One draws objects $A_i$ that contain each 1 of 5 figures ($ F_1, F_2,...F_5$) with an equal probability of $1/5$ . Using the inclusion-exclusion principle, I have to determine the probability to get the figures $F_1, F_2, F_3$, while picking at random 6 such objects.
In the case of 3 events, the inclusion-exclusion principle is: $P(A_1\cup A_2\cup A_3)=P(A_1)+P(A_2)+P(A_3)-P(A_1\cap A_2)-P(A_1\cap A_3)-P(A_2\cap A_3)+P(A_1 \cap A_2 \cap A_3)$
In the case of 6 events, the formula gets very complicated. The other problem is the difficulty for me to formalize the given problem in probabilistic terms by means of the inclusion-exclusion principle.
Can you provide some hint ? Thanks.
Here's a hint:
Let $A_i$ denote the event that we don't draw $F_i$ in any of our six draws, and then use inclusion-exclusion to compute $P(A_1 \cup A_2 \cup A_3)$. Then consider how this relates to the probability you want to find.