Question 1203560
<font color=black size=3>
Problem 1


A,B,C are mutually exclusive means
P(A n B) = 0
P(A n C) = 0
P(B n C) = 0
P(A n B n C) = 0
where the "n" refers to set intersection.


Something like P(A n B) = 0 means events A and B cannot happen simultaneously. They are disjoint events.


Use the inclusion-exclusion principle to say the following:
P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(B n C) - P(A n C) + P(A n B n C)
P(A u B u C) = (1/2) + (1/8) + (1/4) - 0 - 0 - 0 + 0
P(A u B u C) = (1/2) + (1/8) + (1/4)
P(A u B u C) = (4/8) + (1/8) + (2/8)
P(A u B u C) = (4+1+2)/8
P(A u B u C) = <font color=red size=4>7/8</font>


Or as a shortcut, because events A,B,C are mutually exclusive we can use this formula
P(A u B u C) = P(A) + P(B) + P(C)


---------------------------


Problem 2


P(A' n B' n C') = P( (A' n B') n C') 
P(A' n B' n C') = P( (A u B)' n C') ... De Morgans Law
P(A' n B' n C') = P( ((A u B) u C)')  ... De Morgans Law
P(A' n B' n C') = P( (A u B u C)') 
P(A' n B' n C') = 1 - P(A u B u C) ... complement law
P(A' n B' n C') = 1 - (7/8) .... use result of problem 1
P(A' n B' n C') = (8/8) - (7/8)
P(A' n B' n C') = (8-7)/8
P(A' n B' n C') = <font color=red size=4>1/8</font>
</font>