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27 Aug 2009, 17:04
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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guys I think I posted it at wrong place earlier...
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 ????
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 ????
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29 Aug 2009, 02:05
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Mike0530
Mike0530, approach 2 can be taken as:
4c1*(1/6)*3c1(1/5)
4c1=No of ways X can be selected.
1/6=Probability of selection of X
3c1=No of ways Y can be selected.
1/5=Probability of selection of Y.
