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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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    Yes they're independent.2011-09-21

2 Answers 2

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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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    will this work for probabilities for union of sets extending to n?2017-04-10
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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.

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    @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$2011-09-22