Data Sufficiency

DS Question - answer choices are as normal DS choices

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

[spoiler]OA is B i.e. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.[/spoiler]

My doubt is

if y < 8 then the likely hood of picking 2 red marbles > one marble of each color
if y = 8 then the likely hood of picking 2 red marbles = one marble of each color
if y > 8 then the likely hood of picking 2 red marbles < one marble of each color

Hence I marked answer as E without doing any mathematical calculation of prabablity

Can anyone clarify why this is wrong ?

The solution given is in a mathematical form which comes to conclusion that y > 3.5 for likely hood of picking 2 red marbles > one marble of each color, which is fine but I am not able to co-relate it with logical understanding.

BTW, the source of this ques is MGMAT CAT

This is how i solved.

8 Red marbles and y white marbles.
Qn: Is probability of choosing 2 red marbles more than probability of choosing a marble of each color?

Let me go straight to stmt II.

II. y >= 4

1. If y=3, 8 Red and 3 white.
P(2 Red) = 8C2 * 3C0 / 11C2 = 28/55
P(1 Red & 1W) = 8C1 * 3C1 / 11C2 = 24/55

P(2 Red) > P(1 Red & 1W).

2. If y=4, 8 Red and 4 white.
P(2 Red) = 8C2 * 4C0 / 11C2 = 28/55
P(1 Red & 1W) = 8C1 * 4C1 / 11C2 = 32/55

P(2 Red) < P(1 Red & 1W).

From y=4, P(2 Red) will be less than P(1 Red & 1W). So, y < 4 will always have P(2 Red) > P(1 Red & 1W).
Hence this choice is sufficient.

Coming to statement I, you have y <= 8. As you know, you get 2 different answers for y<4 and y>=4 respectively.
Hence it's insufficient.

In total, there are 8 red marbles, y white marbles, and 8 + y total marbles in the jar. The probability of obtaining two red marbles is given by:

P(Red AND Red) =


(8 / 8 + y) X (7 / 7 + y) = 56 / (8 + y) X (7 + y)

The probability of obtaining one red marble and one white marble is given by:

P(Red AND White) =

{(8 / 8 + y) X ( y / 7 + y )} X 2 = 16y / (8 + y) X(7 + y)

....mutliply by 2 ( as W R and R W)

So question is

IS

56 / (8 + y) X (7 + y) > 16y / (8 + y) X (7 + y) ????


56 > 16y ?

3.5 > y ?


(1) in suff y < = 8 ...can be greater than 3 .say 4 or 2.5 less than 3
(2) suff : We know that y is NOT less than 3 ... therefore y atleast 4



B wins...

Simple..method !!