.4
Question 4. Given events E, F, G ✓ ⌦:
(i) Prove that (E C )C = E.
(ii) If F C and G are disjoint, prove that (E \ F )C \ G = E C \ G.
(iii) If E and F are elements of a -field, F, prove that E \ F 2 F.
Answer.
(Must take care with direction of implications)
(i) Definition of complement is: E C = {! 2 ⌦ : ! 2
/ E}. Also, E C and E partition ⌦.
Therefore,
(E C )C = {! 2 ⌦ : ! 2
/ E C } = {! 2 ⌦ : ! 2 E} = E.
Alternatively, show (E C )C ✓ E and E ✓ (E C )C .
Recall definition of subset is: A ✓ B i↵ ! 2 A ) ! 2 B.
We have,
! 2 (E C )C , ! 2
/ EC , ! 2 E
) (E C )C = E.
Alternatively, we know that if A \ B = and A [ B = ⌦ (i.e. A and B partition
⌦), then ! 2 B i↵ ! 2
/ A, so B = AC . Result follows by setting A = E C ad B = E
C
(as E and E partition ⌦). (3 marks).
(ii)
(E \ F )C \ G = (E C [ F C ) \ G
De Morgan
= (E C \ G) [ (F C \ G)
C
= (E \ G)
distributivity
as F C \ G = ,
as required. Proof by subset method also acceptable. (3 marks)
(iii) As F is a -field it is closed under union and complement, so we have that, for
E 2 F and F 2 F then E [ F 2 F; E C 2 F and F C 2 F.
Need to show that E \ F 2 F.
E C 2 F, F C 2 F ) (E C [ F C ) 2 F
) (E C [ F C )C 2 F
)E\F 2F
closed under union/complement
closed under complement
De Morgan,
as required. (4 marks) (2 marks for recalling
and complement)
-field closed under union
.4
Question 1.
(i) For events E, F and G ⊆ Ω, we are given the following information: E and F
partition G; E is four times more probable than F and F is twice as probable as G
not occurring. Determine P(G).
(ii) For events A, B and C ⊆ Ω, prove that
P(A ∩ B ∩ C C ) = P(A ∩ B) − P(A ∩ B ∩ C).
(iii) Out of three events, A, B and C ⊆ Ω, we know that the probability of only A
occurring is 0.1, the probability of only B occurring is 0.2; the probability of A and C
occurring is 0.2; the probability of B and C occurring is 0.3 and the probability of all
three occurring is 0.1. Writing out any probability statements carefully, determine
the probability of exactly one of either A or B occurring.
Answer.
(i) We are given:
P(E) = 4P(F );
P(F ) = 2P(GC ).
E and F partition G, so E ∪ F = G and E ∩ F = φ.
⇒ P(E) + P (F ) = P(G) ⇒ 5P (F ) = P(G) ⇒ 5(2P(GC )) = P(G)
10
⇒ 10(1 − P(G)) = P(G) ⇒ P(G) = .
11
(3 marks)
(ii)
A ∩ B = A ∩ B ∩ Ω = (A ∩ B) ∩ (C ∪ C C )
= (A ∩ B ∩ C) ∪ (A ∩ B ∩ C C )
⇒ P(A ∩ B) = P(A ∩ B ∩ C) + P(A ∩ B ∩ C C )
⇒ P(A ∩ B ∩ C C ) = P(A ∩ B) − P(A ∩ B ∩ C)
as required (as (A ∩ B ∩ C) and (A ∩ B ∩ C C ) are disjoint).
(2 marks)
(iii) Exactly one of either A or B is the event that A and B C occur or that B and AC
occur, i.e.
P((A ∩ B C ) ∪ (AC ∩ B)) = P(A ∩ B C ) + P(AC ∩ B)
P(A ∩ B C ) = P(A ∩ B C ∩ C C ) + P(A ∩ B C ∩ C)
P(AC ∩ B) = P(AC ∩ B ∩ C C ) + P(AC ∩ B ∩ C).
