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Pansi
Joined: 04 Jul 2011
Last visit: 01 Dec 2019
Posts: 42
Given Kudos: 87
Status:Fighting hard
Schools: Kellogg '15 ISB '14 (S)
GMAT Date: 10-01-2012
09 Nov 2012, 01:34
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?
(1) The probability of selecting exactly 2 apples is greater than 1/2.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.
(1) The probability of selecting exactly 2 apples is greater than 1/2.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.
09 Nov 2012, 03:43
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Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?
Say there are \(x\) bananas and \(y\) (\(y=8-x\)) apples. The question is \(P(bb)=\frac{x}{8}*\frac{x-1}{7}=?\). Basically we need to find how many bananas are there.
(1) The probability of selecting exactly 2 apples is greater than 1/2 --> \(\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}\) --> \(y(y-1)>28\) --> \(y\) can be 6, 7, or 8, thus \(x\) can be 2, 1, or 0. not sufficient.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. \(2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}\) --> \(x(8-x)>\frac{28}{3}=9\frac{1}{3}\), thus \(x\) can be 2, 3, 4, 5, or 6. Not sufficient.
(1)+(2) From above x can only be 2. Sufficient.
Answer: C.
Hope it's clear.
Say there are \(x\) bananas and \(y\) (\(y=8-x\)) apples. The question is \(P(bb)=\frac{x}{8}*\frac{x-1}{7}=?\). Basically we need to find how many bananas are there.
(1) The probability of selecting exactly 2 apples is greater than 1/2 --> \(\frac{y}{8}*\frac{y-1}{7}>\frac{1}{2}\) --> \(y(y-1)>28\) --> \(y\) can be 6, 7, or 8, thus \(x\) can be 2, 1, or 0. not sufficient.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. \(2*\frac{x}{8}*\frac{8-x}{7}>\frac{1}{3}\) --> \(x(8-x)>\frac{28}{3}=9\frac{1}{3}\), thus \(x\) can be 2, 3, 4, 5, or 6. Not sufficient.
(1)+(2) From above x can only be 2. Sufficient.
Answer: C.
Hope it's clear.
09 Nov 2012, 03:27
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Pansi
Total No. of ways of selecting = 8C2 = 28
1)No. of apples = 7, No. of bananas = 1, Probability : \(\frac{7C2}{8C2} = \frac{21}{28}\)
No. of apples = 6, No. of bananas = 2, Probability : \(\frac{6C2}{8C2} = \frac{15}{28}\)
No. of apples = 5, No. of bananas = 3, Probability : \(\frac{5C2}{8C2} = \frac{10}{28}\)
So, No. of apples can be 6 or 7. Insufficient
2)No. of apples = 7, No. of bananas = 1, Probability : \(\frac{7C1*1C1}{8C2} = \frac{7}{28}\)
No. of apples = 6, No. of bananas = 2, Probability : \(\frac{6C1*2C1}{8C2} = \frac{12}{28}\)
No. of apples = 5, No. of bananas = 3, Probability : \(\frac{5C1*3C1}{8C2} = \frac{15}{28}\)
So, No. of apples can be 5 or 6. ( More values other than 7 are also possible, but two values are enough to make the statement insufficient.)Insufficient
1 & 2 together. No. Of apples = 6, No. of bananas = 2. Enough info to find what is required. Sufficient.
Although, I'm not very strong at combinatronics and hence I'm not 100% sure of my method. Also since the question states that there ARE bananas, I'm assuming that no. of apples cannot be 8.
Kudos Please... If my post helped.
