1.  disjoint

2.  many possibilities

3.  no

4.  false

5.  dispersion

6.  We see from (2.29)-(2.31) that

P(R = 3, B <= 4) = 4/36 = 1/9

If we falsely assumed independence, we would have

P(R = 3, B <= 4) = P(R = 3) P(B <= 4)

Now 

P(B <= 4) = P(0 <= B <= 4)  # key point!
          = P(B = 0) + P(B is in {1,2,3,4})
          = 5/6 + P(R = 3 and B is in {1,2,3,4})
          = 5/6 + P(R = 3) P(B is in {1,2,3,4} | R = 3)
          = 5/6 + (1/6) (4/6)
          = 34/36
          = 17/18

So, our (wrong) answer would be

(1/6) (17/18) = 17/108, about 0.157

7.  same as 6, not counted

8.  (2/6) 12 + (4/6) 1, about 4.67

9.  Let S denote the square we are on after the first turn.  Its support
is 0,1,2,4,5,6,7, with associated probabilities

a1 = (1/6) * (1/6)
a2 = 1/6 + (1/6) * (1/6)
1/6
a2 = (1/6) + (1/6) * (1/6)
a2 = (1/6) + (1/6) * (1/6)
a2 = (1/6) + (1/6) * (1/6)
a1 = (1/6) * (1/6)

Var(S) = E(S^2) - (ES)^2
E(S^2) = 
a1 * 0^2 +
a2 * 1^2 +
(1/6) * 2^2 +
a2 * 4^2 +
a2 * 5^2 +
a2 * 6^2 +
a1 * 7^2

E(S) = 
a1 * 0 +
a2 * 1 +
(1/6) * 2 +
a2 * 4 +
a2 * 5 +
a2 * 6 +
a1 * 7

about 3.9


10.  

nochange <- function(c,r,k) (r^k)^(c-k)
nochange(4,0.2,3)  # about 0.008