ECE302 – Probability and
Applications
Tutorial #3
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Reza Rafie
[email protected]
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Administrivia
Tutorial website!
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Administrivia
Tutorial website!
www.comm.utoronto.ca/~rrafie/ece302.html
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Question 2.62
A die is tossed twice and the number of dots
facing up is counted and noted in the order of
occurrence. Let A be the event ``number of dots
in first toss is not less than number of dots in
second toss,'' and let B be the event ``number
of dots in first toss is 6.''
Find P[A|B] and P[B|A].
2
.
Question 2.62
The conditional probability is defined as
P[A|B] =
P[A ∩ B]
P[B]
3
.
Question 2.62
The conditional probability is defined as
P[A|B] =
P[A ∩ B]
P[B]
B = {61, 62, . . . , 66} = A ∩ B
3
.
Question 2.62
The conditional probability is defined as
P[A|B] =
P[A ∩ B]
P[B]
B = {61, 62, . . . , 66} = A ∩ B
P[A|B] = 1
3
.
Question 2.62
The conditional probability is defined as
P[B|A] =
P[A ∩ B]
P[A]
4
.
Question 2.62
The conditional probability is defined as
P[B|A] =
P[A ∩ B]
P[A]
A ∩ B = {61, 62, . . . , 66}
4
.
Question 2.62
The conditional probability is defined as
P[B|A] =
P[A ∩ B]
P[A]
A ∩ B = {61, 62, . . . , 66}
A = {11, 21, 22, 31, . . . , 66} =⇒ |A| =
4
.
Question 2.62
The conditional probability is defined as
P[B|A] =
P[A ∩ B]
P[A]
A ∩ B = {61, 62, . . . , 66}
A = {11, 21, 22, 31, . . . , 66} =⇒ |A| = 21
4
.
Question 2.62
The conditional probability is defined as
P[B|A] =
P[A ∩ B]
P[A]
A ∩ B = {61, 62, . . . , 66}
A = {11, 21, 22, 31, . . . , 66} =⇒ |A| = 21
P[B|A] =
6
2
=
21
7
4
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Question 2.69
A number x is selected at random in the
interval [−1, 2]. Let the events A = {x < 0},
B{|x − 0.5| < 0.5}, and C = {x > 0.75}. Find
P[A|B], P[B|C], P[A|Cc ], P[B|Cc ].
5
.
Question 2.69
S = [−1, 2], |S| = 3
A = {x < 0}
B{|x − 0.5| < 0.5}
C = {x > 0.75}
6
Question 2.69
−∞
0
.
+∞
Question 2.69
−∞
S
−1
0
.
2
+∞
Question 2.69
−∞
S
A
−1
0
.
2
+∞
Question 2.69
−∞
S
A
B
−1
0
2
.
1
+∞
.
Question 2.69
−∞
S
A
B
−1
0
2
+∞
.
1
0.75
C
7
.
Question 2.69
0
A
.
0
1
B
P[A|B] =
P[A∩B]
P[B]
=
8
.
Question 2.69
0
A
.
0
1
B
P[A|B] =
P[A∩B]
P[B]
=
P(∅)
8
.
Question 2.69
0
A
.
0
1
B
P[A|B] =
P[A∩B]
P[B]
=
P(∅)
P(B)
8
.
Question 2.69
0
A
.
0
1
B
P[A|B] =
P[A∩B]
P[B]
=
P(∅)
P(B)
=0
8
.
Question 2.69
.
0
B
1
0.75
2
C
P[B|C] =
P[B∩C]
P[C]
=
9
.
Question 2.69
.
0
B
1
0.75
2
C
P[B|C] =
P[B∩C]
P[C]
=
P((0.75,1))
9
.
Question 2.69
.
0
B
1
0.75
2
C
P[B|C] =
P[B∩C]
P[C]
=
P((0.75,1))
P((0.75,2])
9
.
Question 2.69
.
0
B
1
0.75
2
C
P[B|C] =
P[B∩C]
P[C]
=
P((0.75,1))
P((0.75,2])
=
0.25
1.25
=
1
5
9
.
Question 2.69
.
0
B
C
1
0.75
−1
2
0.75
Cc
P[B|Cc ] =
P[B∩Cc ]
P[Cc ]
=
10
.
