A basket contains 5 apples, of which 1 is spoiled and the rest are good. 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 spoiled apple?
(a) 1/5
(b) 3/10
(c) 2/5
(d) 1/2
(e) 3/5
Answer is c.
I am trying to figure out why you add the 2 scenarioes instead of multiplying.
A basket contains apples
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sk8ternite wrote:A basket contains 5 apples, of which 1 is spoiled and the rest are good. 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 spoiled apple?
(a) 1/5
(b) 3/10
(c) 2/5
(d) 1/2
(e) 3/5
Answer is c.
I am trying to figure out why you add the 2 scenarioes instead of multiplying.
4C1 / 5C2 = 2 / 5
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the way I thought about it was:
the probability of picking the bad apple first is 1/5, and then the 2nd pick would be 1, so scenario would is 1/5*1=1/5.
But you could pick the bad apple 2nd, so it would be: 4/5*1/5=4/20.
If either situation can occur, it is an "or" situation and you add the two scenarios. 1/5+4/20= 8/20 or 2/5
Is this correct? Could someone also explain the combinatorics way of doing the problem?
the probability of picking the bad apple first is 1/5, and then the 2nd pick would be 1, so scenario would is 1/5*1=1/5.
But you could pick the bad apple 2nd, so it would be: 4/5*1/5=4/20.
If either situation can occur, it is an "or" situation and you add the two scenarios. 1/5+4/20= 8/20 or 2/5
Is this correct? Could someone also explain the combinatorics way of doing the problem?
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probabilty of not selecting the spoiled one:
4C2/5C2= 3/5
so prob of selecting the spoiled one=1-3/5=2/5
4C2/5C2= 3/5
so prob of selecting the spoiled one=1-3/5=2/5
The powers of two are bloody impolite!!
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4 GOOD APPLES AND 1 BAD APPLE...
No of ways for chosing 1 bad apple is 1C1.
No of ways for chosing 1 Good Apple is 4C1.
Therefore, no of ways for chosing the apples ( as given under problem )are ( 1C1*4C1) = 4.
Total number of ways for chosing 2 apples from 5 apples is ( Denominator part ) = 5C2 = 10.
Therefore the required probability is (4/10) = (2/5).
Hope it is clear enough..
No of ways for chosing 1 bad apple is 1C1.
No of ways for chosing 1 Good Apple is 4C1.
Therefore, no of ways for chosing the apples ( as given under problem )are ( 1C1*4C1) = 4.
Total number of ways for chosing 2 apples from 5 apples is ( Denominator part ) = 5C2 = 10.
Therefore the required probability is (4/10) = (2/5).
Hope it is clear enough..