A certain jar contains only "b" black marbles, "w" white marbles and "r" red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen is red greater than the probability that the marble chosen will be white?

1) r/(b+w) > w/(b+r)

2) b - w > r

The question asks whether r/(b+w+r) > w/(b+w+r) or in other words is r > w?

1. r(b+r) > w(b+w)
br + r^2 > bw + w^2
br - bw > w^2 - r^2
b(r-w) > (w-r)(w+r)
r-w > (w-r)(w+r)/b ----> We know b is positive. So, we can divide both sides without changing the inequality
r-w > k(w-r) ----> Where k > 0 as b,r and w are all positive
This is true only when r > w.
If r < w, left side is -ve and right side is +ve and the inequality doesn't hold.
SUFFICIENT.

2. b - w > r
b > w + r
This doesn't tell us anything about relationship between w and r.
INSUFFICIENT.

Answer is A.

Suppose total is T , T=r+b+w

r/(b+w) > w/(b+r)

r/(b+w+r-r) > w/(b+w+r-w)

r/(t-r) > w/(t-w)

Since t>r and t>w, we can cross multiply.

rt-rw > wt-rw

rt > wt

Since t > 0

r > w

So 1 is sufficient.

2 is obviously not sufficient.

You guys are brilliant! The explanation makes perfect sense and the answer is (A).

Anadi, I like your solution. Good thinking.

You guys are all doing a great job here - you don't even need me! :)

adding +ve no r to lhs and w to rhs
proper frac behave properly
therefore
we have
2r/(b+r+w) > 2 w/(b +r+w)
divinding by +ve no 2

r/(b+r+w)[prob of red bll] > w/(b +r+w)[prob of white ball]

2 in insuffincent

hence ,A

I like Anadi's method better than mine, but...

1) r/(b+w) > w/(b+r)
2)r/(b+w) + (b+w)/(b+w) > w/(b+r) + (b+r)/(b+r) ---> I added 1 to both sides inthe form of (b+w)/(b+w) & (b+r)/(b+r)
3)(r+b+w)/(b+w) > (r+b+w)/(b+r) ---> Next I cross multiply
4)(b+r)(r+b+w) > (r+b+w)(b+w) ---> divide out the (r+b+w)
5)(b+r) > (b+w) ---> It's pretty obvious now, but you can take away the b's if you want
6)r>w

Also I don't agree with crackers' math. Can anyone explain it better?

A certain jar contains only "b" black marbles, "w" white marbles and "r" red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen is red greater than the probability that the marble chosen will be white?

P(red)=r/(b+w+r)
P(white)=w/(b+w+r)

P(red) > P (White) if r > w


1) r/(b+w) > w/(b+r)

gives
r/(b+w+r) > w/(b+w+r) [ this follows from: if A/B > C/D then A/(A+B) > C/(C+D) ]

hence r > w
Sufficient

2) b - w > r

Not Sufficient

I believe this also helps to explain cracker's method