# Using inclusion-exclusion principle to find the probability of events

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.

• I think it wants you to find the probability of $F_1 \cup F_2 \cup F_3$ for any given object, then use that probability for $1$ object to find the probability to find one of those figures in your set of $6$ objects. Apr 17 at 22:28

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.
• Can you expand a little bit more? What is the event one should pay attention to? Is it the fact to find one of the figures or the fact to draw one of the objects? So I am confused if one should look at $P(A_1\cup...\cup A_6)$ or $P(A_1\cup A_2\cup A_3)$. Anyway, the probability that at least one of $F_1,F_2,F_3$ is not drawn in any of six drwas is 124/125.Thus the probability that all of them are drawn should be 1/125. Is it correct ? Apr 17 at 23:27