Question 2.69
.
0
B
C
1
0.75
−1
2
0.75
Cc
P[B|Cc ] =
P[B∩Cc ]
P[Cc ]
=
P((0,0.75])
10
.
Question 2.69
.
0
B
C
1
0.75
−1
2
0.75
Cc
P[B|Cc ] =
P[B∩Cc ]
P[Cc ]
=
P((0,0.75])
P([−1,0.75])
10
.
Question 2.69
.
0
B
C
1
0.75
−1
2
0.75
Cc
P[B|Cc ] =
P[B∩Cc ]
P[Cc ]
=
P((0,0.75])
P([−1,0.75])
=
0.75
1.75
=
3
7
10
.
Question 2.69
−1
A
−1
.
0
0.75
Cc
P[A|Cc ] =
P[A∩Cc ]
P[Cc ]
=
11
.
Question 2.69
−1
A
−1
.
0
0.75
Cc
P[A|Cc ] =
P[A∩Cc ]
P[Cc ]
=
P([−1,0))
11
.
Question 2.69
−1
A
−1
.
0
0.75
Cc
P[A|Cc ] =
P[A∩Cc ]
P[Cc ]
=
P([−1,0))
P([−1,0.75])
11
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Question 2.69
−1
A
−1
.
0
0.75
Cc
P[A|Cc ] =
P[A∩Cc ]
P[Cc ]
=
P([−1,0))
P([−1,0.75])
=
1
1.75
=
4
7
11
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Question 2.77
A nonsymmetric binary communications
channel is shown in the figure below. Assume
the input is 0 with probability p and 1 with
probability 1 − p.
(a) Find the probability that the output is 0.
1 − ϵ1
0
0
ϵ1
ϵ2
1
.
1 − ϵ2
1
12
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Question 2.77
Input: X
Output: Y
13
.
Question 2.77
0
1 − ϵ1
ϵ1
ϵ2
1
0
.
1 − ϵ2
1
P[{Y = 0}] = P[{Y = 0} ∩ S]
= P[{Y = 0} ∩ ({X = 0} ∪ {X = 1})]
= P [({Y = 0} ∩ {X = 0}) ∪ ({Y = 0} ∩ {X = 1})]
= P [({Y = 0} ∩ {X = 0})] + P [({Y = 0} ∩ {X = 1})]
14
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Question 2.77
0
1 − ϵ1
ϵ1
ϵ2
1
0
.
1 − ϵ2
1
P [({Y = 0} ∩ {X = 0})] = P [({Y = 0}|{X = 0})] × P [{X = 0}]
= (1 − ϵ1 ) × p
15
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Question 2.77
0
1 − ϵ1
ϵ1
ϵ2
1
0
.
1 − ϵ2
1
P [({Y = 0} ∩ {X = 0})] = P [({Y = 0}|{X = 0})] × P [{X = 0}]
= (1 − ϵ1 ) × p
P [({Y = 0} ∩ {x = 1})] = P [({Y = 0}|{X = 1})] × P [{X = 1}]
= ϵ2 × (1 − p)
15
.
Question 2.77
0
1 − ϵ1
ϵ1
ϵ2
1
0
.
1 − ϵ2
1
P [({Y = 0} ∩ {X = 0})] = P [({Y = 0}|{X = 0})] × P [{X = 0}]
= (1 − ϵ1 ) × p
P [({Y = 0} ∩ {x = 1})] = P [({Y = 0}|{X = 1})] × P [{X = 1}]
= ϵ2 × (1 − p)
P [{Y = 0}] = (1 − ϵ1 ) × p + ϵ2 × (1 − p)
15
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Question 2.77
Remark:
To evaluate P(Y), one can partition S into
disjoint subsets X1 , X2 , . . . , Xn and use the law
of total probability:
P(Y) =
i=n
∑
P(Y|Xi ) × P(Xi )
i=1
16
.
Question 2.77
(b) Find the probability that the input was 0
given that the output is 1. Find the probability
that the input is 1 given that the output is 1.
Which input is more probable?
17
.
Question 2.77
Probability that the input was 0 given that the
output is 1 = P[X = 0|Y = 1]
18
.