We are given:
P(A ∩ B C ∩ C C ) = 0.1;
P(A ∩ C) = 0.2;
P(AC ∩ B ∩ C C ) = 0.2
P(B ∩ C) = 0.3 P(A ∩ B ∩ C) = 0.1
⇒ P(A ∩ B C ∩ C) = P(A ∩ C) − P(A ∩ B ∩ C) = 0.1
⇒ P(AC ∩ B ∩ C) = P(B ∩ C) − P(A ∩ B ∩ C) = 0.2
⇒ P(A ∩ B C ) = 0.1 + 0.1 = 0.2
⇒ P(AC ∩ B) = 0.2 + 0.2 = 0.4.
So, the required probability is:
(5 marks)
P((A ∩ B C ) ∪ (AC ∩ B)) = 0.6.
.4
Question 2.
You have two coins which look identical. One is fair, and the other comes up Heads with
probability 0.6. Assume that successive flips are independent for both coins. You flip one of
the coins five times and obtain the sequence HHT T H. You flip the other coin three times
and obtain the sequence HT H. Which of the coins is more likely to be the biased one?
Answer.
(i) Let A1 be the event that you obtain the sequence HHT T H and A2 be the event that
you obtain the sequence HT H. Let B be the event that the coin is biased.
We need to compare P(B | A1 ) and P(B | A2 ).
From Bayes we have:
P(A1 | B)P(B)
P(B | A1 ) =
P(A1 )
We have
P(A1 | B) = (0.6)3 (0.4)2 ,
P(A1 | B C ) = (0.5)5 ,
P(B) = 0.5
and
P(A1 ) = P(A1 | B)P(B) + P(A1 | B C )P(B C )
= (0.6)3 (0.4)2 (0.5) + (0.5)5 (0.5)
= (0.6)3 (0.4)2 (0.5) + (0.5)6
Giving
(0.6)3 (0.4)2 (0.5)
(0.6)3 (0.4)2
=
(0.6)3 (0.4)2 (0.5) + (0.5)6
(0.6)3 (0.4)2 + (0.5)5
(4 marks) (2 marks for theorem of total prob, 2 for correct formulation of
Bayes theorem)
P(B | A1 ) =
P(A2 | B) = (0.6)2 (0.4),
P(A2 | B C ) = (0.5)3 ,
P(B) = 0.5
and
P(A2 ) = P(A2 | B)P(B) + P(A2 | B C )P(B C )
= (0.6)2 (0.4)(0.5) + (0.5)3 (0.5)
= (0.6)2 (0.4)(0.5) + (0.5)4
Giving
P(B | A2 ) =
(0.6)2 (0.4)(0.5)
(0.6)2 (0.4)
=
(0.6)2 (0.4)(0.5) + (0.5)4
(0.6)2 (0.4) + (0.5)3
(4 marks)
Note
P(B | A1 ) =
⇣
(0.24)(0.6)2 (0.4)
(0.24) (0.6)2 (0.4) +
(0.5)2
(0.5)3
0.24
⌘=
(0.6)2 (0.4)
< P(B | A2 )
3
(0.6)2 (0.4) + 0.25
(0.5)
0.24
So the coin whose outcome is HT H is more likely to be the biased one. (2 marks)
.4
Question 3.
Each week, the mark that you receive on your weekly test will be chosen independently at
random from the integers between 0 and 10 inclusive. After 10 tests, determine the following:
(i) The total number of possible score patterns (where week order matters).
(ii) The probability that you receive four marks of 4, three marks of 3, two marks of 2 and
one mark of 1.
(iii) The probability that your highest mark is less than 9.
You may leave your answers in terms of powers and/or factorials.
Answer.
(i) In each of the ten weeks you can receive any one of 11 marks ({0, 1, 2, . . . , 10}), so
n⌦ = 1110 . (2 marks)
(ii) Let E = event that you get four 4s, three 3s, two 2s and one 1. Then from combinatorial
partitioning we have n = 10, r1 = 1, r2 = 2, r3 = 3, r4 = 4 and the total number of
score patterns that satisfy event E is
nE =
10!
.
4!3!2!1!
