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How is $$P(A \cap B' \cap C')+P(A' \cap B \cap C')+P(A' \cap B' \cap C) = P(A)+P(B)+P(C)-2[P(A \cap B)+P(B \cap C)+P(C \cap A)]+3P(A \cap B \cap C) $$

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up vote 3 down vote accepted

The Venn diagram looks as follows:

enter image description here

The following identities hold, because probabilities of disjunctive variables sum up to the probability of their union.

$$\begin{align}P(A) & = P(A \cap B' \cap C') + P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A \cap B' \cap C') \\ P(B) & = P(A' \cap B \cap C') + P(A \cap B \cap C') + P(A' \cap B \cap C) + P(A \cap B \cap C) \\ P(C) & = P(A' \cap B' \cap C) + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B \cap C) \\ P(A \cap B) & = P(A \cap B \cap C') + P(A \cap B \cap C) \\ P(B \cap C) & = P(A' \cap B \cap C) + P(A \cap B \cap C) \\ P(A \cap C) & = P(A \cap B' \cap C) + P(A \cap B \cap C) \end{align}$$

Substituting these identities in the right-hand-side of your equation shows the correctness.

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Why do you share his doubts? This diagram clearly illustrates that the equality holds. – Patrick Jul 11 '13 at 16:16
Yes, my doubts have gone by now. I was not sure about the "2". – Axel Kemper Jul 11 '13 at 16:19
Thankyou both the solutions helped :) – chndn Jul 12 '13 at 15:04

Draw a Venn diagram with three circles, all intersecting. There should be eight regions, including the outside. Then go through all eight regions, and decide how many times each region is counted on the left-hand side, and how many times it is counted on the right-hand side. For example, $A\cap B\cap C'$ is counted on the right-hand side once in $P(A)$, once in $P(B)$, minus twice in $2P(A\cup B)$, or zero times in all.

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I want theoretical proof – chndn Jul 11 '13 at 15:56

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