Math 151 Midterm 1 Solutions
1. Suppose that 28 percent of the population drink tea, 35 percent drink coffee, 10 percent drink both. Find: (i) percentage of people who do not drink either. (ii) percentage of people who drink tea but not coffee Soluti
Math 151 Homework 9 Solutions (Winter 2014)
Problem 7.25
Observe that event cfw_N n is same as the event cfw_X1 X2 Xn1 . Since Xi are
independent fX1 ,.,Xn1 (x1 , . . . , xn1 ) = f (x1 ) f (xn1 ) where f is the density function of X1 .
So as in problem 6.
Math 151 Homework 8 Solutions (Winter 2014)
Problem 6.41
(a) First compute marginal densities fX (x) and fY (y). For x > 0
f (x, y)dy =
fX (x) =
0
xex(y+1) dy = ex(y+1)
0
= ex
and for y > 0
fY (y) =
xex(y+1) dx
f (x, y)dx =
0
ex(y+1)
=
x
y+1
+
0
0
x(y+1)
Math 151 Homework 7 Solutions (Winter 2014)
Problem 6.16
(a) A =
n
i=1
Ai
(b) Ai and Aj are disjoint because in Ai all points, in particular Pj , lie on the same semicircle
beginning from Pi and going clockwise. So assuming that Ai golds, Aj doesnt hold.
Math 151 Homework 2 Solutions (Winter 2014)
Problem 3.14
(a) Let A1 and A2 be the events that the rst and the second selected balls, respectively, are white.
Also let B1 and B2 be the events that the third and the fourth selected balls, respectively,
are
Math 151 Homework 5 Solutions (Winter 2014)
Problem 5.4
(a)
P (X > 20) =
f (x)dx =
20
20
10
dx =
x2
(b) If a 10 then
10
x
=
20
1
2
a
FX (a) = P (X a) =
0dx = 0
and if a > 10 then
a
a
f (x)dx =
FX (a) = P (X a) =
10
So
FX (a) =
0
10
dx =
x2
10
x
a
10
=1
10
Math 151 Homework 1 Solutions (Winter 2014)
Problem 2.2
Let (x1 , x2 , . . . , xk ) be the event that the ith roll has outcome xi for 1 i k. Then the event
that the game stops in ith roll is Ai = cfw_(x1 , . . . , xi1 , 6) : x1 , . . . , xi1 cfw_1, 2, 3,
Math 151 Homework 4 Solutions (Winter 2014)
Problem 4.62
()
Let Eij denote the event that trials i and j both have outcome and let Eij =
event that trials i and j have the same outcome. Then
be the
k
k
()
p2
P (Eij ) =
P (Eij ) =
()
k
=1 Eij
=1
=1
We know
Math 151 Homework 3 Solutions (Winter 2014)
Problem 4.6 For k = 0, 1, 2, 3 if there are k heads then there must be 3 k tails and in this
case X = k (3 k) = 2k 3. So X can get values 3, 1, 1, 3 and if H is the number of heads
then
3
0
1
2
3
P (X = 3) = P (
Math 151 HW 2 Solutions
Problems 4. Let A be the probability that at least on of a pair of dice lands on 6. Easy computaion gives: P (A|i = 2) = 0, P (A|i = 3) = 0, P (A|i = 4) = 0, P (A|i = 5) = 0, P (A|i = 6) = 0, P (A|i = 7) = 2/6, P (A|i = 8) = 2
Math 151 Homework 6 Solutions (Winter 2014)
Problem 5.41
Assume A positive. Then:
- if x < A then FR (x) = P (A sin x) = P (sin x/A) = 0 since x/A < 1 sin .
- if A x A then
arcsin(x/A) + /2
FR (x) = P (A sin x) = P (sin x/A) = P (/2 arcsin(x/A) =
- if A <