Please note: these are only answers, not
solutions. You are welcome to disagree with everything that follows.

HW1.

1. 8 � 2 � 9 = 144 possible codes. 1 � 2 � 9 = 18 that started with a 4.
2. (a) 6! (b) 3!2!3! (c) 3!4! 3. 10 � 9 � 8 � � � 5 � 4 = 604,800
4. (a)There are 4 steps to the right and three up for the total of 7.
At each step you have a choice of going either up or to the right.
So each path consists in selecting the three up steps (or four right
steps) out of seven, for the total number (7 choose 4)=35.
(b) By a similar argument, there are 4!/(2!2!) paths from (0,0) to (2,2) and
 3!/(2!1!)
paths from  (2,2)  to (4,3). This gives 18 paths total.
(c) This is an example where the Catalan numbers C_n appear. The answer in this
case is C_4=14, and you can see all the paths at

 http://www.answers.com/topic/catalan-number-1

 where you can read more on the subject.
5. (a) 12
choose 3 (b) 15 choose 2 + 14 choose 2 + 13 choose 2 + 13 choose
2 + 12 choose 3 = 552.
6. EF = {(1, 2), (1, 4), (1, 6), (2, 1), (4, 1),
(6, 1)}.
 E or F occurs if the sum is odd or if at least one of the dice lands on 1.
 FG = {(1, 4), (4, 1)}.
EF^c is the event that neither of the dice lands on 1 and the sum is odd. EFG = FG.

HW2.

1. (a) 0.4 (b) 0.1
2. (a) 0.5 (b) 0.32 (for example, by Venn diagram) (c) (50 choose 2) / (100 choose 2)
3. 5/12 (for example, from equation 1=2x+(1/6) )
4. (a) 2(N-1)!/N!=2/N (b) 2/(N-1)
5. 12/35 (for example, by inclusion-exclusion)
6 Of course, i should go from 7 to 12. The corresponding probabilities are
1/6, 1/5, 1/4, 1/3, 1/2, 1.
7. (a) No, otherwise, P(A or B)=1.1; (b) 1.1-0.3=0.8.

HW3.

1. (a) 0.36*0.22 (b) 0.36*0.22/0.3
2. (a) 6*5*4/(6^3) (b) 1/3! (c) the product of the probabilities in (a) and (b)
3. 0.05p/(0.05p+0.0025(1-p)), where p is the proportion of males
4. (1/2)*0.4/((1/2)*0.4+(2/5)*0.6)
5. (a) p_A(1-p_B)/(1-(1-p_A)(1-p_B)) (b)  p_Ap_B/(1-(1-p_A)(1-p_B))
      (c), (d), (e)  ((1-p_A)(1-p_B))^{n-1}(1-(1-p_A)(1-p_B)) (f) 0
6. (a) 2p^2+2p^3-5p^4+2p^5 (inclusion/exclusion).
 (b) (4p-4p^2+p^3)/(2+2p-5p^2+2p^3).

HW4.

1. X takes values from 1 to 6 with probabilities 1/2, 5/18, 5/36, 5/84, 5/252, 1/252
  X cannot be bigger than 6.
2. n - 2i, i = 0, 1, �, n
3. (a) P(X = 3} = 1/8, P{X = 1} = 3/8, P{X = -1} = 3/8, P{X = -3} = 1/8 (b) you can do it.
4. (b) P{x > 0} = P{win first bet} + P{lose, win, win}
= 18/38 + (20/38)(18/38)2, which is approximately  .5918
(c) E[X] = [18/38 + (20/38)(18/38)^2] - [(20/38)2(20/38)(18/38)] - 3(20/38)^3,
which is approximately  -0.108. One can argue that this is a "win-win" situation: the house
wins on average, and the gambler wins with probability bigger than 1/2.
5. (a) 2 + 2p(1 - p) (b) 6p^4 - 12p^3 + 3p^2 + 3p + 3 (p is probability of win for the first team)
6. (a) q=(5 choose 3)(0.2)^3(0.8)^2+(5 choose 4)(0.2)^4(0.8)+(0.2)^5=0.0579 (b) 1-(1-q)^3.
7. Formulas are long, the ideas are simple.
P(correct)=P(convict|guilty)P(guilty)+P(acquit|innocent)P(innocent);
P(convict)=P(convict|guilty)P(guilty)+P(convict|innocent)P(innocent).
Conditional probability are sums of obvious binomial probabilities, and P(guilty)=0.65.
One thing to keep in mind: you need 9 or 10 or 11 or 12 ``guilty'' votes for conviction;
 you need 4 or 5 or ... 12 ``innocent'' votes for acquittal.
 I never had the patience to work out the numbers...


