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4 out of 15 apples are rotten. They are taken out one by one at random

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4 out of 15 apples are rotten. They are taken out one by one at random [#permalink] 4 out of 15 apples are rotten. They are taken out one by one at random   31 Mar 2020, 07:46
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4 out of 15 apples are rotten. They are taken out one by one at random and examined. The ones which are examined are not replaced. What is the probability that the 9th one examined is the last rotten one?

A. \(\frac{4c_3*11c_5}{15c_8*7c_1}\)
B. \(\frac{11c_5*4c_4 }{15c_9}\)
C. \(\frac{4c_3*11c_5}{15c_9}\)
D. \(\frac{11c_5}{15c_9*7C_1}\)
E. None of these
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Re: 4 out of 15 apples are rotten. They are taken out one by one at random [#permalink] 4 out of 15 apples are rotten. They are taken out one by one at random   01 Apr 2020, 01:20
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Dillesh4096 wrote:
4 out of 15 apples are rotten. They are taken out one by one at random and examined. The ones which are examined are not replaced. What is the probability that the 9th one examined is the last rotten one?

A. \(\frac{4c_3*11c_5}{15c_8*7c_1}\)
B. \(\frac{11c_5*4c_4 }{15c_9}\)
C. \(\frac{4c_3*11c_5}{15c_9}\)
D. \(\frac{11c_5}{15c_9*7C_1}\)
E. None of these


Quote:
4 out of 15 apples are rotten

So Normal - 11 and rotten -4

Quote:
What is the probability that the 9th one examined is the last rotten one

So 8 have been picked. These 8 should be 5 normal and 3 rotten - ways = 11C5*4C3
Total ways = 15C8
Probability = \(\frac{11C5*4C3}{15C8}\)
BUT now the next can be any of the 7 remaining. However, it is the rotten one which we are looking at so probability is \(\frac{1}{7}=\frac{1}{7C1}\)

Final answer = \(\frac{11C5*4C3}{15C8}*\frac{1}{7C1}\)

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4 out of 15 apples are rotten. They are taken out one by one at random [#permalink] 4 out of 15 apples are rotten. They are taken out one by one at random   01 Apr 2020, 01:39
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Dillesh4096 wrote:
4 out of 15 apples are rotten. They are taken out one by one at random and examined. The ones which are examined are not replaced. What is the probability that the 9th one examined is the last rotten one?

A. \(\frac{4c_3*11c_5}{15c_8*7c_1}\)
B. \(\frac{11c_5*4c_4 }{15c_9}\)
C. \(\frac{4c_3*11c_5}{15c_9}\)
D. \(\frac{11c_5}{15c_9*7C_1}\)
E. None of these



Probability of 9th one being last rotten picked = Probability of 3 rotten to be pocked among 8 picks * Probability of last one being rottem among remaining 7

i.e. Probability of 9th one being last rotten picked = \(\frac{4c_3*11c_5}{15c_8}\) * \(\frac{1}{7}\)

Answer: Option A
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Re: 4 out of 15 apples are rotten. They are taken out one by one at random [#permalink] 4 out of 15 apples are rotten. They are taken out one by one at random   24 Jul 2023, 07:37
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Re: 4 out of 15 apples are rotten. They are taken out one by one at random [#permalink]
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