Question 2.77
Probability that the input was 0 given that the
output is 1 = P[X = 0|Y = 1]
P[X = 0|Y = 1] =
P({X = 0} ∩ {Y = 1})
P(Y = 1)
(1)
18
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Question 2.77
We have P(Y|X) and P(Y = 0), so we want to
write everything in terms of P(Y|X) and P(Y = 0)
19
.
Question 2.77
We have P(Y|X) and P(Y = 0), so we want to
write everything in terms of P(Y|X) and P(Y = 0)
Denominator:
P(Y = 1) = 1 − P(Y = 0)
19
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Question 2.77
We have P(Y|X) and P(Y = 0), so we want to
write everything in terms of P(Y|X) and P(Y = 0)
Denominator:
P(Y = 1) = 1 − P(Y = 0) = 1 − (1 − ϵ1 ) × p − ϵ2 × (1 − p)
19
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Question 2.77
We have P(Y|X) and P(Y = 0), so we want to
write everything in terms of P(Y|X) and P(Y = 0)
Denominator:
P(Y = 1) = 1 − P(Y = 0) = 1 − (1 − ϵ1 ) × p − ϵ2 × (1 − p)
Numerator:
By definition of conditional probability,
P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0}) × P(X = 0)
19
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Question 2.77
We have P(Y|X) and P(Y = 0), so we want to
write everything in terms of P(Y|X) and P(Y = 0)
Denominator:
P(Y = 1) = 1 − P(Y = 0) = 1 − (1 − ϵ1 ) × p − ϵ2 × (1 − p)
Numerator:
By definition of conditional probability,
P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0}) × P(X = 0) =
ϵ1 × p
19
.
Question 2.77
Probability that the input was 0 given that the
output is 1 = P[X = 0|Y = 1]
20
.
Question 2.77
Probability that the input was 0 given that the
output is 1 = P[X = 0|Y = 1]
P({X = 0} ∩ {Y = 1})
(2)
P(Y = 1)
ϵ1 × p
=
1 − (1 − ϵ1 ) × p − ϵ2 × (1 − p)
(3)
P[X = 0|Y = 1] =
20
.
Question 2.77
Similarly, probability that the input was 1 given
that the output is 1 = P[X = 1|Y = 1]
21
.
Question 2.77
Similarly, probability that the input was 1 given
that the output is 1 = P[X = 1|Y = 1]
P({X = 1} ∩ {Y = 1})
(4)
P(Y = 1)
(1 − ϵ2 ) × (1 − p)
=
1 − (1 − ϵ1 ) × p − ϵ2 × (1 − p)
(5)
P[X = 1|Y = 1] =
21
.
Question 2.77
Which input is more probable?
22
.
Question 2.77
Which input is more probable? Same
denominators, only need to compare
numerators
22
.
Question 2.77
Which input is more probable? Same
denominators, only need to compare
numerators
?
ϵ1 × p ⋛ (1 − ϵ2 ) × (1 − p)
22
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Question 2.80
A computer manufacturer uses chips from
three sources. Chips from sources A, B, and C
are defective with probabilities 0.005, 0.001, and
0.01, respectively. If a randomly selected chip is
found to be defective, find the probability that
the manufacturer was A; that the manufacturer
was C. Assume that the proportions of chips
from A, B, and C are 0.5, 0.1, and 0.4, respectively.
23
.
Question 2.80
What do we know?
24
.
Question 2.80
What do we know?
P(A) = 0.5
P(B) = 0.1
P(C) = 0.4
24
.
Question 2.80
What do we know?
P(A) = 0.5
P(B) = 0.1
P(C) = 0.4
P(Defective|A) = 0.005
24
.
Question 2.80
What do we know?
P(A) = 0.5
P(B) = 0.1
P(C) = 0.4
P(Defective|A) = 0.005
P(Defective|B) = 0.001
P(Defective|C) = 0.01
24
.
Question 2.80
Remark:
Bayes' rule is a way to relate P(X|Y) to P(Y|X):
P(X|Y) =
P(Y|X)P(X)
P(Y)
Most of the time, the law of total probability is
used to find the denominator in Bayes'
formula.
25
.
Question 2.80
P(A|Defective) =
26
.
