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 (2) There were originally 5 glazed doughnuts in the box. Why is 1) Not sufficient? If P of second donut selected = 2/5 isn't P of first donut 2+1 / 5+1 = 3/6? (i.e. we put donut back in the box). Here is what Kaplan said, but I don't get why 1) is not suff. 1)+2): If there were originally 5 glazed doughnuts in the box, and one was selected, then there are 4 glazed doughnuts in the box. Since the probability of picking another glazed doughnut is 2/5, you know that 2/5 = 4 / num of d left, so there must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total. Thank you kindly.

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doughnuts = d glazed = g Statement 1: Not sufficient The box could have 6d, of which 3g. Minus 1, 2/5 The box could also have 16d, of which 7g. Minus 1, 6/15 = 2/5 Statement 2: Not sufficient 5d, but what about the rest? Both Statements: Sufficient 5g - 1g = 4g if 4g / xd = 2/5, then xd = 10 So originally there were 11doughnuts and 5glazed

So ,uhm, from solving the equation with the information from S1 I find that in g.(g-1)/t.(t-1) = 2/5 or the numerator part of the equation would equal a multiple of 2. i.e. g^2-g- (multiple of 2) would have an integer root equal to the number of glazed doughnuts. After trial and error I found it to be 20 and this gave me an amount of apples equal to 5, as provided in Statement 2. So, maybe s1 alone is sufficient A

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