A basket contains 5 apples of which one is rotten. If Henry is to select 2 apples from the basket simultaneously and at random what is the probability that the 2 apples selected will include the rotten apple?
A. 1/5
B. 3/10
C. 2/5
D. 1/2
E. 3/5
The OA is C.
Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer.
I know that I have an universe of 5 apples where there is one rotten. Then, if I selected 2 simultaneously the probability that one of these will be the rotten apple should be 2/5.
Is correct my analisys? Thanks.
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A basket contains 5 apples of which one is rotten...
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Hi swerve,
We're told that a basket contains 5 apples (4 normal apples and 1 rotten apple). We're asked for the probability that selecting 2 apples from the basket simultaneously and at random will include the rotten apple. There are a number of different ways to approach this question. Since the number of apples is so small though, you can actually do a bit of 'brute force' math and get the answer.
Let's call the 4 normal apples A, B, C and D and the rotten apple Z. There are only 10 possible 'pairings' of apples:
A and B
A and C
A and D
A and Z
B and C
B and D
B and Z
C and D
C and Z
D and Z
Of those 10 possibilities, 4 of them include the rotten apple. Thus, the probability of pulling the rotten apple under these circumstances is 4/10 = 2/5
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that a basket contains 5 apples (4 normal apples and 1 rotten apple). We're asked for the probability that selecting 2 apples from the basket simultaneously and at random will include the rotten apple. There are a number of different ways to approach this question. Since the number of apples is so small though, you can actually do a bit of 'brute force' math and get the answer.
Let's call the 4 normal apples A, B, C and D and the rotten apple Z. There are only 10 possible 'pairings' of apples:
A and B
A and C
A and D
A and Z
B and C
B and D
B and Z
C and D
C and Z
D and Z
Of those 10 possibilities, 4 of them include the rotten apple. Thus, the probability of pulling the rotten apple under these circumstances is 4/10 = 2/5
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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Here's another way of thinking about it: Henry could select the rotten apple FIRST or SECOND.
The probability that he selects it first = p(rotten apple) * p(fresh apple) = 1/5 * 4/4 = 1/5
The probability that he selects it second = p(fresh apple) * p(rotten apple) = 4/5 * 1/4 = 1/5
Adding those up, we have 1/5 + 1/5 = 2/5.
The probability that he selects it first = p(rotten apple) * p(fresh apple) = 1/5 * 4/4 = 1/5
The probability that he selects it second = p(fresh apple) * p(rotten apple) = 4/5 * 1/4 = 1/5
Adding those up, we have 1/5 + 1/5 = 2/5.
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Matt@VeritasPrep
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Another approach:
p(getting the rotten apple) + p(not getting the rotten apple) = 1, since one of those two things MUST happen.
Subtracting from both sides, we have
p(getting the rotten apple) = 1 - p(not getting the rotten apple)
and
p(not getting the rotten apple) = p(apple 1 fresh) * p(apple 2 fresh) = 4/5 * 3/4 = 3/5
So our answer is 1 - 3/5, or 2/5.
p(getting the rotten apple) + p(not getting the rotten apple) = 1, since one of those two things MUST happen.
Subtracting from both sides, we have
p(getting the rotten apple) = 1 - p(not getting the rotten apple)
and
p(not getting the rotten apple) = p(apple 1 fresh) * p(apple 2 fresh) = 4/5 * 3/4 = 3/5
So our answer is 1 - 3/5, or 2/5.
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Matt@VeritasPrep
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Another approach could use combinatorics:
p(rotten) = (# of pairs with the rotten apple) / (total # of pairs)
# of pairs with the rotten apple = 4, since the rotten apple could be paired with any of the other four apples
total # of pairs = (5 choose 2) = 5!/(3! * 2!) = 10
So our ratio is 4/10, or 2/5.
p(rotten) = (# of pairs with the rotten apple) / (total # of pairs)
# of pairs with the rotten apple = 4, since the rotten apple could be paired with any of the other four apples
total # of pairs = (5 choose 2) = 5!/(3! * 2!) = 10
So our ratio is 4/10, or 2/5.
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Jeff@TargetTestPrep
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We need to determine the probability of selecting a spoiled apple and a non-spoiled apple when selecting two apples.swerve wrote:A basket contains 5 apples of which one is rotten. If Henry is to select 2 apples from the basket simultaneously and at random what is the probability that the 2 apples selected will include the rotten apple?
A. 1/5
B. 3/10
C. 2/5
D. 1/2
E. 3/5
The number of ways to select the spoiled apple is 1C1 = 1. The number of ways to select a good apple is 4C1 = 4. Thus, the spoiled apple and a good apple can be selected in 1 x 4 = 4 ways.
The number of ways to select 2 apples from 5 is 5C2 = (5 x 4)/2! = 20/2 = 10.
Thus, the probability of selecting the spoiled apple and a good apple is 4/10 = 2/5.
Alternate Solution:
There are two outcomes that satisfy the requirement that the spoiled apple (S) is chosen along with a good apple (G), either S-G or G-S. The probability of S-G is (1/5)(4/4) = 1/5. The probability of G-S is (4/5)(1/4) = 1/5. Since either outcome satisfies our requirement, we add these two probabilities: 1/5 + 1/5 = 2/5.
Answer: C
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