Question 2.80
P(A|Defective) =
P(Defective|A)P(A)
P(Defective)
26
.
Question 2.80
P(A|Defective) =
Denominator:
P(Defective|A)P(A)
P(Defective)
P(Defective) =P(Defective|A)P(A)
+ P(Defective|B)P(B)
+ P(Defective|C)P(C)
P(Defective) = 0.005×0.5+0.001×0.1+0.01×0.4 = 0.0066
26
.
Question 2.80
P(A|Defective) =
0.005 × 0.5
25
=
0.0066
66
P(Defective|A)P(A)
P(Defective)
27
.
Question 2.80
Similarly,
P(C|Defective) =
0.01 × 0.4 40
=
0.0066
66
P(Defective|C)P(C)
P(Defective)
28
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Question 2.81
A ternary communication system is shown on
the blackboard .
Suppose that input symbols 0, 1, and 2 occur
with probability 1/3 respectively. (a) Find the
probabilities of the output symbols. (b)
Suppose that a 1 is observed at the output.
What is the probability that the input was 0? 1?
2?
29
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Question 2.82
Let S = {1, 2, 3, 4} and A = {1, 2}, A = {1, 3},
A = {1, 4}. Assume the outcomes are
equiprobable. Are A, B, and C independent
events?
30
.
Question 2.82
A, B, C are independent if
P(A ∩ B) = P(A)P(B)
P(B ∩ C) = P(B)P(C)
P(C ∩ A) = P(C)P(A)
P(A ∩ B ∩ C) = P(A)P(B)P(C)
31
.
Question 2.82
A and B are independent.
32
.
Question 2.82
A and B are independent.
B and C are independent.
C and A are independent.
32
.
Question 2.82
A and B are independent.
B and C are independent.
C and A are independent.
But A, B and C are not independent!
32
.
Question 2.87
Let A, B, and C be events with probabilities P[A],
P[B], and P[C].
(a) Find P[A ∪ B] if A and B are independent.
33
.
Question 2.87
P[A ∪ B] = P[A] + P[B] − P[A ∩ B]
34
.
Question 2.87
P[A ∪ B] = P[A] + P[B] − P[A ∩ B] =
P[A] + P[B] − P[A]P[B]
34
.
Question 2.87
(b) Find P[A ∪ B] if A and B are mutually
exclusive.
35
.
Question 2.87
P[A ∪ B] = P[A] + P[B] − P[A ∩ B]
36
.
Question 2.87
P[A ∪ B] = P[A] + P[B] − P[A ∩ B] = P[A] + P[B]
36
.
Question 2.87
(c) Find P[A ∪ B ∪ C] if A, B, and C are
independent.
37
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
38
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
First term:
P[A ∪ B] = P[A] + P[B] − P[A ∩ B]
38
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
First term:
P[A ∪ B] = P[A] + P[B] − P[A ∩ B] =
P[A] + P[B] − P[A]P[B]
38
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
39
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
Third term:
P[(A ∪ B) ∩ C] = P[(A ∩ C) ∪ (B ∩ C)]
39
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
Third term:
P[(A ∪ B) ∩ C] = P[(A ∩ C) ∪ (B ∩ C)] =
P[(A ∩ C)] + P[(B ∩ C)] − P[(A ∩ C) ∩ (B ∩ C)]
39
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
Third term:
P[(A ∪ B) ∩ C] = P[(A ∩ C) ∪ (B ∩ C)] =
P[(A ∩ C)] + P[(B ∩ C)] − P[(A ∩ C) ∩ (B ∩ C)]
= P[(A ∩ C)] + P[(B ∩ C)] − P[(A ∩ B ∩ C)]
39
.
Question 2.87
P[A ∪ B ∪ C] = P[A ∪ B] + P[C] − P[(A ∪ B) ∩ C]
Third term:
P[(A ∪ B) ∩ C] = P[(A ∩ C) ∪ (B ∩ C)] =
P[(A ∩ C)] + P[(B ∩ C)] − P[(A ∩ C) ∩ (B ∩ C)]
= P[(A ∩ C)] + P[(B ∩ C)] − P[(A ∩ B ∩ C)] =
P[A]P[C] + P[B]P[C] − P[A]P[B]P[C]
39
.
