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COMP11120: MATHEMATICAL TECHNIQUES
FOR COMPUTER SCIENCE LECTURE NOTES
1
1
Axioms of probability
1. Sample space of an experiment is the set of all possible outcomes, denotes by S.
2. An event E is any subset of S.
3. E ∪ F (read as E union F ) is the set of all outcomes belonging to E or to F or both.
4. E ∪ F = F ∪ E.
5. E ∩ F (read as E intersection F ) is the set of all outcomes belong to both E and F .
6. E ∩ F = F ∩ E.
7. E c (read as E complement) is the set of all outcomes in the sample space S that are not in
E.
8. De Morgan’s laws:
(E ∪ F )c = E c ∩ F c
(1)
(E ∩ F )c = E c ∪ F c .
(2)
and
9. Two events E and F are mutually exclusive if E and F have no outcomes in common.
10. If an experiment can result in any one of n different, equally likely outcomes, and if exactly
m of these outcomes correspond to event E, then the probability of event E is
m
(3)
P (E) = .
n
11. The odds ratio of an event E is:
P (E)
.
1 − P (E)
(4)
12. Axiom 1: the probability assigned to each event takes a value between 0 and 1 (inclusive).
13. Axiom 2: the probability of the entire sample space S is 1.
14. Axiom 3: if E and F are mutually exclusive events then
P (E ∪ F ) = P (E) + P (F ).
(5)
15. Additive law: if E and F are any events then
P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F ).
(6)
P (E c ) = 1 − P (E).
(7)
P (E c ∩ F ) = P (F ) − P (E ∩ F ).
(8)
P (E ∩ F c ) = P (E) − P (E ∩ F ).
(9)
16. Complementary law 1:
17. Complementary law 2:
18. Complementary law 3:
2
1.1
Problems
1. The number 1, 2, 3, and 4 are written separately on four slips of paper. The slips are then
put into a hat and stirred. A blindfolded person draws two slips from the hat, one after the
other, without replacement. Describe a sample space for the experiment.
2. A card is drawn at random from an ordinary bridge deck. Find the probability and the odds
that the card is an “honor” (that is, an ace, king, queen, jack, or ten).
3. For a chronic disease, there are five standard ameliorative treatments: a, b, c, d and e. A
doctor has resources for conducting a comparative study of three of these treatments. If he
chooses the three treatments for study at random from the five, what is the probability that
(i) treatment a will be chosen,
(ii) treatments a and b will be chosen,
(iii) treatments a or c will be chosen,
(iv) treatments a and b or b and c will be chosen?
4. The integers 1, 2, 3, . . . , 20 are written on slips of paper which are placed in a bowl and
thoroughly mixed. A slip is drawn from the bowl at random. What is the probability that
the number on the slip is either prime or divisible by 3?
5. Let E and F be events in a sample space S, such that P (E) = 0.4, P (F ) = 0.3 and P (E∩F ) =
0.2. Find the following:
(i) P (E ∪ F );
(ii) P (E c );
(iii) P (F c );
(iv) P (E c ∩ F );
(v) P (E c ∩ F c ).
6. Three coins are tossed. Find the probability of getting
(i) no heads;
(ii) at least one head.
1.2
Solutions
1. The sample space is:
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}.
2. The probability that the card is an honor is: (5 × 4)/52 = 5/13. The odds is: (5/13)/(1 −
(5/13)) = 5/8.
3. First find the sample space as:
S = {abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde}.
Then,
3
(i) 6/10 = 3/5;
(ii) 3/10;
(iii) 9/10;
(iv) 5/10 = 1/2.
4. Let E be the event that the number is prime and F be the event that it is divisible by 3.
Then,
P (E) = 9/20
P (F ) = 6/20,
P (E ∩ F ) = 1/20.
From the additive law,
P (E ∪ F ) = 9/20 + 6/20 − 1/20 = 14/20 = 0.7.
5.
(i) by the additive law, P (E ∪ F ) = 0.4 + 0.3 − 0.2 = 0.5;
(ii) by complementary law 1, P (E c ) = 1 − 0.4 = 0.6;
(iii) by complementary law 1, P (F c ) = 1 − 0.3 = 0.7;
(iv) by complementary law 2, P (E c ∩ F ) = 0.3 − 0.2 = 0.1;
(v) by de Morgan’s law and complementary law 1, P (E c ∩ F c ) = P ((E ∪ F )c ) = 1 − P (E ∪
F ) = 1 − 0.5 = 0.5.
6. First find the sample space as:
S = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }.
(i) The probability of no heads is 1/8;
(ii) By complementary law 1, the probability of at least one head is 1 minus the probability
of no heads which is 1 − (1/8) = 7/8.
