3 From the records of Mulcaster United Football Club the following distribution was suggested as a probability model for future matches. \(X\) and \(Y\) denoted the numbers of goals scored by the home team and the away team respectively.
| \(X\) | |
| \cline { 2 - 5 }
\multicolumn{1}{c}{} | 0 | 1 | 2 | 3 |
| 0 | 0.11 | 0.04 | 0.06 | 0.08 |
| 1 | 0.08 | 0.05 | 0.12 | 0.05 |
| 2 | 0.05 | 0.08 | 0.07 | 0.03 |
| 3 | 0.03 | 0.06 | 0.07 | 0.02 |
Use the model to find
- \(\mathrm { E } ( X )\),
- the probability that the away team wins a randomly chosen match,
- the probability that the away team wins a randomly chosen match, given that the home team scores.
One of the directors, an amateur statistician, finds that \(\operatorname { Cov } ( X , Y ) = 0.007\). He states that, as this value is very close to zero, \(X\) and \(Y\) may be considered to be independent.
- Comment on the director's statement.