Question 1163230: Assume three events A, B and C are such that P(A)=0.5, P(B)=0.6,
P(C)=0.4, P(A∩B)=0.3, P(A∩C)=0.1, P(B∩C)=0.2 and P(A∩B∩C)=0.05.
Find
(i) P(A∩B∩C') (ii) P(AUBUC) (iii) p(A'∩B∩C')
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
(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.
-------------
The formula (ii) is used VERY OFTEN.
It is one of the basic formula of the elementary probability theory.
So it makes sense to memorize it.
To make memorizing easier, pay your attention that the formula is an alternate sum.
First three addends are the probabilities of the original events (with the sign " + ").
Next three terms are the probabilities of the in-pair intersections of the events (with the sign " - ").
The last term is the probability of the triple intersection of the events, with the sign " + ".
Below is a short proof of this formula.
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.
|
|
|
