)

Here the order in which we take 2 marbles is not important, so why do we have to multiply by 2??

When you do a probability calculation, it doesn't matter whether the order is considered.
Now you can go two ways about it.

1. Use the permutations approach using different ordered cases, like Mitch and I did here. Then you need to consider what's picked first. Whenever you are using FCP, (Fundamental Counting Principle aka the slot method) you are inherently considering order in the options. Hence we multiply by 2!.

The other way, of course is the basic selection probability approach. Then:
2. Use the combinatorics approach without the consideration of order. Then the probability is

P (one red, one white) = 8C1*yC1/( (8+y)C2) = 8*y / ((8+y)(8+y-1)/2!) = 2! *8*y /((8+y)(8+y-1)) = 2* 8/(8+y) * y/(8+y-1) which is the same as with the FCP approach.

Let me know if the solution/explanation helped

It is still not 100% clear to me..
Maybe I am unable to state my doubt precisely.

Here's the gist of what I did while solving the question

P(RR) = 8/(8+y) * 7/(7+y)
P(RW) = 8/(8+y) * y/(7+y) Here I thought that P(RW) = P(WR), so considering any one of them will solve the purpose. This is where I got wrong

Now , what I understand from yours and Mitch's posts is that

P(one of each color) = P(WR) + P(RW) or 2!*P(WR) (2! -- since there are 2 colors to chose from)

If there were 3 different colors -- red, white and green, then
would P(one of each color) be 3!*P(WRG) ?

I suggest that you check out my posts here:

https://www.beatthegmat.com/mgmat-probabilty-t87745.html

https://www.beatthegmat.com/rain-prob-t88396.html

https://www.beatthegmat.com/select-exact ... 88786.html

https://www.beatthegmat.com/kids-probabi ... 04374.html

https://www.beatthegmat.com/baseball-t69721.html

https://www.beatthegmat.com/probability- ... 91490.html

https://www.beatthegmat.com/princeton-re ... 78922.html