Question 2.87
P[A ∪ B ∪ C] =P[A] + P[B] + P[C]
− P[A]P[B] − P[B]P[C] − P[C]P[A]
+ P[A]P[B]P[C]
40
.
Inclusion–exclusion principle:
P[A ∪ B ∪ C] =P[A] + P[B] + P[C]
− P[A ∩ B] − P[B ∩ C] − P[C ∩ A]
+ P[A ∩ B ∩ C]
41
.
Question 2.87
(d) Find P[A ∪ B ∪ C] if A, B, and C are pairwise
mutually exclusive.
42
.
Question 2.87
P[A ∪ B ∪ C] =P[A] + P[B] + P[C]
− P[A ∩ B] − P[B ∩ C] − P[C ∩ A]
+ P[A ∩ B ∩ C]
=⇒ P[A ∪ B ∪ C] = P[A] + P[B] + P[C]
43
.
Question 2.88
An experiment consists of picking one of two
urns at random and then selecting a ball from
the urn and noting its color (black or white). Let
A be the event ``urn 1 is selected'' and B the
event ``a black ball is observed.''
Under what conditions are A and B
independent?
44
.
Question 2.88
If P[B] = 0, A and B are independent.
45
.
Question 2.88
If P[B] = 0, A and B are independent.
Assume P[B] ̸= 0. Then, A and B are
independent iff
45
.
Question 2.88
If P[B] = 0, A and B are independent.
Assume P[B] ̸= 0. Then, A and B are
independent iff
P[A|B] = P[A]
45
.
Question 2.88
If P[B] = 0, A and B are independent.
Assume P[B] ̸= 0. Then, A and B are
independent iff
P[A|B] = P[A] = P[A|B]P[B] + P[A|Bc ]P[Bc ]
45
.
Question 2.88
If P[B] = 0, A and B are independent.
Assume P[B] ̸= 0. Then, A and B are
independent iff
P[A|B] = P[A] = P[A|B]P[B] + P[A|Bc ]P[Bc ] =
P[A|B]P[B] + P[A|Bc ] (1 − P[B])
45
.
Question 2.88
If P[B] = 0, A and B are independent.
Assume P[B] ̸= 0. Then, A and B are
independent iff
P[A|B] = P[A] = P[A|B]P[B] + P[A|Bc ]P[Bc ] =
P[A|B]P[B] + P[A|Bc ] (1 − P[B])
⇐⇒ P[A|B] (1 − P[B]) = P[A|Bc ] (1 − P[B]).
45
.
Question 2.88
If P[B] = 0, A and B are independent.
Assume P[B] ̸= 0. Then, A and B are
independent iff
P[A|B] = P[A] = P[A|B]P[B] + P[A|Bc ]P[Bc ] =
P[A|B]P[B] + P[A|Bc ] (1 − P[B])
⇐⇒ P[A|B] (1 − P[B]) = P[A|Bc ] (1 − P[B]).
If P[B] = 1, A and B are independent.
45
.
Question 2.88
Therefore, if P[B] ̸= 0 and P[B] ̸= 1, A and B are
independent iff P[A|B] = P[A|Bc ].
46
.
Question 2.97
A block of 100 bits is transmitted over a binary
communication channel with probability of bit
error p = 10−2 .
(a) If the block has 1 or fewer errors then the
receiver accepts the block. Find the probability
that the block is accepted.
47
.
Question 2.97
A = The event that the block is accepted
48
.
Question 2.97
A = The event that the block is accepted
A = there is no bit error (E0 ) or there is exactly 1
error (E1 )
48
.
Question 2.97
A = The event that the block is accepted
A = there is no bit error (E0 ) or there is exactly 1
error (E1 )
E0 ∩ E1 = ∅
48
.
Question 2.97
A = The event that the block is accepted
A = there is no bit error (E0 ) or there is exactly 1
error (E1 )
E0 ∩ E1 = ∅
A = E0 ∪ E1
48
.
Question 2.97
A = The event that the block is accepted
A = there is no bit error (E0 ) or there is exactly 1
error (E1 )
E0 ∩ E1 = ∅
A = E0 ∪ E1 =⇒ P[A] = P[E0 ] + P[E1 ]
48
.
