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Difficult Math Question #30 - Probability

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800guy Master | Next Rank: 500 Posts
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Difficult Math Question #30 - Probability

Fri Oct 20, 2006 10:37 am
OA coming after some people answer:

5 girls and 3 boys are arranged randomly in a row. Find the probability that:

A) there is one boy on each end.

B) There is one girl on each end.

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800guy Master | Next Rank: 500 Posts
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Mon Oct 23, 2006 2:25 pm
OA:

For the first scenario:
A) there is one boy on each end.

The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3
the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2
the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways
Total possible outcome = 8!
Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28

For the second scenario:
A) there is one girl on each end.

The first seat can be filled in 5C1 (5 girls 1 seat) ways = 5
the last seat can be filled in 4C1 (2 girls 1 seat) ways = 4
the six seats in the middle can be filled in 6! (3 boys and 3 girls) ways
Total possible outcome = 8!
Probability= (5C1 * 4C1 * 6!)/ 8! = 5/14

rajs.kumar Senior | Next Rank: 100 Posts
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09 Jun 2006
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87 messages
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Mon Oct 23, 2006 7:37 am

1. 3P2 x 6!/8! = 3/28

2. 5P2 x 6!/8! = 5/14

800guy Master | Next Rank: 500 Posts
Joined
27 Jun 2006
Posted:
354 messages
Followed by:
5 members
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Mon Oct 23, 2006 2:25 pm
OA:

For the first scenario:
A) there is one boy on each end.

The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3
the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2
the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways
Total possible outcome = 8!
Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28

For the second scenario:
A) there is one girl on each end.

The first seat can be filled in 5C1 (5 girls 1 seat) ways = 5
the last seat can be filled in 4C1 (2 girls 1 seat) ways = 4
the six seats in the middle can be filled in 6! (3 boys and 3 girls) ways
Total possible outcome = 8!
Probability= (5C1 * 4C1 * 6!)/ 8! = 5/14

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