Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
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).