A certain jar contains only "b" black marbles

[Harish Dorai]

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

[givemeanid]

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.

[Anadi]

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.

[Harish Dorai]

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

[givemeanid]

Anadi, I like your solution. Good thinking.

[StaceyKoprince]

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

[cracker]

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

[k0nc3pt10n]

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?

[Guest]

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

[MBA Action]

Ok, I have another approach

Can we say "safely" that since the total is fixed then
if: r/(rest(r)) > w/(rest(w)) as given in statement 1 then definitely r>w?
(rest(r) means rest of the marbles without r)

[RonPurewal]

Ok, I have another approach

Can we say "safely" that since the total is fixed then
if: r/(rest(r)) > w/(rest(w)) as given in statement 1 then definitely r>w?
(rest(r) means rest of the marbles without r)

yes, you can.

if you have a fixed total, then, as the quantity of something increases, the quantity of everything else will decrease accordingly. therefore, the higher the quantity of the "something", the higher the ratio of that "something" to "everything else" will be.

nice observation.

[pddtruong]

1)
a) r(b+r) > w(b+w)
b) r(b+r) + rw > w (b+w) + rw [Add same number to both sides]
c) r[(b+r)+w] > w[(b+w)+ r] [factor out]
d) divide both sides: r/(b+w+r) > w/(b+w+r). Sufficient.

2) Ins.

[ishkaran]

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?

Found this approach the best. Thanks

[RonPurewal]

good stuff

[sudeepkapoor]

Taking statement (1)

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

taking reciprocal ,

(b+w)/r < (b+r)/w

[take the example of 1/2 and 1/3 ; 1/2 > 1/3 but if one takes the reciprocal , 2<3 ]

now, add 1 to both sides,
(b+w)/r +1 < (b+r)/w +1 [inequality holds good when a positive constant is added]

This implies , (b+w+r)/r < (b+r+w)/w

Again take the reciprocal and the sign changes

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

also we know that :

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


therefore P(red) > P(white)

Therefore statement 1 is sufficient

Statement (2) does not give any relation between red and white marbles and is obviously not sufficient ;

Answer is A.