Math 302 Solutions to assignment 6
1. We have that
a
cx4 =
1=
0
and
ca5
,
5
a
cx5 =
1 = E[X] =
0
ca6
,
6
so we deduce that a = 6/5 and c = 5/a5 . Hence
a
ca7
5a2
36
2
E[X ] =
cx6 =
=
=
,
7
7
35
0
so Var(X) = 1/35.
2. (a) This is the probability that a un
Math 302 Solutions to Assignment 5
1. (a) Write p for the number p = P (X = 1). Then EX = p and EX 2 = p so V ar(X) = pp2 .
The equality EX = 3V ar(X) then turns to be p = 3p 3p2 and so p = 2/3, hence
P (X = 0) = 1/3.
(b) Write n, p for the parameters of
Math 302 Assignment 8
1. For any a 0 we have
ay
FX/Y (a) = P (X/Y a) =
1 2 e
0
0
= 1
2 e(2 +1 a)y dy = 1
1 x 2 x
e
2
2 +1
=
1
2 +1
dxdy =
(1 e1 ay )2 e2 y
0
2
.
2 + 1 a
0
So P (X < Y ) = FX/Y (1) = 1
.
2.
P (|X Y | L/4) =
L/4 L
2
(L/2)
0
L/2
= 1/2 +
Math 302 Solutions to assignment 4
1. (a) We have that
P (X > m + n, X > m)
(1 p)m+n
=
= (1 p)n = P (X > n) .
P (X > m)
(1 p)m
(b) We have that
E[rX ] =
(1 p)k1 prk = rp
[r(1 p)]j =
j=0
k=1
rp
,
1 r(1 p)
where in the last equality we used the geometric su
Math 302 solutions to assignment 3
1. Let A be the event that the two dice land on dierent numbers and B the event that one
of the dice is 6. Then P (A) = 30/36 = 5/6 because there are 30 pairs of dierent numbers
in cfw_1, . . . , 6. Also, P (A B) = 10/36