Question 1163230
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(i)   P(A ∩ B ∩ C') = P(A ∩ B) - P(A ∩ B ∩ C) = 0.3 - 0.05 = 0.25.


(ii)  P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C) = 0.5 + 0.6 + 0.4 - 0.3 - 0.1 - 0.2 + 0.05 = 0.95.
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The formula &nbsp;(ii) &nbsp;is used &nbsp;VERY &nbsp;OFTEN. 

It is one of the basic formula of the elementary probability theory.

So it makes sense to memorize it.


To make memorizing easier, &nbsp;pay your attention that the formula is an alternate sum.


First three addends are the probabilities of the original events &nbsp;(with the sign &nbsp;" + ").

Next three terms are the probabilities of the in-pair intersections of the events &nbsp;(with the sign &nbsp;" - ").

The last term is the probability of the triple intersection of the events, &nbsp;with the sign &nbsp;" + ".


Below is a short proof of this formula.



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    It is totally clear to you why I add the first three addends in the formula (ii).

    But when I add them, I count twice the probabilities of each in-pair intersection.

    Therefore, I subtract the probabilities of  each in-pair intersection.

    Next, when I add three first addends, I count thrice the probability of  the triple intersection;

    and when I subtract in-pair intersections, I cancel these terms thrice.

    Therefore, I must add the probability of the triple intersection one more time to restore the balance.
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