Nevermind guys. That's wrong too.
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 (1/3)^3(2/3)^5=32/6561, and there are C(8,3)=56 ways of selecting three pieces of fruit to be the oranges so the probability that 3 will be oranges is 56*(32/6561)=1792/6561. Similarly, the probability that 6 particular pieces of fruit will be apples and the other two won't be is given by (1/3)^6*(2/3)^2=4/6561 and there are C(8,6)=28 ways of selecting which ones will be the apples, so multiplying again gives us a probability of 28*(4/6561)=112/6561. Adding those two probabilities give us our final answer: 1792/6561+112/6561=1904/6561.
+88 