Probability for 6 and 11 sets

Question

2 sets formulas

P(A∪B)=P(A)+P(B)−P(A∩B)

3 sets formula

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

4 sets formula

P (A U B U C U D) = P(A) + P(B) + P(C) +P(D) - P(A ∩ B) - P(A ∩ C) - P(A ∩ D)- P(B ∩ C) - P(B ∩ D) - P(C ∩ D) + P(A ∩ B ∩ C) + P(A ∩ B ∩ D) + P(A ∩ C ∩ D) + P(B ∩ C ∩ D) - P(A ∩ B ∩ C ∩ D).

What is 6 sets formula?

What is 11 sets formula?

Please help me

Answer 1

This is the inclusion-exclusion principle. add ones, subtract twos, add threes, subtract 4s, add 5s, subtract 6s, add 7s, subtract 8s, add 9s, subtract 10s, add 11s. I leave it to you to write all of the elements. For 11 groups, you will have 2^11 - 1 = 2047 groups to show. You might see if you can write the solution without explicitly writing the 2047 groups.