math
posted by tina .
jerome morgan and bly compete in a 10 kilometer race with 5 other athletes. what are the odds that they finish in the top 3 places? i got 3 to 5 but my teacher marked it wrong and wanted us to find out why it was wrong on our own. help would be nice!

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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