A fair coin is tossed 3 times....

A fair coin is tossed three times.
The events A, B, and C are defined as follows:

A: {At least one head is observed}

B: {At least two heads are observed}

C: {The number of heads observed is odd}

Find the following probabilities
(note: 0 is an even number; "and", "or", "not" have their usual meaning of intersection, union, complement).

(a) P (C)

(b) P (A or (not B))

(c) P ((not A) or B or (not C))

probability is my weakness. So these questions make no sense.. I understand complement, intersection, and union though; but how to add them with coin tossing? no idea.

(a)

P(C)

= P(The number of heads is odds)

= P(1 head or 3 heads)

= P(1 head) + P(3 heads)

= [(1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2)] + P(3 heads)

= [3(1/2)^3] + P(3 heads)

= 3/8 + P(3 heads)

= 3/8 + [(1/2)(1/2)(1/2)]

= 3/8 +1/8

= 4/8

= 1/2

(b)

P(A or (not B))

= P(A) + P(not B)

= P(at least 1 head) + P(not at least 2 heads)

= P(1 head or 2 heads or 3 heads) + P(0 heads or 1 head)

= P(1 head) + P(2 heads) + P(3 heads) + P(0 heads) + P(1 head) note: Since A and notB overlap when exactly 1 head is tossed, we aren't really adding P(1 head) twice, but it is left in this step for clarity. It is removed in the next step.

= P(0 heads) + P(1 head) + P(2 heads) + P(3 heads) note: Now it is important to notice that the answer is 1 (i.e. 100% chance) since we can only toss 0, 1, 2 or 3 heads.

= 1

(c)

P((not A) or B or (not C))

= P(notA) + P(B) + P(notC)

= P(not at least 1 head is observed) + P(at least 2 heads are observed) + P(the number of heads is not odd)

= P(0 heads) + P(2 heads or 3 heads) + P(the number of heads is even)

= P(0 heads) + P(2 heads) + P(3 heads) + P(0 heads or 2 heads) note: notC is actually contained by the union of notA and B (kinda like the first note), so we will remove it in the next step.