2
Conditional probability and independence
1. Conditional probability of an event E, given F , is denoted by P (E | F ) and is defined by
P (E | F ) =
P (E ∩ F )
P (F )
(10)
provided that P (F ) > 0. Similarly, the conditional probability of an event F , given E, is
denoted by P (F | E) and is defined by
P (F | E) =
provided that P (E) > 0.
4
P (E ∩ F )
P (E)
(11)
2. Multiplicative laws:
P (E ∩ F ) = P (E)P (F | E)
(12)
P (E ∩ F ) = P (F )P (E | F ).
(13)
P (E) = P (E | F )P (F ) + P (E | F c )P (F c ).
(14)
and
3. A useful identity:
4. Bayes’s theorem:
P (F | E) =
P (E | F )P (F )
.
P (E | F )P (F ) + P (E | F c )P (F c )
(15)
5. Two events E and F are independent if and only if anyone of the following is true:
(i) P (E ∩ F ) = P (E)P (F );
(ii) P (E | F ) = P (E);
(ii) P (F | E) = P (F ).
6. Let E and F be independent events. Then E and F c are independent events, as are E c and
F , and E c and F c .
2.1
Problems
1. Two a’s and two b’s are arranged in order. All arrangements are equally likely. Given that
the last letter, in order, is b, find the probability that the two as are together.
2. A coin is tossed until a head appears, or until it has been tossed three times. Given that the
head does not occur on the first toss, what is the probability that the coin is tossed three
times?
3. Jimmy likes to go shopping with his mother because he can sometimes get her to buy him a
toy. The probability that she takes him along on her shopping trip this afternoon is 0.4, and
if she does, the probability that she gets a toy is 0.8. What is the probability that she takes
him shopping and buys him a toy?
4. An urn contains 5 black balls and 10 red balls. Two balls are drawn at random, one after
the other, without replacement. Set up a sample space for the possible outcomes of the
experiment, with appropriate probabilities.
5. Two coins are tossed. Show that event “head on first coin” and event “coins fall alike” are
independent.
6. Sixty percent of the students in a school are boys. Eighty percent of the boys and 75% of the
girls have activity tickets for all the school activities. A ticket is found and turned in to the
school’s lost and found department. What is the probability that it belongs to a girl? To a
boy?
5
2.2
Solutions
1. The sample space for this experiment is:
S = {aabb, abab, abba, bbaa, baba, baab}.
Let E denote the event that the two as are together and let F denote the event that the last
letter, in order, is b. The question is to find the conditional probability that P (E | F ). From
the sample space,
P (F ) = 3/6 = 1/2
and
P (E ∩ F ) = 2/6 = 1/3.
Thus by definition of the conditional probability
P (E | F ) = P (E ∩ F )/P (F ) = 2/3.
2. The sample space for this experiment is:
S = {H, T H, T T H, T T T }.
Let E denote the event that the coin is tossed three times and let F denote the event that
the head does not occur on the first toss. The question is to find the conditional probability
that P (E | F ). Clearly P (F ) = 1/2 and from the sample space P (E ∩ F ) = 1/4. Thus, by
the definition of the conditional probability P (E | F ) = 1/2.
3. Let E denote the event that Jimmy’s mother takes him along on her shopping trip this
afternoon. Let F denote the event that she buys Jimmy a toy. The question is to find
P (E ∩ F ). We are given that
P (E) = 0.4
and
P (F | E) = 0.8.
So by the multiplicative law the required probability
P (E ∩ F ) = 0.4 × 0.8 = 0.32.
4. The possible outcomes are: (First Red, Second Red), (First Red, Second Black), (First Black,
Second Red) and (First Black, Second Black). The corresponding probabilities are:
P (RR) = P (Second Red | First Red) × P (First Red) =
9
3
10
×
= ,
15 14
7
10
5
5
×
= ,
15 14
21
5
10
5
P (BR) = P (Second Red | First Black) × P (First Black) =
×
= ,
15 14
21
4
2
5
×
= .
P (BB) = P (Second Black | First Black) × P (First Black) =
15 14
21
P (RB) = P (Second Black | First Red) × P (First Red) =
6
5. The sample space for this experiment is:
S = {HH, HT, T H, T T }.
Let E denote the event that head on first coin and let F denote the event that coins fall alike.
From the sample space,
P (E) = 1/2,
P (F ) = 1/2
and
P (E ∩ F ) = 1/4.
Thus E and F must be independent.
6. Let E be the event that a student has an activity ticket. Let F be the event a student is a
girl. The question is to find P (F | E). We are given that
P (F ) = 0.4,
P (E | F ) = 0.75,
P (F c ) = 1 − P (F ) = 0.6
and
P (E | F c ) = 0.8.
