# Thread: probability question

1. ## probability question

wanted to check my answers for this question.

a team played 50 games this year. the results are shown in the table.

before the season started the coach chose a fixture at random to visit the club. what is the probability that the fixture he attended was

a) an away win
b)a match the team did not loose
c)a win given it was a home game

for a) $\displaystyle \frac{13}{17+13}$ = $\displaystyle \frac {13}{30}$

b) $\displaystyle \frac {17+2+13+5}{17+2+7+13+5+6}$ = $\displaystyle \frac{37}{50}$

c) $\displaystyle \frac {17}{17+2+7}$ = $\displaystyle \frac{17}{26}$

2. Originally Posted by sigma1
wanted to check my answers for this question.

a team played 50 games this year. the results are shown in the table.

before the season started the coach chose a fixture at random to visit the club. what is the probability that the fixture he attended was

a) an away win
b)a match the team did not loose
c)a win given it was a home game

for a) $\displaystyle \frac{13}{17+13}$ = $\displaystyle \frac {13}{30}$

b) $\displaystyle \frac {17+2+13+5}{17+2+7+13+5+6}$ = $\displaystyle \frac{37}{50}$

c) $\displaystyle \frac {17}{17+2+7}$ = $\displaystyle \frac{17}{26}$
Hi sigma1,

you have answered part a) in terms of finding the probability that the fixture was
an away win, given that the game was won.

This is different to the probability that the fixture was an away win

$\displaystyle P(home\ win)=\frac{17}{number\ of\ fixtures}$

$\displaystyle P(home\ win\ given\ that\ game\ is\ won)=\frac{17}{number\ of\ won\ games}$

and so on.