Bunuel wrote:

Official Solution:

(1) 20% of all chips in the basket are green. Therefore 80% of all chips in the basket are red, which means that the ratio of the number of red chips to the number of green chips is 4:1 (80:20). Now, if there are total of 5 chips in the basket (4 red + 1 green) then the probability that both chips will be green will be 0 (since there are NOT two chips in a basket) but if there are total of 10 chips in the basket (8 red + 2 green) then the probability that both chips will be green will be more than 0. Not sufficient.

(2) The ratio of the number of red chips to the number of green chips is 4:1. The same info as above. Not sufficient.

(1)+(2) Both statements tell the same thing, so we have no new info. Not sufficient.

Answer: E

BunuelHi Bunuel can we solve it this way?

total number of chips = r + g

no of ways 2 green can be drawn = gC2 = g(g-1)

no of ways 2 chips can be drawn from the combined pool =(r+g)C2 = (r+g)(r + g -)

probability that both chips will be green = \frac{g(g-1)}{(r+g)(r+g-1)}

per statement 1: g = 20% of (r + g)

therefore it comes \frac{g}{r+g}=\frac{1}{5} therefore not sufficient

per statement 2: \frac{r}{g} = \frac{4}{1} not sufficient

combined same situation not sufficient

_________________

Resources

GMATNinja Webinars

GMATNinja Chats

Quant

Mixtures