Exercise 11 :
A child collect soccer player photos, that he nds in cereal boxes. Each cereal
box contain one photo. Each photo is equally likely to be found in a cereal box. The
complete collection consists in n different photos. Let Xk be the number of c

ball, . . . , nth ball.
(a) Compute the generating function of Xi for i 2 [|1, n|].
(1 + t )2
.
22
(b) Let S be the total score that we get after picking the n ball. Find the generating
function of S and its distribution.
(1 + t)2n
1
Answer :
, B(2n, ).
2

Fall 2015 Math 130A
Review problems
Exercise 1 : A pharmaceutical company decides to cut its advertising budget by only
sending three fifth of its mail as priority mail and the other two fifth as standard
mail.
(a) Four letters are sent to four different

Exercise 13 :
A plane has three engines : one central and one on each wing. Each engine can
malfunction independently of the others. The central engine has a probability p to
malfunction and each wing engine has a probability q to malfunction. The plane c

Fall 2015 Math 130A
Review problems
Exercise 1 : Let X be an uniform random variable over [ a, b] with 0 < a < b. Let
Y = X2.
(a) Compute E(Y ) and Var(Y ).
Answer :
b3
3( b
a3 4( b a )4
,
.
a)
45
(b) Give the cumulative distribution function and the prob

2
n
n!
Answer : Let n = 2k.
.
k
2k
Exercise 17 :
In a calling center, a client has a probability 0.25 to wait until someone picks up.
(a) A client calls 4 times. Let X be the number of calls where he had to wait
before talking to someone. What distributi

X
The probability that the satellite system will function corresponds to the event
k. We write the law of total probability
n
n
n
P( X k) = P( X = i) =
P( X = i | R) P( R) + P( X = i | Rc ) P( Rc )
i
i=k
i=k
n
n i
n i
n i
=
p1 (1 p1 ) a +
p (1 p2 ) n

Let X be the number of Heads that we obtain when we flip a coin n times. We
have X B(n, p). The probability to get an even number of Heads is then
b n2 c
b n2 c
n 2i
P(X is even) = P( X = 2i ) =
p (1 p)n 2i .
2i
i =0
i =0
We use the previous formula wit

We consider the sum
n
k p k (1
k =0
n
k p k (1
k =0
p)n
p)n
k k
x . By the binomial theorem we have
k k
p)n .
x = ( xp + 1
We integrate both sides to get
n
k p k (1
k =0
p)n
k
x k +1
( xp + 1 p)n+1
=
+ C.
k+1
p ( n + 1)
To obtain C we evaluate th

Fall 2015 Math 130A
Homework solutions
Due Date
Sections
Problems
Thu, Nov. 12
Chapter 4 - Random Variable
71 , 72 , 78, 79
Theoretical : 25, 30 , 35
Exercise 71 :
(a) Let X be the number of bets that he wins. X follows a binomial distribution
6
with para

We make the change of variable l = k + 1 in the last sum to obtain
k
n =
k=n
N
N +1
l = n +1
1
l
n
Therefore we get
E (Y ) =
N +1
=
l = n +1
+1
n( N
n +1 )
( Nn )
=
l 1
n+1 1
=
N+1
.
n+1
n ( N + 1)
.
n+1
Exercise 35 : (a) Let Rk be the event that the k

Theoretical exercises
Exercise 25 : Let Y be the number of events that occur. Let X be the number of events
that are counted. Let n 2 N. Given that Y = n, X follows a binomial distribution
with parameters n and p. Thus we have for all k 2 [|0, n|]
n k
P

We now suppose that a team wins the series if it wins 2 games. X follows a negative binomial distribution with parameter 2 and 0.6. The probability mass function
of X is given by
i 1
P( X = i) =
0.62 0.4i 2 = (i 1)0.62 0.4i 2 .
1
Thus we have
2
P( X = 2)

Theoretical exercises
Exercise 1 : We verify that f is a density function. We perform an integration by
parts and get
Z
f ( x ) dx =
=
Let I =
I2
R +
0
=
=
e
Z
0
bx2
0
h
bx2
ax2 e
a
xe
2b
dx. We have
Z +
+
bx2
e
dx
e
bx2
0
0
r
p
and
4b
e
Z
We have then

Let Y be the number of devices among 6 of them that work more than 15 hours. If
we suppose that each device work independently of the others, Y B(6, p). We
thus get
P (Y
3) =
3 3 4 2 5 1 6
6
2
1
6
2
1
6
2
1
2
+
+
+
.
3
3
3
4
3
3
5
3
3
3
Exercise 6 : (a)

