1. (a) 560.
(b) 12.
2. (a) 3/11.
(b) 3/44.
3. 8%. (The probability of smoking is 0.1).
4. (a) rs + 1 + rq - r - q.
(b) rs/(1 + rs + rq - r - q).
5. (a) 7.
(b) 11.
(c) a = 1/sqrt{11} and b = -7/sqrt{11} is one solution.
6. (a) 2 2/3.
(b) 16/27.
7. Notation: The complement of B will be written here as B';
the intersection of A and B will be written A n B;
the union of A and B will be written A u B.
We seek to show that P(A n B') = P(A)P(B').
Here is one way to do this problem.
For independent events we have P(B|A) = P(B) and hence
P(A n B') = P(A)P(B'|A)
= P(A)[1 - P(B|A)]
= P(A)[1 - P(B)] by independence
= P(A)P(B').
Here is another way to do this problem.
P(A n B') = P(A) - P(A n B) (see below)
= P(A) - P(A)P(B) by independence of A and B
= P(A)[1 - P(B)] by algebra
= P(A)P(B'), which shows independence of A and B'.
The reason for the first step is that
A is the union of the two disjoint sets
A n B' and A n B (imagine a Venn diagram here), and
hence P(A) = P(A n B') + P(A n B).