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.
Difficult Math Question #30 - Probability
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- ajith
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Total no: of ways to arrange them = 8!
No: of ways in which boys are the end
we can select 2 boys from 3 in 3 ways ...
and arrange them in 2 ways
we can arrange the remaining 6 in 6! ways
So the total no of ways = 2*3*6!
the probability = 6*6!/8!= 6/7*8=3/28
Similarly no of ways in which gals are in the end = 5c2*2*6!
=20*6!
The probability= 20*6!/8!=20/56=5/14
There can be mistakes ...
No: of ways in which boys are the end
we can select 2 boys from 3 in 3 ways ...
and arrange them in 2 ways
we can arrange the remaining 6 in 6! ways
So the total no of ways = 2*3*6!
the probability = 6*6!/8!= 6/7*8=3/28
Similarly no of ways in which gals are in the end = 5c2*2*6!
=20*6!
The probability= 20*6!/8!=20/56=5/14
There can be mistakes ...
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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
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