As each score pattern occurs with equal probability, we have
P(E) =
nE
10!
= 10
.
n⌦
11 4!3!2!
(5 marks)
(iii) Let F = event that your highest mark is less than 9.
Then all of your marks must be chosen from {0, 1, 2, . . . , 8} and nF = 910 , and
nF
910
P(F ) =
= 10 =
n⌦
11
(3 marks)
✓
9
11
◆10
Question.Thenumberofmistakes,X,thatProfessorCarelessmakesduringapar-
ticularlecturehas pmf:
fX (x) =
e
20
20x
,
x!
x 2 {0, 1, 2, 3, . . .}.
The number of mistakes, Y , that Professor Nearly Perfect makes during a particular
lecture has pmf:
✓ ◆1 y ✓ ◆y
9
1
fY (y) =
, y 2 {0, 1}.
10
10
(i) Determine the probability generating function of X.
(ii) Determine the probability generating function of Y .
(iii) Given that X and Y are independent, determine the probability generating function
of the total number, Z, of mistakes made in both lectures.
(iv) Calculate P(Z = 0) directly and verify that GZ (0) = P(Z = 0).
.4
Answer.
(i)
GX (t) =
1
X
t
xe
x=0
(3 marks)
20
20x
=e
x!
(ii)
GY (t) =
1
X
y=0
(3 marks)
t
y
✓
20
1
X
(20t)x
x!
x=0
9
10
◆1
y
✓
1
10
= e20(t
◆y
=
1)
,
|t|  1.
9+t
.
10
(iii) Let Z = X + Y . As X and Y are independent
GZ (t) = GX (t)GY (t) =
e20(t
1)
(9 + t)
.
10
(2 marks)
(iv) From independence:
P(Z = 0) = P(X = 0 \ Y = 0) = P(X = 0)P(Y = 0) = fX (0)fY (0) = e
Also,
GZ (0) = e
20
9
,
10
as required. (2 marks)
Note that, in this case getting at fZ (z) is straightforward:
for z > 0,
1 \ Y = 1) [ (X = z \ Y = 0))
1
9
1)fY (1) + fX (z)fY (0) = fX (z 1) + fX (z).
10
10
fZ (z) = P((X = z
= fX (z
Could use this results to calculate GZ (t) directly.
20
9
.
10
Question 2. The continuous random variable X has the following pdf:
fX (x) = 2(1
x),
0 < x < 1.
(i) Find the moment generating function (mgf), MX (t), of X.
(ii) Show that the mgf can be expressed as
MX (t) =
1
X
cr t r ,
where you should determine the form of cr .
r=0
(iii) Hence determine EfX (X r ), r = 1, 2, 3, . . .
.4
Answer.
(i)
tX
Z
1
MX (t) = EfX (e ) =
etx 2(1 x) dx
0

Z 1 tx
tx 1
e
e
= 2(1 x)
+
2 dx
t 0
t
0
 tx 1
2
e
=
+ 2 2
t
t 0
t
2 2e
2
2(et t 1)
=
+ 2
=
.
t
t
t2
t2
(4 marks)
(ii)
t2 t3 t4
et = 1 + t + + + + . . .
✓✓ 2! 3!2 4!3
◆
2
t
t
t4
) MX (t) = 2
1 + t + + + + ...
t
2! 3! 4!
✓ 2
◆
3
2 t
t
t4
= 2
+ + + ...
t 2! 3! 4!
✓
◆
1
t
t2
=2
+ + + ...
2! 3! 4!
1
1
X
X
tr
2
=
2
=
tr .
(r + 2)!
(r + 2)!
r=0
r=0
As required with cr =
2
.
(r+2)!
(4 marks)
(iii) We know that
MX (t) = 1 +
Hence, for r = 1, 2, . . .:
1
X
tr
EfX (X r ) .
r!
r=1
EfX (X r )
2
=
r!
(r + 2)!
2
) EfX (X r ) =
.
(r + 2)(r + 1)
(r)
(2 marks) (could also determine form of MX (0).)
t
1
◆