A box of doughnuts contains glazed, cream-filled, and jelly

The answer is C.

Nsentra,

For the 1 statement we cannot add 1 to the den and num coz we don't know what the fraction would be it can be 2/5 or 4/10 or 6/15 which will give different answers every time.

Ah, ok I got it. I see, this goes in hand with a type of question that asks if it's possible to find a new ratio when something was added/subtracted to some part of current ratio.

Thank you!

C.

Stmt2 gives number of glazed doughnuts and stmt1 gives total number of doughnuts.

I am not sure how stmt1 is giving the total number of DN's.

The probability of selecting a second glazed DN is 2/5. We need P of selecting the first DN.

Are you saying that the total number of DN's is 10 because 4/10 is 2/5?

I picked E because the problem does not tell us whether the first DN is put back or not. I see where you are getting at but some how I cannot explain

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5

I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5

(1)
say there are x glazed donuts, and "n" total donuts
for the first glazed donut prob = x/n
prob for second = x-1/n-1 = 2/5
But as this is ratio we can not conclude anything about x and n (it can be anythign 2/5, 10/25 etc)
insuff

(2)
we know that x = 5 but insuff. alone

together,
x-1/n-1=2/5 and x = 5
we get, 10=n-1 or n = 11

Suff to calculate prob the doughnut selected is glazed

C

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.

IMHO, if the question does not specifically mention replacement => no replacement. I am not sure but it just occurred to me that I have read this somewhere......

Actually, the question would have speicified that the second probability is after the replacement, then there was no value in the question. But again.. I think it is responsibility of the question maker to avoid any ambiguity to us.

IMO C too. Agree with you, guys.

I think that first statement gives the probability of selecting second doughnut without putting it back. Thus, together with statement 2 it gives us the number of total doughnuts and the number of glazed doughnuts.

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

This does NOT mean that the first chance is 3/6 = 1/2 because if there were, for example, 4 glazed and 6 other donuts after the first draw (4/10 = 2/5 chance), then before that there were 5 glazed and 6 others (5/11 chance). Insufficient.

(2) There were originally 5 glazed doughnuts in the box.

Obviously insufficient on its own since we need to know the total number BUT, with the first one it is sufficient:

If there were 5 glazed and one was drawn, there are now 4 glazed left. There's a 2/5 chance to draw another glazed so there are 4 glazed and 6 others (10 total). So before there were 5 glazed and 6 others (5/11 chance).

C

Let 'g' be the number of glazed doughnuts and let 'n' be the total number of doughnuts.
Thus, we need to find the value of (g / n).

Statement 1:
After the first glazed doughnut is selected, the number of remaining glazed doughnuts = (g - 1)
Number of remaining total doughnuts = (n - 1)
Thus this statement tells us that (g - 1)/(n - 1) = 2/5 => (5g - 5) = (2n - 2) => 5g = 2n + 3
As there are two unknowns, we cannot get a value for (g/n). INSUFFICIENT.

Statement 2: g = 5
This does not give a value for n. So INSUFFICIENT.

Combining these two statements, we have the value of 'g' from (2), which can be substituted in the equation obtained from (1) to find the value of 'n'.
Thus, we can find out (g/n). SUFFICIENT.

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5. We could have total of 6 doughnuts out of which 3 were glazed or 11 doughnuts out of which 5 were glazed or 16 doughnuts out of which 7 were glazed ... Not sufficient.

(2) There were originally 5 glazed doughnuts in the box. We don't know the total number doughnuts. Not sufficient.

(1)+(2) If there were originally 5 glazed doughnuts, then after selecting 1 of them there would be 4 left, so according to (1) 4/(total - 1) = 2/5 --> total = 11 --> P(glazed) = (glazed)/(total) = 5/11. Sufficient.

Answer: C.

pretty straight forward question..looks like a 600-700 than a 700 lvl...

we need to find: G/(G+C+J)

1. G-1/(G+C+J-1) = 2/5
it could have been 3 G, and probability is 1/2
or it could be 5 G, and 11 total, thus probability is 5/11.
1 alone is insufficient.

2 alone is insufficient.

1+2
we know that originally, there were 5 G.
so 2x/5x => 2x=5-1 = 4, so x=2. or total there were 11 donuts, out of which 5 were G.

sufficient.
C

Bunuel

I think I have this problem consistently. Why do we assume the probability of drawing second donut as X-1/n-1.

If the first donut is picked , then the prob of picking second shouldn't be x/n*x-1/n-1 =2/5 VeritasPrepKarishma any help would be appreciated.

Thanks


Posted from my mobile device

Hi..

If we are asked "Probability of selecting two glazed doughnuts in two picks without replacement", the answer would be correct...
But here it is ALREADY given that first selected is glazed. so now it is not a probability but a certainity.
hence we work with the remaining numbers which are
TOTAL = n-1
and glazed = g-1
so Prob will be g-1/n-1
hope it helps