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27 Aug 2009, 14:50
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Question: There are 6 team members. 4 have to be selected. What is probability that 2 particular members, say X and Y, are both selected.
Approach 1:
Total possible outcomes = 6! / (4! x 2!) = 15 teams
Winning outcomes where 2 particular members are on team= 4! /(2! x 2!) = 6
Therefore probability that 2 particular members on team = 6/15 = 2/5
Approach 2:
let X be selected first, Y second, "Some other" third, "Some other" fourth - to make a four member team out of 6.
Therefore probability = 1/6 x 1/5 x 4/4 x 3/3 = 1/30
But since order does not matter, actual probability = 1/30 x 4! = 4/5
OA : 2/5
Whats wrong with approach 2
Approach 1:
Total possible outcomes = 6! / (4! x 2!) = 15 teams
Winning outcomes where 2 particular members are on team= 4! /(2! x 2!) = 6
Therefore probability that 2 particular members on team = 6/15 = 2/5
Approach 2:
let X be selected first, Y second, "Some other" third, "Some other" fourth - to make a four member team out of 6.
Therefore probability = 1/6 x 1/5 x 4/4 x 3/3 = 1/30
But since order does not matter, actual probability = 1/30 x 4! = 4/5
OA : 2/5
Whats wrong with approach 2
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27 Aug 2009, 16:00
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Hi,
I'm not familiarized with this method (2) but I think you just have to do the following:
\(\frac{1}{30} * \frac{4!}{2!} = \frac{12}{30} = \frac{2}{5}\)
Good luck tomorrow!
I'm not familiarized with this method (2) but I think you just have to do the following:
\(\frac{1}{30} * \frac{4!}{2!} = \frac{12}{30} = \frac{2}{5}\)
Good luck tomorrow!
