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29 Mar 2005, 05:25
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29 Mar 2005, 15:07
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Total ways in which 5 (3+2) goals can be scored = 5! = 120.
Total permutations of L1, L2, W1, W2 and W3.
[Assuming all individual goals are distinct]
If the losing team scores first, say L1, then L2 can be placed in 2, 3, 4 or 5 positions. Number of possible permutations = 4!.
Ditto for L2. If L2 is the first goal that the loser team scored, then L1 can be placed in 2, 3, 4, or 5 positions, and number of possible permutations = 4!.
Therefore, probability = 2x4!/5! = 48/120 = 4/10=2/5.
Please note that its not necessary that L1 and L2 are in chronological order. They just represent the two goals scored by the losing team.
Hope that helps.
Total permutations of L1, L2, W1, W2 and W3.
[Assuming all individual goals are distinct]
If the losing team scores first, say L1, then L2 can be placed in 2, 3, 4 or 5 positions. Number of possible permutations = 4!.
Ditto for L2. If L2 is the first goal that the loser team scored, then L1 can be placed in 2, 3, 4, or 5 positions, and number of possible permutations = 4!.
Therefore, probability = 2x4!/5! = 48/120 = 4/10=2/5.
Please note that its not necessary that L1 and L2 are in chronological order. They just represent the two goals scored by the losing team.
Hope that helps.
29 Mar 2005, 15:16
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choose a losing team = 2 ways then number ways to order the goals
2(1*4*3*2*1) , where the first "1" says that the outcome is fixed that is only 1 way for that goal i.e. losing team goals first.
p(e) = 2(1*4*3*2*1) / 5! = 2/5
2(1*4*3*2*1) , where the first "1" says that the outcome is fixed that is only 1 way for that goal i.e. losing team goals first.
p(e) = 2(1*4*3*2*1) / 5! = 2/5
