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Is c/ (a+b+c) = + 1? Ok, ok... here's the problem behind the [#permalink]

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09 Dec 2006, 04:55

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Is c/ (a+b+c) = [c/(a+b)] + 1?

Ok, ok... here's the problem behind the question:

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

first, clearly statement 2 is not sufficient - so lets consider statement 1:
the left side has r/(b+w) the right side has w/(b+r)

there are three possible cases:
a) r=w
b) r>w
c) r<w

if r=w, both sides of statment 1 are equal - so given statment 1 is true (a) cannot be true.
if r<w, the numerator of the left side fraction is less than the numerator of the right side fraction. furthermore, the denominator of the left side (b+w) is greater than the denominator of the right side (b+r).... since its numerator is smaller and the denominator is greater - the left side fraction is less than the right side.
hence, given the opposite (e.g. statement 1) - it is clear that r<w cannot hold.

so if statement 1 is true we are left with only one possibility which is r>w.
hence statement 1 is sufficient. no need to do any algebra.

when comparing complex expressions (fractions, long multiplications, polynomes) it is worthwhile to look at parts of them and check its effect on the overall value.

do whatever strategy you can think of to avoid algebra on complex exporessions. do algebra only if there is no other way... that's my suggestion.

Hobbit, smart thinking! Thanks so much for that tip! I'll keep that in mind. I tend to just dive into the arithmetic and get lost in computation = wasted time.

But if we were to solve statement (1) to prove that it is sufficient, how would it work out?
_________________

now... if r<w (and both positive.. we must remember that)
the left side is negative and right side is positive... which constradicts it.
if r=w also constradiction as everything becomse 0
hence r>w.

note that at some stage i abandoned the algebra again (i just can't help it...)

another option to continue algebra:
r^2-w^2>b(w-r)
(r+w)(r-w)>b(w-r)
(r+w)(r-w) + b(r-w) >0
(r-w)(r+w+b) > 0
and since r+w+b is positive
r-w>0
r>w

that's the full algebra way (which as i said is not needed... and requires some sophistication...)

hobbit solved it in too complicated way .
.check this approch out ...

probability of chosen marble being red = r/b+w+r
probability of chosen marble being white = w/b+w+r

from 1:
r/(b+w) > w/(b+r)
=> r(b+r)>w(b+w)
=> (b+r)/w > (b+w)/r
=> (b+r)/w + 1 > (b+w)/r + 1 (add 1 both sides)
=> b+r+w/w > b+w+r/r
=> w/b+r+w <r/b+w+r
=> probability of being white < probability of being red

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

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

(2) b - w = r

question is r > w or w > r?

From 1)
r/(b+w) > w/(b+r)
=> r/(b+w)+1 > w/(b+r)+1
=> (r + b + w)/(b+w) > (r + b + w)/(b+r)
=> 1/(b+w) > 1 /(b+r)
=> b + r > b + w
=> r > w
Suff

From 2)
b - w = r
Here r + w = b, here we get b > r or w but don't get any info of r and w