Probability of Union of 4 or More Elements 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?
 A: 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:
$$\mathbb{P}( A \cup B \cup C \cup D) = 1 - (1-0.2)(1-0.4)(1-0.8)(1-0.5) = 0.952$$
A: The question you pose cannot be solved; all that can be said is that
$$
\max\{P(A), P(B), P(C), P(D)\} = 0.8 \leq P(A \cup B \cup C \cup D) \leq 1
$$
However, the answer $0.952$ that you give corresponds to the case when $A$, $B$, $C$, and $D$ are independent events.  Did you leave out this important piece of information when you typed in your question?
Generally, if $A_i, 1 \leq i \leq n$ are independent events, then
using DeMorgan's laws
$$
P(A_1 \cup \cdots \cup A_n) = 1 - P((A_1 \cup \cdots \cup A_n)^c) = 1 - P(A_1^c\cap\cdots \cap A_n^c) = 1 - P(A_1^c)\cdots P(A_n^c)
$$
In order to avoid unnecessary arithmetic calculations, it is of the utmost importance that the expression on the right  not
be expressed as
$$
1 - [1-P(A_1)]\cdots[1-P(A_n)]
$$
and multiplied out to get
$$P(A_1) + \cdots + P(A_n) - [P(A_1)P(A_2) + P(A_1)P(A_3) + \cdots P(A_{n-1})P(A_n)] + \cdots
$$
The latter is an expression based on the principle of inclusion and exclusion and is a lot more work to evaluate.
