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ritula
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Gatprep 29 May 2008, 23:03
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Pls help with explanation



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Gatprep 29 May 2008, 23:31
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This took me a little while, but here is my approach

picking 2 out of 5, the number of combinations is 10 (5!/2!3!)

for 1 bad apple, you have 4 good apples, so there are 4 possible combinations of 1 bad/1 good apples.

4/10 = 2/5
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ritula
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Gatprep 30 May 2008, 00:16
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can u pls explain in a more elaborate way. Actually a probability question is not my cup of tea.



gmatnub
This took me a little while, but here is my approach

picking 2 out of 5, the number of combinations is 10 (5!/2!3!)

for 1 bad apple, you have 4 good apples, so there are 4 possible combinations of 1 bad/1 good apples.

4/10 = 2/5
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Gatprep 30 May 2008, 01:53
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This is quite straight forward.

Just enumerate the various possibilities :
Probability 1st rotten 2nd fine :
\(1/5 * 4/4\)
PLUS
Probability that 1st fine 2nd rotten
\(4/5 * 1/4\)

= \(1/5 + 1/5 = 2/5\)
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ritula
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Gatprep 30 May 2008, 02:37
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Thanks u made things very simple :lol:
bsd_lover
This is quite straight forward.

Just enumerate the various possibilities :
Probability 1st rotten 2nd fine :
\(1/5 * 4/4\)
PLUS
Probability that 1st fine 2nd rotten
\(4/5 * 1/4\)

= \(1/5 + 1/5 = 2/5\)
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Gatprep 30 May 2008, 06:29
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Another way to solve the problem

1 - (probability of NOT GETTING THE SPOILED APPLE IN THE 2 simultaneous attempts)

= 1 - ( 4/ 5 * 3 / 4)

4/ 5 = probability of getting good apple in the first draw
3/ 4 = probability of getting good apple in the second draw

= 1 - 12 / 20
= 8/20
=2/5
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GMATBLACKBELT
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Gatprep 30 May 2008, 08:10
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ritula
Pls help with explanation
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Just calculate the probability that there will be no spoiled apple.

4/5*3/4= 12/20 --> 3/5

Now

1-3/5 = 2/5.
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Gatprep 30 May 2008, 08:46
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we have 4 fresh apples and one rotten apple.
we pick two apples. Out of which one has to be rotten.
How do we do that?
1 out of 4 normal apples and the other rotten one
Hence, the number of favourible cases = (4C1)*(1C1)
Total number of ways to pick apples
5C2.
Hence
Probability= (4C1)*(1C1)/(5C2) = (2/5)

Cheers !



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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