Question 2.97
P[A] = P[E0 ] + P[E1 ]
49
.
Question 2.97
P[A] = P[E0 ] + P[E1 ]
P[E0 ] =
49
.
Question 2.97
P[A] = P[E0 ] + P[E1 ]
P[E0 ] = (1 − p)100
49
.
Question 2.97
P[A] = P[E0 ] + P[E1 ]
P[E0 ] = (1 − p)100
P[E1 ] =
49
.
Question 2.97
P[A] = P[E0 ] + P[E1 ]
P[E0 ] = (1 − p)100
( )
99
P[E1 ] = 100
1 p(1 − p)
49
.
Question 2.97
P[A] = P[E0 ] + P[E1 ]
P[E0 ] = (1 − p)100
( )
99
P[E1 ] = 100
1 p(1 − p)
P[A] = (1 − p)99 (1 + 99p) ≈ 0.7358
49
.
Question 2.97
(b) If the block has more than 1 error, then the
block is retransmitted. Find the probability that
M retransmissions are required.
50
.
Question 2.97
P[retransmission] = 1 − P[A] ≈ 0.2642
51
.
Question 2.97
P[retransmission] = 1 − P[A] ≈ 0.2642
P[having M retransmissions] = P[A](1 − P[A])M
51
.
Question 2.97
P[retransmission] = 1 − P[A] ≈ 0.2642
P[having M retransmissions] = P[A](1 − P[A])M
P[having at leastM retransmissions] =
51
.
Question 2.97
P[retransmission] = 1 − P[A] ≈ 0.2642
P[having M retransmissions] = P[A](1 − P[A])M
P[having at leastM retransmissions] = (1 − P[A])M
51
.
Question 2.100
Each of n terminals broadcasts a message in a
given time slot with probability p.
52
.
Question 2.100
(a) Find the probability that exactly one
terminal transmits so the message is received
by all terminals without collision.
53
.
Question 2.100
(a) Find the probability that exactly one
terminal transmits so the message is received
by all terminals without collision.
np(1 − p)n−1
53
.
Question 2.100
(b) Find the value of p that maximizes the
probability of successful transmission in part
(a).
54
.
Question 2.100
(b) Find the value of p that maximizes the
probability of successful transmission in part
(a).
d
n−1
=0
dp np(1 − p)
54
.
Question 2.100
(b) Find the value of p that maximizes the
probability of successful transmission in part
(a).
d
n−1
= 0 =⇒ p = n1
dp np(1 − p)
54
.
Question 2.100
(c) Find the asymptotic value of the maximum
probability of successful transmission as n
becomes large.
55
.
Question 2.100
(c) Find the asymptotic value of the maximum
probability of successful transmission as n
becomes large.
1
lim P[success] = lim n (1 − n)n−1
n→∞
n→∞ n
1
= lim (1 − )n−1
n→∞
n
1
= ≈ 0.3679
e
(6)
55
.
Question 2.100
Remark:
)n
(
1
=e
lim 1 +
n→∞
n
56
.
Question 2.106
A biased coin is tossed repeatedly until heads
has come up three times. Find the probability
that k tosses are required.
Hint: Show that {``k tosses are required''} =
A ∩ B, where
A ={``kth toss is head''} and
B ={``2 head occur in k − 1 tosses''}.
57
.
Question 2.106
Probability of head = p
58
.
Question 2.106
Probability of head = p
P[A] = p
58
.
Question 2.106
Probability of head = p
P[A] = p
( ) 2
k−3
P[B] = k−1
2 p (1 − p)
58
.
Question 2.106
Probability of head = p
P[A] = p
( ) 2
k−3
P[B] = k−1
2 p (1 − p)
A and B are two independent experiments
58
.
Question 2.106
Probability of head = p
P[A] = p
( ) 2
k−3
P[B] = k−1
2 p (1 − p)
A and B are two independent experiments
( ) 3
k−3
P(A ∩ B) = P(A, B) = k−1
for k > 2
2 p (1 − p)
58
.
Administrivia
Tutorial website!
www.comm.utoronto.ca/~rrafie/ece302.html
59
.
Questions?
59