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Two pieces of fruit are selected out of a group of 8 pieces [#permalink]
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29 Sep 2010, 15:39
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37% (02:53) correct
63% (02:24) wrong based on 83 sessions
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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 ½. (2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. OE To know the probability of selecting 2 bananas, we need to know exactly how many bananas are in the original group of 8. Thus, we can rephrase the question as “How many of the 8 pieces of fruit are bananas?”
Statement (1): INSUFFICIENT. We should work out the scenarios in which this condition is true.
Obviously, if all 8 pieces of fruit are apples, the probability is 1. So let’s start from this end.
If there are 7 apples and 1 banana, then the probability of 2 apples is (7/8)(6/7) = 42/56, which is greater than ½.
If there are 6 apples and 2 bananas, then the probability of 2 apples is (6/8)(5/7) = 30/56, which is still greater than ½. (28/56 would be exactly ½.)
If there are 5 apples and 3 bananas, then the probability of 2 apples is (5/8)(4/7) = 20/56, which is NOT greater than ½.
So we know from this statement that the number of apples could be 6, 7, or 8, and the number of bananas could be 0, 1, or 2. This narrows down the possibilities, but we do not know exactly how many bananas there are.
Statement (2): INSUFFICIENT. Again, we should work out the scenarios in which this condition is true. Let’s start with the scenario in which it would be most likely to pick one of each type of fruit: 4 apples and 4 bananas.
If there are 4 apples and 4 bananas, then the probability of picking 1 of each can be found this way:
1) Pick an apple, then a banana: (4/8)(4/7) = 16/56.
2) Pick a banana, then an apple: (4/8)(4/7) = 16/56.
Add: 16/56 + 16/56 = 32/56. This is obviously greater than 1/3.
Alternatively and more easily, we could have just calculated the probability of “apple then banana,” then multiplied by 2! = 2 to get our result. This is much faster.
Let’s now figure out the other scenarios:
3 apples and 5 bananas:
The probability of picking 1 of each is (3/8)(5/7)×2 = 30/56, also bigger than 1/3. (Notice that we have not wasted time simplifying these fractions.)
By symmetry, the probability must be the same for 5 apples and 3 bananas.
2 apples and 6 bananas:
The probability of picking 1 of each is (2/8)(6/7)×2 = 24/56, also bigger than 1/3. (We can compare 24/56 to 20/60 by eye.)
The same again is true for 6 apples and 2 bananas.
1 apple and 7 bananas:
The probability of picking 1 of each is (1/8)(7/7)×2 = 14/56, which is ¼. Too small.
Thus, this statement says that the number of bananas is 6, 5, 4, 3, or 2.
Statements (1) and (2) together: SUFFICIENT. The number of bananas must be 2.
The correct answer is (C). OPEN DICUSSION OF THIS QUESTION IS HERE: twopiecesoffruitareselectedoutofagroupof8pieces142146.html
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Last edited by zisis on 05 Oct 2010, 15:46, edited 1 time in total.



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Re: MGMAT Challenge Problem Showdown [#permalink]
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29 Sep 2010, 15:39
zisis wrote: 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 ½.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. IMO A no OA yet  will post as soon as it is provided along with OE



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Re: MGMAT Challenge Problem Showdown [#permalink]
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29 Sep 2010, 15:53
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zisis wrote: 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 ½.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. Let there be x apples, so (8x) bananas P(AA) = x(x1)/8*7 P(AB) = P(BA) = x(8x)/8*7 P(BB) = (8x)(7x)/8*7 (1) : \(\frac{x^2x}{56} > 0.5\) \(x*(x1) > 28\) So x=8,7,6 Not sufficient to know P(BB) (2) : \(\frac{2x(8x)}{56} > 1/3\) \(x * (8x) > 9.333\) x = 2,3,4,5,6 Not sufficient to know P(BB) (1) + (2) : Only solution satisfying both is x=6, which is enough to know P(BB) So I think answer is (C)
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Re: MGMAT Challenge Problem Showdown [#permalink]
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29 Sep 2010, 18:34
zisis wrote: 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 ½.
(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3. I think it's (C) 1) Testing a few numbers of apples: the probability of selecting 2 apples if there are 5 is (5/8)(4/7) = 20/56, which is less than 1/2. If there are 6 apples, the probability is (6/8)(5/7) = 30/56, which is greater than 1/2. So there are 6, 7, or 8 apples. If there are 6 apples, then the probability of selecting 2 bananas is (2/8)(1/7) = 1/28. But if there are 7 or 8 apples, the probability of selecting 2 bananas is 0. So insufficient. 2) The probability of selecting 1 apple and 1 banana is (a/8)(b/7) + (b/8)(a/7) = (ab/56) + (ab/56) = ab/28. Since (ab/28) = (1/3), we know that ab > (28/3) = 9.333... This doesn't tell us how many of each fruit there is, though. Together: We know there are 6, 7, or 8 apples, and that a*b > 9.333... If there are 8 apples, then there are 0 bananas, and 8*0 < 9.333... Likewise, if there are 7 apples and 1 banana, 7*1 < 9.333... However, if there are 6 apples and 2 bananas, 6*2 > 9.333... So there are 6 apples and 2 bananas, and the probability of selecting both bananas is 1/28.



