| Date | May 2009 | Marks available | 4 | Reference code | 09M.2.sl.TZ1.7 |
| Level | SL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
In any given season, a soccer team plays 65 % of their games at home.
When the team plays at home, they win 83 % of their games.
When they play away from home, they win 26 % of their games.
The team plays one game.
Find the probability that the team wins the game.
If the team does not win the game, find the probability that the game was played at home.
Markscheme
appropriate approach (M1)
e.g. tree diagram or a table
\({\rm{P(win)}} = {\rm{P}}(H \cap W) + {\rm{P}}(A \cap W)\) (M1)
\( = (0.65)(0.83) + (0.35)(0.26)\) A1
\( = 0.6305\) (or 0.631) A1 N2
[4 marks]
evidence of using complement (M1)
e.g. \(1 - p\) , 0.3695
choosing a formula for conditional probability (M1)
e.g. \({\rm{P}}(H|W') = \frac{{{\rm{P}}(W' \cap H)}}{{{\rm{P}}(W')}}\)
correct substitution
e.g. \(\frac{{(0.65)(0.17)}}{{0.3695}}\) \(\left( { = \frac{{0.1105}}{{0.3695}}} \right)\) A1
P(home) = 0.299 A1 N3
[4 marks]
Examiners report
Part (a) was nearly always correctly answered by those who attempted the question, but part (b) (conditional probability) was poorly done. A surprisingly small number of students drew a tree diagram in part (a) and those who did answered this part and part (b) well. Many found the correct complement in part (b) but could not make any further progress.
Part (b) (conditional probability) was poorly done. A surprisingly small number of students drew a tree diagram in part (a) and those who did answered this part and part (b) well. Many found the correct complement in part (b) but could not make any further progress.