Fall 2015 Math 130A
Homework solutions
Due Date
Sections
Problems
Thu, Nov. 24
Chapter 5 - Continuous RV
1, 4 , 6 , 7
Theoretical : 1, 8
Exercise 1 : (a) Since f is a probability density function, we have
Moreover we have
Z +
3
Thus c = .
4
f ( x )dx = c

Answer : E(Yn ) = 1 +
1
n
(0.85)n , n ln(0.85) >
ln n.
Exercise 4 : In a company, 60% of the objects are produced by a machine M1 and
the others by a machine M2. An object produced by M1 (resp. M2) is defective with
probability 0.1 (resp. 0.2).
(a) We ran

Exercise 6 :
An urn contains n balls with the numbers 1 to n. We pick two balls without
replacement. Let X be the largest number of the two balls that we pick. For all k n
give P( X k) and P( X = k ).
Answer : P( X k ) =
k(k
n(n
1)
2( k
, P( X = k) =
1)
n

(a) We pick without replacement six balls out of the urn. Let R and G be respectively the number of red balls and green balls that we pick. What distributions do R
and V follow? Give their expected values and variances.
2
6
Answer : R H(15, 6, ), E( R) =

(b) Find the probability density function of X.
8
> 0
>
for x 0
<
Answer : f X ( x ) =
.
1
> p e ln2 x/2 for x > 0
>
:
x 2p
(c) Compute E( X ).
p
Answer : e.
Exercise 4 : Let f : R ! R dened by
f (x) =
e x
,
(1 + e x )2
for all x 2 R.
(a) Show that f is a

Answer : n
n
1
.
k
k =1
Exercise 20 :
Cows have a certain disease D with probability p = 0.15. A farmer owns n cows.
To nd out if they have the disease D, the farmer has two possibilities:
Either he tests the milk of each cow,
Or he rst tests a mix of t

Fall 2015 Math 130A
Homework solutions
Due Date
Sections
Problems
Thu, Oct 29.
Chapter 4 - Random Variable
1 , 13 , 17, 18, 20 , 22, 25 , 27, 29
Theoretical : 2 , 3
Exercise 1 : The possible values of X : W ! F are
F = cfw_ 2,
X=
2 corresponds to the prob

Exercise 20 : Let X denote your winnings when you quit. X : W ! F has the
following possible values
F = cfw_ 3, 1, 1.
X > 0 corresponds to X = 1. There are two possibilities for that. Either you win the
rst bet or lose it and win the two next bets. Thus w

X = 500 corresponds to the probability to sell 1 standard encyclopedia. Thus
P( X = 500) = 0.3 0.5 (1
0.6) + (1
0.3) 0.6 0.5 = 0.27.
X = 1000 corresponds to the probability to sell 2 standard encyclopedias or 1
deluxe. Thus
P( X = 1000) = 0.3 0.5 0.6 0.5

(b) We rst check the cause 2, and then we check the cause 1. The possible values
for X are
cfw_ R2 + C2 , C2 + C1 + R1 .
X = C2 + R2 corresponds to the event that the second cause is the one responsible for the breakdown.
P( X = C2 + R2 ) = 1 p.
X = C2 +

Thus
E ( X ) = 3( p3 + (1
= 3(2p4
p)3 ) + 12( p(1
p )3 + (1
p) p3 ) + 30( p2 (1
p )3 + (1
p )2 p3 )
4p3 + p2 + p + 1).
Exercise 25 : Let X be the total number of heads. (a) X = 0 corresponds to both
coins landing on tails.
P ( X = 0) = (1
0.6)(1
0.7) = 0.

Fall 2015 Math 130A
Homework solutions
Due Date
Sections
Problems
Thu, Nov. 5
Chapter 4 - Random Variable
33, 35 , 38, 39, 42, 44 , 49 , 57 , 60
Theoretical : 6, 10 , 15, 17
Exercise 33 : Let N be the number of purchased newspaper. Let S be the number of

Moreover E( X ) = 1 and E( X 2 ) = Var( X ) + E( X )2 = 6. Thus we get
E(2 + X )2 ) = 14.
(b) We have
Var(4 + 3X ) = 9Var( X ) = 45.
Exercise 39 : Let X be the number of black balls in the rst four balls drawn. When
1
we pick one ball from the urn, the pr

Exercise 57 : (a) Let X be the number of accidents that occurred today. X follows a
Poisson distribution with parameter 3. Thus we have
P( X
3) = 1
P ( X 2) = 1
=1
3 (1 + 3 +
e
P ( X = 0)
9
)=1
2
17
e
2
P ( X = 1)
3
P ( X = 2)
.
(b) We use the conditional

MATH 130A (Fall 2016)
Worksheet 2016.M.1a.s
Worksheet Solutions
Problem 1. Given a finite nonempty set X, prove that the number of subsets of X
with an even number of elements is equal to the number of subsets with an odd number
of elements.
(Hint: you ca