The following are what I think are the answers to the practice test.

1a. 52 choose 13

1b.  (26 choose 8) * (26 choose 5) / (52 choose 13)

1c. Look at complements. Answer 1-x where x=44!*39!/31!*52!

2. x = # red balls in urn. There are 5 choose 2 blue draws.
There are 5+3+x choose 2 red draws. Solve 10/(8+x choose 2) = 2/9, get x=2.

3. 28

4. 1- (1-0.3)(1-0.5)(1-0.25)

5. F is (1+x)^2/2 between -1 and 0, and 1/2 + (1-x)^2/2 between 0 and
1.
E(X) is 0. V(X)=E( (X-E)^2) = E(X^2) is 2* the integral from 0 to 1 of
(1-x)*x^2, which is 2*(1/3-1/4).. 

6. P(H)=7/15 with Bayes. Then P(I|H)=5/14

7. k is in 1..50 and none of 1..k-1 was chosen. 
By product rule, P(A_k)=(100-k choose 49)/(100 choose 50).

8. a. Geometric series. a=2/3.

b. P(N<12|N>9) = (P(N=10)+P(N=11))/1-P(N=9) = 8/9.

9. f_X(t)=e^{-t/4}. This measures "breaks", not "lives".
So, integrate between 0
and 3 and subtract result from 1, for part 1. For part 2,
P(2 more| already 3) = P(at least 5)/P(at least
3)=[1-integrate f_X(0..5)]/[1- integrate f_X(0..3)].  

10. f_X(t)=f_Y(t)=e^{-t}. P(Z<a)=P(X<aY). If X=alpha, then
P(X<ay)=P(alpha/a<Y)= integral f_Y(alpha/a..infinity)=e^{-alpha/a}. 
Since P(X=alpha) is f_X(alpha), we need to look at the integral
over e^{-t}e^{-t/a} from 0 to infinity. That gives F_Z. Then
differentiate to get f_Z.

11. Let a, b be the two arrival times. Draw a picture in the plane and
look at the area that counts the good cases. Divide by total area.

12. We don't know that yet.

13. A=number of aces in first 1000, B=number of aces in second 1000.

A is Bin(1000, 4/52). B is Bin(1000, 3/51).
We want P(A+B>139). Approximate A and B by normals. Then we have a
rule how to add two normals to make another normal from them. Use this
new normal to estimate P(A+B>139) by de Moivre Laplace.

14. P(W>a)=P(Y^3>a)=P(Y>b) =integral of e^{-t} from 0 to b where b^3=a
. Calculatethe integral and then take derivative to get f_W.