HW5.

1. (a) 1 - exp(-80000/365^2) (b) 1 - exp(-80000/365)
2.(a) exp(-L), L=2.2 (b) 1-\sum_{i=0}^4 exp(-L) L^i/i!, L=2.2*k, k=28/7 or 30/7 or 31/7 (depending on the month)
3. exp(-3)(9/2)(3/4)/(exp(-3)(9/2)(3/4)+exp(-5)(25/2)(1/4))
4. exp(-lambda t) + (1-exp(-lambda t))p, p is the probability of heads.
5. (18/35) (17/35)^(n-1).
6. F((x- beta)/ alpha), if alpha > 0;
1-F(((x- beta)/ alpha)-), if alpha < 0 (here you are taking the limit from the left)

HW6.

1. (a) No: not increasing (b) could be if C=3/4.
2. 2/(1+a)
3. F(0.8333); 2F(1)-1; 1 - F(.3333); F(1.6667); 1-F(1) (F - standard normal distrib.)
4. 0.2=P(Z>4/sigma); sigma = 4.76
5. (a) P{Z > - .3.25} (b) 2P{Z < 1.15} - 1 (c) P{Z < 9.5/4.77}
6. (a) 1/e (b) 1/3
7. e^y exp(- e^(y))
8. 1/y, 1<y<e.

HW7.

1. Too much to write (but you can do it)
2. (a) 1/8 (b) f_X(x)=(1/4)(1+|x|)e^(-|x|), f_Y(y)=(1/6)y^3e^(-y) (c) E(X)=0.
3. 1/6; 1/2
4. (a) A is the union of A_i (b) A_i and A_j are mutually exclusive (c) P(A)=n2^(-n+1)
because P(A_i)=2^(-n+1) for all i.
5. 7/9
6. (a) yes (b) no
7. (a) P{Z > .42} (b) P{Z > .82} continuity correction is used because X and Y are
clearly discrete. Thus, for example,  {Y>X}={Y-X>=1}={Y-X>0.5}.

HW8.

1. the density is (3/4)(1/x^3)(x^2-y^2)
2. (1-(2d/L))^3 (see example 6a)
3. in polar coordinates,  f(r,\theta)=r/\pi, 0<r<1, 0<\theta<\pi
4. you know the answer
5. just compute the Jacobians

HW9.

1. 105/24
2. 35
4. (a) 9/10 (b) 49/10 (c) 42/10
5. 1 (do not put an upper bound on age!)

HW10.

1. 365 (100 choose 3) 365^{-3} (364/365)^{97}
2. 6(1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6))
3. P{N >= n}= P{X1 >= X2 >= �... >= Xn} = 1/n!, EN=e.
4. (a) 1/2 (b) 0
5. (a)  binomial with parameters n - 1 and p. (b) 1/(n-1)
6. (a) 6 (b) 7 (c) 1(1/5)+2(4/25)+3(16/125)+4(64/625)+11(256/625)

HW11.

1. 2y^2 (given Y = y, X is exponential with mean y)
2. 12, by conditioning on the initial choice of the door.
3. N(1-exp(-10/N))
4. 5*2.5=12.5
5. (a) p mu1+(1-p) mu2 (b) p(sigma1^2+mu1^2)+(1-p)(sigma2^2+mu2^2)-
               (p mu1+(1-p) mu2)^2
6. (a) 0.6e^{-2}+0.4e^{-3} (b) 0.6(8/6)e^{-2}+0.4(27/6)e^{-3}
     (c) (0.6(8/6)e^{-4}+0.4(27/6)e^{-6})/(0.6e^{-2}+0.4e^{-3})

HW12.

1. (a) 20/15 (b) P{Z > -1.006}
2. P(Z< 1.58)
3. 117 or more
4. P(Z>1.58)

HW13.

1. (a) 1/9 (b) 5/9
2. (1+(s/20))e^{-s/20}
3. you have the answer
4. (b) 1/6

HW14.

1. X=F^{-1}(U)=... (write the cases)
2. X=(-(1/a) ln U)^{1/beta}
3. X=(-(1/a) ln U)^{1/beta} with a= c/(n+1) and beta=n+1; n is the power of t.
4. (a) U^{1/n} (c) max(U1,...,Un)