- #1
wanchosen
- 19
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I have been looking at the following problem but have doubts about my solution:
Suppose you are told that a particular team has won w games, drawn d games and lost l games during the course of a season. Calculate the probability that
a) the team won both the first and last games of the season,
b) the team lost both the first and the last games of the season.
I have tried to solve part a) using two methods and both seem to give me the same answer;
a) 1st Method using possible combinations/total combinations
Total no. of games played = w+d+l
Total no. of combinations =
(w+d+l)!
--------
w!d!l!
No. of combinations if 1st and last is a win =
(w+d+l-2)!
----------
(w-2)!d!l!
1) Probability of winning 1st and Last game) :-
(w+d+l-2)!
----------
(w-2)!d!l!
--------------
(w+d+l)!
--------
w!d!l!
where (w-2)! =
w!
-------
w(w-1)
and (w+d+l-2)! =
(w+d+l)!
---------------
(w+d+l)(w+d+l-1)
Equation 1) then simplifies to:-
w(w-1)
-----------------
(w+d+l)(w+d+l-1)
2nd Method
Probability of winning 1 game:-
P(w) =
w
-------
(w+d+l)
Probability of winning a second game
P(w2) =
w-1
---------
(w+d+l-1)
Therefore probability of winning the last game, given they won the first game :-
w(w-1)
----------------
(w+d+l)(w+d+l-1)
b)
Using the same logic I calculated that:-
Probability team loses the 1st and last game:-
l(l-1)
----------------
(w+d+l)(w+d+l-1)
I keep thinking I've missed something here. I'm not sure about the probability of the last win/loss and also if any wins/losses between the first/last game are relevant. Can someone please help?
Suppose you are told that a particular team has won w games, drawn d games and lost l games during the course of a season. Calculate the probability that
a) the team won both the first and last games of the season,
b) the team lost both the first and the last games of the season.
I have tried to solve part a) using two methods and both seem to give me the same answer;
a) 1st Method using possible combinations/total combinations
Total no. of games played = w+d+l
Total no. of combinations =
(w+d+l)!
--------
w!d!l!
No. of combinations if 1st and last is a win =
(w+d+l-2)!
----------
(w-2)!d!l!
1) Probability of winning 1st and Last game) :-
(w+d+l-2)!
----------
(w-2)!d!l!
--------------
(w+d+l)!
--------
w!d!l!
where (w-2)! =
w!
-------
w(w-1)
and (w+d+l-2)! =
(w+d+l)!
---------------
(w+d+l)(w+d+l-1)
Equation 1) then simplifies to:-
w(w-1)
-----------------
(w+d+l)(w+d+l-1)
2nd Method
Probability of winning 1 game:-
P(w) =
w
-------
(w+d+l)
Probability of winning a second game
P(w2) =
w-1
---------
(w+d+l-1)
Therefore probability of winning the last game, given they won the first game :-
w(w-1)
----------------
(w+d+l)(w+d+l-1)
b)
Using the same logic I calculated that:-
Probability team loses the 1st and last game:-
l(l-1)
----------------
(w+d+l)(w+d+l-1)
I keep thinking I've missed something here. I'm not sure about the probability of the last win/loss and also if any wins/losses between the first/last game are relevant. Can someone please help?