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Bunuel
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Ten girls to be arranged around the table. What is the probability tha 04 Mar 2021, 04:15
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55% (hard)

Question Stats:

55% (01:19) correct 45% (01:45) wrong based on 130 sessions

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Ten girls to be arranged around the table. What is the probability that 2 particular girls are NOT together ?

A. 1/10
B. 1/9
C. 2/9
D. 7/9
E. 8/9
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D
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Schools: IIM (A) ISB '24
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Ten girls to be arranged around the table. What is the probability tha 04 Mar 2021, 04:31
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Bunuel
Ten girls to be arranged around the table. What is the probability that 2 particular girls are NOT together ?

A. 1/10
B. 1/9
C. 2/9
D. 7/9
E. 8/9
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Total arrangement of 10 girls around the table = (10-1)! = 9!

Total arrangement of 10 girls around the table such that two particular girls are together = 8!*2!

Probability \(= \frac{8!*2!}{9!} = \frac{2}{9}\)

Answer: Option C
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Ten girls to be arranged around the table. What is the probability tha 04 Mar 2021, 04:39
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Quote:


Total arrangement of 10 girls around the table such that two particular girls are together = 8!*2!

Probability \(= \frac{8!*2!}{9!} = \frac{2}{9}\)
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Isn't the question asking the probability that 2 particular girls are NOT together?

So, I am wondering if the answer should be 7/9?
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Ten girls to be arranged around the table. What is the probability tha 04 Mar 2021, 08:31
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sahilvermani
Quote:


Total arrangement of 10 girls around the table such that two particular girls are together = 8!*2!

Probability \(= \frac{8!*2!}{9!} = \frac{2}{9}\)
Isn't the question asking the probability that 2 particular girls are NOT together?

So, I am wondering if the answer should be 7/9?
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yus we need to minus that value from 1
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Ten girls to be arranged around the table. What is the probability tha 14 Mar 2021, 23:48
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Bunuel
Ten girls to be arranged around the table. What is the probability that 2 particular girls are NOT together ?

A. 1/10
B. 1/9
C. 2/9
D. 7/9
E. 8/9
Show more

Two particular girls are together

Let us put the two girls together as \(1\) unit now we have a total of \(9\) units

In a circle the total number of arrangements is \((n-1)! \) for n units.

Hence here for total \(9\) units total arrangements is \(8!\) ways

The two girls among themselves can be arranged in \(2!\) ways

So total ways \(2\) girls are together = \(8!2!\)

Total ways there are NOT together = Total ways without any restriction - total ways 2 girls are together

So required ways = \(9!- 8!2!\)

Probability=Required ways / Total ways

\(\frac{ 9!- 8!2!}{9!} =\frac{ 7}{9} \)

Ans-D

Hope it's clear.

sahilvermani, yes you are correct.
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Ten girls to be arranged around the table. What is the probability tha 14 Aug 2024, 10:28
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