doclkk 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)

Question

[spoiler]Can someone explain this question via picking #'s rather than algebra?

[/spoiler]Thanks!

I can´t be B) the answer as if in 2) y is greater than or equal to 4 y can be 8 as in 1) y less then or equal to 8
Source?

Let's take first case where y<=8

Put y = 8
Case A: probability of picking 2 Red Marble = 1/2*7/15 = 7/30
Case B: probability of picking 1 red and 1 white = 2*(1/2*8/15) = 16/30
Implies A<B

Put y = 1
Case A: probability of picking 2 Red Marble = 8/9*7/8 = 7/9
Case B: probability of picking 1 red and 1 white = 2*(8/9*1/8) = 2/9
Implies A>B

1 is Insufficient

Let's take second case where y>=4
Put y = 4
Case A: probability of picking 2 Red Marble = 8/12*7/11 = 14/33
Case B : probability of picking 1 red and 1 white = 2*(8/12*4/11) = 16/33

Implies A<B
Even for y = 8 as we saw in first case, A<B
It means, as y is increasing, we are still seeing that case A < case B. So statement 2 is sufficient to answer

El Cucu wrote:

doclkk 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 &#8804; 8
(2) y &#8805; 4

OA: (B)

Question

[spoiler]Can someone explain this question via picking #'s rather than algebra?

[/spoiler]Thanks!

I can´t be B) the answer as if in 2) y is greater than or equal to 4 y can be 8 as in 1) y less then or equal to 8
Source?

It's MGMAT ... very very very low likelihood that its wrong. I'm gonna go out on a limb and say that it is impossible that its wrong.

IMO B
probability of getting 2 red balls= 8C2/(8+Y)C2=28/(8+Y)C2

probability of getting one marble of each color=8Y/(8+Y)C2

1)y<=8
for Y=4, 28/(8+Y)C2<8Y/(8+Y)C2
but for Y=3,28/(8+Y)C2>8Y/(8+Y)C2

2)Y>=4, SUFFICIENT
28/(8+Y)C2<8Y/(8+Y)C2, always

Ritesh, why are you multiplying by 2 for Case B?

" Case B: probability of picking 1 red and 1 white = 2*(8/9*1/8) = 2/9 "

Hi tom4lax,
I am multiplying this with 2 because of following reason:

Case B: Probability of picking 1 red and 1 marble:
Case (i): Pick first red and second white = (8/9)*(1/8) = 1/9
Case (ii): Pick first white and second red = (1/9)*(8/8)=1/9
Case(i) + Case (ii) = 2/9

So when we have such equations, we should multiply by 2 because we have 2 similar cases of probability.


Thanks,
Ritesh

Ans: B