I have the following problem:
Given $P(A)=0.2$, $P(B)=0.4$, $P(C)=0.8$, $P(D)=0.5$, find $P(A\cup B\cup C\cup D)$
And the final answer should be 0.952
I know how to find the union of two and three elements (for 2, its: $A+B-AB$), but the formula becomes clumsy after 3. The best things I've found says that to find the union for n elements, I add as follows $$0.2-(0.2\times0.4)+(0.2\times0.4\times0.8)-(0.2\times0.4\times0.8\times0.5) = 0.152$$ which is wrong.
What is a good general rule for n events?
Best Answer
Use the following identity:
$$\mathbb{P}( A \cup B \cup C \cup D) = 1 - \mathbb{P}( (A \cup B \cup C \cup D)^c ) = 1 - \mathbb{P}( A^c \cap B^c \cap C^c \cap D^c )$$
Here $A^c$ means complement of set $A$.
Given independence of events $\mathbb{P}( A^c \cap B^c \cap C^c \cap D^c ) = \mathbb{P}( A^c ) \mathbb{P}( B^c ) \mathbb{P}( C^c ) \mathbb{P}( D^c )$. Now:
$$\mathbb{P}( A \cup B \cup C \cup D) = 1 - (1-0.2)(1-0.4)(1-0.8)(1-0.5) = 0.952$$