# soccer problem

• Apr 26th 2010, 06:08 AM
kingman
soccer problem
Dear Sir,
I need help some help with the below problems and would appreciate very much if can offer some help.
thanks
Kingman

1) In a competition, 2 teams (A and B) will play each other in the best of 3 games. That is, the final team to win 2 games will be the winner. In the first games, both teams have equal chance of winning. In subsequent games, the probability of team A winning team B given that team A won in the previous game is p and the probability of team A winning team B given that team A lost in the previous game is 1/3.

a) Illustrate the information with an appropriate tree diagram.
b) Find the value of p such that the team A has equal chances of winning and losing the competition.

2) A soccer team of 30 players has 18 Americans and 12 foreigners. Of the
Americans, 10 are midfielders and 8 are defenders. Of the foreigners, 5 are
goalkeepers and 7 are strikers. Four players are chosen at random.
a) If exactly one striker is chosen, find the probability that 3 American are
chosen.
b) Find the probability that there is exactly one goalkeeper or exactly one defender (or both)
• Apr 28th 2010, 03:11 PM
atalay

a)$\displaystyle P(X=3) = \binom{18}{3}$$\displaystyle \binom{11}{0} / \displaystyle \binom{29}{3} b) exactly one goalkeeper \displaystyle P(goalkeeper=1) = \binom{5}{1}$$\displaystyle \binom{18}{2}$$\displaystyle \binom{6}{0} / \displaystyle \binom{29}{3} exactly one defender \displaystyle P(d=1) = \binom{8}{1}$$\displaystyle \binom{15}{2}$$\displaystyle \binom{6}{0} / \displaystyle \binom{29}{3} both \displaystyle P(both=1) = \binom{8}{1}$$\displaystyle \binom{5}{1}$$\displaystyle \binom{10}{1}$$\displaystyle \binom{6}{0}$ / $\displaystyle \binom{29}{3}$

(if exactly one striker chosen)
• Apr 29th 2010, 04:53 AM
kingman
Dear atalay,
Thanks very much for the solution but I wonder why it does not tally with the answer given to question 2 as
a).461
b).671
Thanks very much
Kingman
• Apr 29th 2010, 08:59 AM
atalay

a) $\displaystyle P(X=3) = \binom{18}{3}$$\displaystyle \binom{7}{1}$ / $\displaystyle \binom{25}{4}$

b) which one calculated?
• Apr 29th 2010, 12:07 PM
shenanigans87
I have only looked at problem 2a so far.

You have 18 Americans, 12 Foreigners.

1 of those Foreigners is chosen (striker). You now randomly select 3 more people. What are the chances that all three are American?

So now you have 18 Americans, 11 Foreigners.

the chance of first being an American is 18/(18+11) = 18/29
the chance of the second being an American is 17/(17+11) = 17/28
the chance of the third being an American is 16/(16+11) = 16/27

The chances of those happening consecutively?

(18/29)*(17/28)*(16/27) = (4896/21924) = 22.31%

Proof? I just got done making a simulation that would choose 3 players from 29 total. 18 of them are americans, 12 foreigners. Once one is chosen, it cant be chosen again.

Here were the results of the first test with 100,000 trials.

http://img256.imageshack.us/img256/7904/probab01.png

which is 22.188%

and here's the code.

http://img706.imageshack.us/img706/2118/probab02.png
• Apr 30th 2010, 06:16 AM
atalay
Quote:

Originally Posted by shenanigans87
I have only looked at problem 2a so far.

You have 18 Americans, 12 Foreigners.

1 of those Foreigners is chosen (striker). You now randomly select 3 more people. What are the chances that all three are American?

So now you have 18 Americans, 11 Foreigners.

the chance of first being an American is 18/(18+11) = 18/29
the chance of the second being an American is 17/(17+11) = 17/28
the chance of the third being an American is 16/(16+11) = 16/27

The chances of those happening consecutively?

(18/29)*(17/28)*(16/27) = (4896/21924) = 22.31%

Proof? I just got done making a simulation that would choose 3 players from 29 total. 18 of them are americans, 12 foreigners. Once one is chosen, it cant be chosen again.

Here were the results of the first test with 100,000 trials.

http://img256.imageshack.us/img256/7904/probab01.png

which is 22.188%

and here's the code.

http://img706.imageshack.us/img706/2118/probab02.png

• Apr 30th 2010, 07:18 AM
shenanigans87
You are right, let me change it a bit.

I was counting the rest of the strikers in the probability, I forgot about the part which states that exactly 1 striker is chosen, not more.

So

$\displaystyle \frac{18}{29}*\frac{17}{28}*\frac{16}{27}=23.3317$%

becomes

$\displaystyle \frac{18}{23}*\frac{17}{22}*\frac{16}{21}=46.0757$%
• Apr 30th 2010, 02:43 PM
kingman
sample space of soccer problem
Dear shenanigans87
Thanks very much for providing great insight into the
soccer problem but I wonder whether you can enumerate
the sample space in the form of tree diagram or
ordered n-tuplets in spreadsheet thus enabling one to see how
the events look like.
Thanks
Kingman
• Apr 30th 2010, 02:47 PM
shenanigans87
Quote:

Originally Posted by kingman
Dear shenanigans87
Thanks very much for providing great insight into the
soccer problem but I wonder whether you can enumerate
the sample space in the form of tree diagram or
ordered n-tuplets in spreadsheet thus enabling one to see how
the events look like.
Thanks
Kingman

I don't understand what you are asking for...?
• May 11th 2010, 05:28 AM
kingman
Dear Sir,