| Yet one more way to think about ace or king appears is to note that it can happen on either the first or second pull. So on pull 1, p(A or K)=2/13 (4 aces, 4 kings=8/52=2/13). Or, we can get something other than an ace or king on pull 1 (p=11/13) and an ace or king on pull 2 (again, p=2/13). So the total is p=2/13+(11/13)(2/13)=2/13+22/169=26/169+22/169=48/169. |