Gprep2-Probabilty Jar/Marbles

I solved it this way :

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


so

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

= (b+w)/r +1 <(b+r)/w +1 = (b+w+r)/r <(b+r+w)/w = r/(b+w+r) > w/(b+r+w) and so A is sufficient .

But ain't both the statements say the same thing?

From Stmt 1) r.(b+r) > w.(b+w) => rb + r^2 > wb + w^2.
=> rb + r^2 + rw > wb + w^2 + rw [add rw on both sides]
So, r.(b+w+r) > w.(b+w+r). Thus, sufficient

From Stmt 2) b > r + w = b.(r-w) > (r+w).(r-w) [multiply by r-w]
=> rb - bw > r^2 - w^2
=> r.(b+r) > w(b+w), which is actually Stmt1

So if Stmt1 is Suff, then Stmt2 shd also be Suff. Can anybody explain?

The question is \(\frac{R}{R+B+W}>\frac{W}{R+B+W}\) true? Or is \(R>W\) true?

(1) \(\frac{R}{B+W} > \frac{W}{B+R}\) --> \(\frac{R}{B+W} +1> \frac{W}{B+R}+1\) --> \(\frac{R+B+W}{B+W}> \frac{W+B+R}{B+R}\) --> \(\frac{1}{B+W}> \frac{1}{B+R}\) --> \(B+R>B+W\) --> \(R>W\). Sufficient.

(2) \(B-W>R\), not sufficient to determine whether \(R>W\) or not.

Answer: A.

As for your solution: you can not multiply inequality by \(R-W\), as you don't know whether this expression is positive or not, thus you don't know whether you should flip the sign of inequality or not. Actually when you multiply by \(R-W\) and not flipping the sign, you wrongly assume that \(R-W>0\), or \(R>W\), but this is exactly what we are asked to determine.

Hope it's clear.

Thanks. Missed it...

Here's an open request all - it would be very helpful to get a compilation of different types of solved probability problems. Is there a source?

A

R/(B+W) > W(B+R)
R/(B+W)+1 > W(B+R)+1
R/(B+W)+R/R > W(B+R)+W/W
R/(B+W+R) > W/(B+R+W)

hmm, I took me more than 5 minutes to solve this question.