probability - red and white marbles!

[vscid]

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

[Stuart@KaplanGMAT]

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

Let's dive right into the statements and use trial and error.


(1) if y=1, then:

prob(2 red) = 8/9 * 7/8 = 56/72
prob(1 each) = 8/9 * 1/8 + 1/9 * 8/8 = 8/72 + 8/72 = 16/72

(for 1 each we need to consider the case of red then white or vice-versa).

So, if y=1 we get a "yes" answer.

if y=4 (4 looks like an important number, since it will also help with statement (2)), then:

prob(2 red) = 8/12 * 7/11 = 56/123

and

prob(1each) = 8/12 * 4/11 + 4/12 * 8/11 = 32/123 + 32/123 = 64/123

So, for y=4 we get a "no" answer.

We can get both a "yes" and "no": insuffcient, eliminate A and D.

(2) based on our work from (1), we know that y=4 generates a "no" answer.

If we increase y we decrease the probability of getting double red, so if y ≥ 4 we're always getting a "no" answer: sufficient.

(2) is sufficient, (1) isn't: choose B.

[ajith]

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

Probability that she will have 2 red marbles = 8/(8+y)*7/(7+y)

Probability that she will get red first and yellow second=8/8+y*y/y+7
Probability that she will get yellow and then red = y/(y+8)*8/(y+7)

it is more likely that she will have 2 red marbles when 8*7>2*y*8
y<3.5

1) is not sufficient
2) sufficient
[spoiler]
b[/spoiler]