Posted by tina on Thursday, April 9, 2009 at 10:43am.
Make a table
--B1 B2 B3 B4 B5 B6 B7
J1 --y- y- n- n- n- n-
J2 y -- y- n- n- n- n-
J3 y y- -- n- n- n- n-
J4 n n- n- -- n- n- n-
J5 n n- n- n- -- n- n-
J6 n n- n- n- n- -- n-
J7 n n- n- n- n- n- --
The diagonal elemnts are not part of our possibilities because they can no both finish in the same position.
That leaves us with 42 possible outcomes
of those 42, six are y which means six of the 42 outcomes have both B and J in the first 3 places.
6/42 =3/21
Make a table
--b1 b2 b3 b4 b5 b6 b7
J1 --y- y- n- n- n- n-
J2 y -- y- n- n- n- n-
J3 y y- -- n- n- n- n-
J4 n n- n- -- n- n- n-
J5 n n- n- n- -- n- n-
J6 n n- n- n- n- -- n-
J7 n n- n- n- n- n- --
The diagonal elements are not part of our possibilities because they can no both finish in the same position.
That leaves us with 42 possible outcomes
of those 42, six are y which means six of the 42 outcomes have both B and J in the first 3 places.
6/42 =3/21
I can not get columns to line up no matter what I do using -- or - for spacers
you have 7 rows and seven columns
the diagonals, marked with -- are out because J and B can not finish in the same place.
The y solutions are the good ones.
The n solutions are no good.
there are 3 people not only 2
you left out morgan
the number of ways for J, M and B and 5 others can be arranged is
3!5!.
The number of ways all 8 can be arranged is 8!
so the prob that they will be the first 3 finishers
= 3!5!/8! = 1/56
but you asked for the odds
odds of some event = prob the event will happen : prob the event will not happen
= (1/56):(55/56)
= 1 : 55
For Jerome the chances are 3/8 (if all of the runners had equal ability). For Morgan, after Jerome had gotten first (let's say he did) then his chances would be 2/7. And for Bly his chances are 1/6. Now you have got to multiply them all together.
(3/8) (2/6) (1/6) = 1/56
i feel like this is the easy way of doing it...
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