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Re: MGMAT Challenge Problem Showdown [#permalink]
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14 May 2011, 09:50
C is 2:55 min, without any writing. i think this could have been faster, it is clear that there are either 7 apples or 6 apples, so either there is 1 banana or 2 bananas. now plugging in the values and checking out for C is easier.
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Re: MGMAT Challenge Problem Showdown [#permalink]
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14 May 2011, 10:13
Statment 1: Tells that there are either 3 or 2 or 1 or 0 bananas ==> Not sufficient alone Statment 2: Tells that A and B are in a fixed particular ratio (what ever it is )but it can be ether way round. ==> Not sufficient alone Both together tells that the apples are more the given ratio. So gives counts of A and B. Hence sufficient find all kind of probabilities.
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Re: MGMAT Challenge Problem Showdown [#permalink]
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15 May 2011, 00:17
a. gives either 6 or 7 apples. Not sufficient. b gives either 5 or 6 apples. Not sufficient. a+b gives 6 apples and 2 bananas. C
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Re: MGMAT Challenge Problem Showdown [#permalink]
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15 May 2011, 09:42
Here the question asks the probability of getting exactly 2 bananas, so doesn't that imply that there are atleast 2 bananas ? If there were less than 2 bananas the answer would be zero straight away. Also question says group of 8 pieces of fruit consisting apples and bananas, so banana can't be zero ! If there are atleast 2 bananas, then A is sufficient! Can someone clarify?
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Re: MGMAT Challenge Problem Showdown [#permalink]
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17 May 2011, 01:58
Bananas  x Apples  8x P(Exactly 2 Bananas)  x/8 * (x1)/7 (1) Exactly 2 Apples (8x)/8 * (8x1)/7 > 1/2 => (8x)/8 * (7x)/7 > 1/2 => (8x)(7x) > 28 X can be 2 or 1 or 0 Insufficient (2) x/8 * (8x)/7 + (8x)/8 * x/7 > 1/3 2 * x/8 * (8x)/7 > 1/3 x/8 * (8x)/7 > 1/6 x(8x) > 28/3 > 9.something => x = 2, 3, 4 etc. Not Sufficient (1) + (2) x = 2 Answer  C
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Re: MGMAT Challenge Problem Showdown [#permalink]
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17 May 2011, 02:15
subhashghosh wrote: Bananas  x
Apples  8x
P(Exactly 2 Bananas)  x/8 * (x1)/7
(1)
Exactly 2 Apples
(8x)/8 * (8x1)/7 > 1/2
=> (8x)/8 * (7x)/7 > 1/2
=> (8x)(7x) > 28
X can be 2 or 1 or 0
Insufficient
(2)
x/8 * (8x)/7 + (8x)/8 * x/7 > 1/3
2 * x/8 * (8x)/7 > 1/3
x/8 * (8x)/7 > 1/6
x(8x) > 28/3 > 9.something
=> x = 2, 3, 4 etc.
Not Sufficient
(1) + (2)
x = 2
Answer  C That I understood. See according to your explanation of statement 1: X can be 2, 1 or 0. Now the question says the group has bananas and apples, so bananas cannot be 0 from what I understand. Leaving X = 2 or 1 Now the question is probability of getting exactly 2 bananas  doesn't that imply that the group has atleast 2 bananas?If that is the case then X has to be 2, so A would become sufficient. If that is not the case then your explanation is correct. I needed a clarification on the above!
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Re: MGMAT Challenge Problem Showdown [#permalink]
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17 May 2011, 02:19
It's not implied that group has at least 2 bananas. (1) implies there are < 4 bananas. So number of bananas can be 2 or 3.
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Re: MGMAT Challenge Problem Showdown [#permalink]
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22 Jun 2011, 05:57
As per me, statement 2 should not use "either". Using "either" gives a wrong meaning to the assertion.



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Re: MGMAT Challenge Problem Showdown [#permalink]
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27 Jun 2011, 17:44
1. Not sufficient A(A1)/(8*7) > 1/2 => A can be 6 or 7 B can be 2 or 1 2. Not sufficient Either in second statement tells us that order does not matter. let A be the number of Apples , B be the number of bananas. So selecting 1 apple and 1 banana in either order = Ac1 * Bc1 = AB Further statement 2 states this probability of selecting 1 apple and 1 banana > 1/3 AB/8c2 > 1/3 =>AB >9 => A =2 or 3 or 4 or 5 or 6 B = 6 or 5 or 4 or 3 or 2 Hence not sufficient Together , It sufficient Answer is C. neel1982 wrote: As per me, statement 2 should not use "either". Using "either" gives a wrong meaning to the assertion.




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