Question 1105636
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(a) The events are mutually exclusive, so P(A union B union C) is just P(A)+P(B)+P(C) = 0.2+0.3+0.2 = 0.7<br>
(b) Again since the events are mutually exclusive, (A complement) intersect (B union C) is just B union C; so P[(A complement) intersect (B union C)] is P(B union C) = P(B)+P(C) = 0.3+0.2 = 0.5<br>
(c) Once more, since the events are mutually exclusive, B union C complement is just C complement; the complement of that set is C.  So P((B union C complement)complement) = P(C) = 0.2