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The bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

1. 20% of all chips in the basket are green 2. The ratio of the number of red chips to the number of green chips is 4:1

The bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

1. 20% of all chips in the basket are green 2. The ratio of the number of red chips to the number of green chips is 4:1

Shldnt the answer be D. bcos from both (1) and (2), we know the probabiliy of taking out a green chip=1/5 and hence the probability can be claculated.

A bowl contains red and green chips and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(1) 20% of all chips in the basket are green --> 80% of all chips in the basket are red --> 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 bowl (4 red + 1 green) then the the probability that both chips will be green will be 0 (since there are NOT two chips in a bowl) but if there are total of 10 chips in the bowl (8 red + 2 green) then the 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.

The bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(1)20% of all chips in the basket are green (2)The ratio of the number of red chips to the number of green chips is 4:1

Shldnt the answer be D. bcos from both (1) and (2), we know the probabiliy of taking out a green chip=1/5 and hence the probability can be claculated.

Percentage is not suffiecient. What if there is 1 green chip and 4 red chips? You have a 1/5 chance of hitting green on the first try, and a 0% chance of hitting green on the second try. 1 & 4 fulfill the requirements for both (1) and (2). Picking different numbers for greens and reds will obviously yield a different result than 0%.

Both statements provides same ratio or persent. If we plug in value according to ratios or percent given, probability will change every time. Total number of chips are required to answer.

Another tip - When both statements gives same info (they are deadly twins) then answer is surely E
_________________

Will go with E , as even with both the statements we will not be able to identify the exact number. Question takes in terms of ration(%) and requires more information to calculate exact numbers
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green to red: (i)4 : 1 P(G) x P(G) = 1/4 x 0 = 0 (ii)8 : 2 P(G) x P(G) = 1/10 x 1/9 = 1/90 (iii)12 : 3 P(G) x P(G) = 3/15 x 2/14 = 1/35

in all instances the probabilities are the same for the first pick; however, the second picks produced different results depending on the initial number of balls. answer is E.
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KUDOS me if you feel my contribution has helped you.

The bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

1. 20% of all chips in the basket are green 2. The ratio of the number of red chips to the number of green chips is 4:1

Stmt1: Let there be 100 chips. Green chips = 20 Probability of picking two green chips = 20/100 * 19/99 = 0.2 * 19/99 Let there be 200 chips Green chips= 40 Probability of picking two green chips = 40/200 * 39/199 = 0.2 * 39/199 Hence two different values depending on total number of chips. Not sufficient

Stmt2: The ratio of the number of red chips to the number of green chips is 4:1 Red=8 Green=2 Total = 10 Probability of picking two green chips = 2/10 * 1/9 = 0.2 * 1/9

Red=16 Green=4 Total =20 Probability of picking two green chips=4/20 * 3/19 = 0.2 * 3/19. Again different probability for different number of of total chips. Insufficient.

Together, also, total number of chip is still missing which is key data needed to answer the question. OA E.

P.S: Note if question had asked probability of picking 1 green chip, answer would have been D. as in each statement we would have been able to tell probability (0.2) in cases above.
_________________

My dad once said to me: Son, nothing succeeds like success.

Never attempt to solve successive draws probability if you only have ratios and not the actual number. That said answer is E, since both statements say the same thing.

The bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(1)20% of all chips in the basket are green (2)The ratio of the number of red chips to the number of green chips is 4:1

Shldnt the answer be D. bcos from both (1) and (2), we know the probabiliy of taking out a green chip=1/5 and hence the probability can be claculated.

I did go in for D. But, realised (1) & (2) are the same, if you equate it.

yeah, i made the same mistake of opting for D. didn't realize it was a tricky question where both the statements provided the same information. should have tried plugging in numbers. that would prove it's E

I think the answer is either one. The answer here should be 1/5 as whether you take 2 or 3 or more upto 20 balls it should be same set of choices with you. Above 20 the prob will become 0.

If there are R red and G green, then probably of 2 greens = P(1st Green) * P(2nd Green) = [G/(G+R)] * [(G-1)/(G+R-1)]

1st part can be factored if ratio R:G is given, BUT NOT the 2nd part.

Good question though. My instinct was D as soon as I read the problem but when I processed the equation in mind I realized it's not trivial and is a trap question

First of all both the statements say the same thing, i.e. green balls are 20% and red balls are 80% of the total balls in the bag. So if we evaluate whether one's sufficient, we can safely assume the same about the other statement too.

now, we know that balls are in the ratio of G:R = 1:4, hence the actual probability that the first ball will be green is 1x/5x. However, the probability that the second ball will be green is 1x-1/5x-1. By multiplying these two, we can not come to a conclusive answer. Hence E

Ans is [E]. Such DS problems in which each statement provides the same info by twisting them a little bit (in this case ratios and percents) are quite common. I reckon such type of problems come in the 600-700 category

So 1) Using 10 as the number 20% of 10 is 2 so 2/10. Insufficient for A 2) 1:4 no matter what number used is insufficient for B. 1 and 2 conflict 20% vs 25% C is insufficient 1 and 2 both don't stand alone so D is insufficient. That leaves E.