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There are 5 horses (A,B,C,D,E).
Each has an equal chance of placing first in a race (20%).
What is the probability that a punter will win a trifector in a bet (that is, the punter will successfully guess who will place 1, 2, and 3.)?

It would help me greatly if you could explain the formula.

I know that the probability of the punter guessing who will place 1 & 2 is just "P(A n B) = P(A)xP(B)." I thought maybe that means the formula for "P(A n B n C) = P(A)xP(B)xP(C)" but this doesn't seem to be correct. Thanks for any help.

2007-11-13 21:13:35 · 3 answers · asked by michael 2 in Science & Mathematics Mathematics

3 answers

Break it up into parts. The probability of guessing who is first is 20%. Then there is a 25% chance of guessing who is second. Finally there is a 33% chance of guessing who is third.

Another way to do it would be to know that there are 5x4x3x2x1 possibilites for outcomes of the race. A fifth of those would have the horse you chose as first. A fourth of those (or a twentieth of the total) would have the correct horse in second. Finally, a third of those (or one sixtieth of the total) would have the correct third place finisher.

So your answer would be 1/60.

2007-11-13 21:22:16 · answer #1 · answered by KJH 4 · 0 0

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2017-01-05 11:18:08 · answer #2 · answered by solem 3 · 0 0

There are 5C3 = 10 combinations of trifectas.
p(w) = (1/10)(1/5)(1/4)(1/3) = 1/600

2007-11-13 21:46:47 · answer #3 · answered by Helmut 7 · 0 0

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