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 Pls explain
Marbles
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pick 2 marbles out of 8. 8C2 = 28.crackgmat007 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
OA - B Pls explain
pick 1 red, 1 white. 8C1 * y C1 (where y is the total white marbles)
you want to find out which of the two is big.
(B) gives the answer. even you take y = 4, 8*4 = 32, which is greater than 28. therefore any value greater than 4 gives a number greater than 32.
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I thought this problem is about probability because it says 'which one is more likely'. Let me know what you think.
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This is a probability problem. Here's how I did it:
The probability of picking 2 red marbles is
56/[(8+y)(7+y)]
Here's how:
There are nCr(8, 2) ways to pick 2 red marbles, and nCr(8+y, 2) ways to choose marbles in general. Now, nCr(8,2) = 8!/(2!6!) = 56, and nCr(8+y,2) = (7+y)(8+y)/2, so we get the above.
Similarly, for choosing 1 red and 1 white marble, we have 8 ways of choosing a red marble and y ways of choosing a white marble, so our probability is
8y/nCr(8+y,2) = 16y/(8+y)(7+y)
We need to be able to definitely compare
16y/(8+y)(7+y) and 56/(8+y)(7+y)
Multiplying both out by denominators, we can just compare 16y and 56. If y <= 8, this doesn't give us anything... y could be 4, in which case 16y > 56, or y could be 2 in which case 16y < 56. So, cross off A. However, looking at B, if y >= 4, 16y > 56 always. So, our answer is B.
The probability of picking 2 red marbles is
56/[(8+y)(7+y)]
Here's how:
There are nCr(8, 2) ways to pick 2 red marbles, and nCr(8+y, 2) ways to choose marbles in general. Now, nCr(8,2) = 8!/(2!6!) = 56, and nCr(8+y,2) = (7+y)(8+y)/2, so we get the above.
Similarly, for choosing 1 red and 1 white marble, we have 8 ways of choosing a red marble and y ways of choosing a white marble, so our probability is
8y/nCr(8+y,2) = 16y/(8+y)(7+y)
We need to be able to definitely compare
16y/(8+y)(7+y) and 56/(8+y)(7+y)
Multiplying both out by denominators, we can just compare 16y and 56. If y <= 8, this doesn't give us anything... y could be 4, in which case 16y > 56, or y could be 2 in which case 16y < 56. So, cross off A. However, looking at B, if y >= 4, 16y > 56 always. So, our answer is B.
Have you wondered how you could have found such a treasure? -T
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IMO - E
1. y <=8
The probability will change based on the value of y. If y=8 then the probability of having bith read marbles will be less and this probability will increase as the vlaue of y reduces.
However we cannot determine the value of this probability since y is not known
2. y>=4
Again here , the probability of choosing both red will be maximum when y=4 and will keep reducing as we increase the vlaue of y
Since value of y is not known we cannot determiine
Even using both statements y can have values from 4 to 8. Not sufficient
For me a sufficient soln would be something like
y<=7
Since we know the maximum value of the number of white marbles.
I am still not able to comprehend the OA !
1. y <=8
The probability will change based on the value of y. If y=8 then the probability of having bith read marbles will be less and this probability will increase as the vlaue of y reduces.
However we cannot determine the value of this probability since y is not known
2. y>=4
Again here , the probability of choosing both red will be maximum when y=4 and will keep reducing as we increase the vlaue of y
Since value of y is not known we cannot determiine
Even using both statements y can have values from 4 to 8. Not sufficient
For me a sufficient soln would be something like
y<=7
Since we know the maximum value of the number of white marbles.
I am still not able to comprehend the OA !