jmc

It is impossible for Pat to select both 3 oranges and 6 apples, so we can calculate the probabilities of these mutually exclusive cases separately and then add to get our final answer. The probability that 3 particular pieces of fruit will be oranges and the rest will not be is given by ${\left(\frac{1}{3}\right)}^3{\left(\frac{2}{3}\right)}^5=\frac{32}{6561}$, and there are $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ ways of selecting three pieces of fruit to be the oranges so the probability that 3 will be oranges is $56\cdot \frac{32}{6561}=\frac{1792}{6561}$. Similarly, the probability that 6 particular pieces of fruit will be apples and the other two won't be is given by ${\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^2=\frac{4}{6561}$ and there are $\left(\genfrac{}{}{0ex}{}{8}{6}\right)=28$ ways of selecting which ones will be the apples, so multiplying again gives us a probability of $28\cdot \frac{4}{6561}=\frac{112}{6561}$. Adding those two probabilities give us our final answer: $\frac{1792}{6561}+\frac{112}{6561}=\boxed{\frac{1904}{6561}}$.