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?

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Are A, B, C, and D independent events? If not, you need more data. – Henning Makholm Sep 21 '11 at 22:00
Yes they're independent. – Clarkson Sep 21 '11 at 22:01

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:

$$\mathbb{P}( A \cup B \cup C \cup D) = 1 - (1-0.2)(1-0.4)(1-0.8)(1-0.5) = 0.952$$

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And note that it's very important (as Sasha mentions) that the events be independent. Otherwise the answer could be anywhere from 0.8 to 1, depending on the specifics. – Greg Martin Sep 21 '11 at 22:12
Is this something that could work with n elements? (I won't need this for my studies, but I'm curious). – Clarkson Sep 21 '11 at 22:12
Yes, it would, because $( \cup_{i=1}^n A_i )^c = \cap_{i=1}^n A_i^c$. – Sasha Sep 21 '11 at 22:14
Thank You for the help! – Clarkson Sep 21 '11 at 22:15

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 $$P(A_1) + \cdots P(A_n) - [P(A_1A_2) + P(A_1)P(A_3) + \cdots P(A_{n-1}P(A_n)] + \cdots$$ The latter is an expression based on the principle of inclusion and exclusion and is a lot more work to evaluate.

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...lot more work and would need lot more data too :-) – Srivatsan Sep 21 '11 at 22:46
@Srivatsan Yes to lot more work, but no to lot more data. The general inclusion-exclusion principle needs lot more data since it needs, for example, the value of $P(A_iA_j)$. But for independent events, $P(A_iA_j) = P(A_i)P(A_j)$, and similarly for all the compound events, and so the inclusion-exclusion principle can be used if we know just the values of $P(A_1), \ldots, P(A_n)$. But, as I pointed out and we agree, it is a lot more work than just computing the product of the probabilities of the complementary events and subtracting from $1$ – Dilip Sarwate Sep 22 '11 at 0:54