The number of all possible selections in \({100 \choose m}\). Now we will count how many of those have \(n\) lemons. First we choose \(n\) out of \(k\) lemons by \({k \choose n}\); and then decide on the remaining cars on the batch. The batch size is \(m\), of which \(n\) are lemons, therefore the number of ways to pick non-lemons for the reamining slots in the batch is \({100 - k \choose m - n}\). Finally, the probability of picking \(n\) lemons in a \(m\) car batch is,

\[\frac{ { k \choose n } \cdot { 100 - k \choose m - n } }{100 \choose m}\]