Using Bayes’ theorem, we get the probability:
P (E | F )P (F )
P (E | F )P (F ) + P (E | F c )P (F c )
0.75 × 0.4
=
0.75 × 0.4 + 0.8 × 0.6
= 0.3846154.
P (F | E) =
Redefine F as the event a student is a boy. Then we have
P (F ) = 0.6,
P (E | F ) = 0.8,
P (F c ) = 1 − P (F ) = 0.4
and
P (E | F c ) = 0.75.
Using Bayes’ theorem again, we get the probability:
P (E | F )P (F )
P (E | F )P (F ) + P (E | F c )P (F c )
0.8 × 0.6
=
0.8 × 0.6 + 0.75 × 0.4
= 0.6153846.
P (F | E) =
7
3
Random variables
1. A random variable is a function that assigns a numerical value to each outcome of a particular experiment. A random variable is denoted by an uppercase letter, such as X and a
corresponding lower case letter such as x is used to denote a possible value of X.
2. A discrete random variable is a random variable with a finite or countably infinite number of
possible values, e.g. number of accidents, number of applicants interviewed, number of power
plants, etc.
3. If the possible values of a random variable contains an interval of real numbers then it is a
continuous random variable, e.g. temperature, breaking strength, failure time, etc.
4. The probability mass function (pmf) of a discrete random variable X is defined by
p(x) = P (X = x).
(16)
Its properties are:
(i) p(x) ≥ 0 for all possible values of x;
(ii)
P
all x p(x) = 1.
5. The probability density function (pdf) of a continuous random variable X is a non-negative
function f (x) ≥ 0 such that
b
Z
f (x)dx = P (a < X < b)
(17)
a
for all a and b. Two consequences are:
Z
∞
−∞
f (x)dx = P (−∞ < X < ∞) = 1
(18)
and
Z
a
f (x)dx = P (X = a) = 0.
(19)
a
6. The cumulative distribution function of a discrete random variable X is defined by
F (x) = P (X ≤ x) =
X
p(z).
(20)
for all z ≤ x
That of a continuous random variable X is defined by
F (x) = P (X ≤ x) =
Its properties are:
(i) 0 ≤ F (x) ≤ 1;
(ii) If a ≤ b then F (a) ≤ F (b);
(iii) F (−∞) = 0;
8
Z
x
−∞
f (y)dy.
(21)
(iv) F (∞) = 1;
(v) if X is a continuous random variable F (b) − F (a) = P (a < X < b);
(vi) if X is a continuous random variable then its pdf can be obtained as
∂F (x)
.
∂x
f (x) =
(22)
7. The expected value of a discrete random variable X is:
X
E(X) =
xp(x).
(23)
for all x
That of a continuous random variable X is defined by
E(X) =
Z
∞
xf (x)dx.
(24)
−∞
Its properties are:
(i) E(c) = c, where c is a constant;
(ii) E(cX) = cE(X), where c is a constant;
(iii) E(cX + d) = cE(X) + d, where c and d are constants.
8. Let g be any real-valued function. The expected value of g(X) for a discrete random variable
X is:
X
E(g(X)) =
g(x)p(x).
(25)
for all x
That of a continuous random variable X is defined by
E(g(X)) =
Z
∞
g(x)f (x)dx.
(26)
−∞
9. The variance of a random variable X (discrete or continuous) is:
V ar(X) = E(X 2 ) − (E(X))2 .
(27)
Its properties are:
(i) V ar(c) = 0, where c is a constant;
(ii) V ar(cX) = c2 V ar(X), where c is a constant;
(iii) V ar(cX + d) = c2 V ar(X), where c and d are constants.
10. The standard deviation of a random variable X (discrete or continuous) is:
SD(X) =
q
V ar(X).
(28)
11. Let X be a continuous random variable with pdf fX . Set Y = g(X) where g is a strictly
monotone differentiable function. Then the pdf of Y is:
fY (y) = fX (g
−1
dg −1 (y) dg(y) −1
,
(y)) = fX (g (y))/ dy dy where g−1 is the inverse function of g.
9
(29)
3.1
Problems
1. Verify that


0


 1/3
if
if
F (x) =

2/3 if



1
if
x < −3,
−3 ≤ x < −1,
−1 ≤ x < 0,
x≥0
is a cumulative distribution function and derive the corresponding probability mass function.
2. A random variable X has the probability mass function:
p(x) =
(
1/3 if x = 0, 1, 2,
0
otherwise.
What is the cumulative distribution function of X?
3. An urn contains 4 balls numbered 1, 2, 3, 4, respectively. Let X denote the number that
occurs if one ball is drawn at random from the urn. What is the probability mass function of
X?
4. Consider the urn defined in question 3. Two balls are drawn from the urn without replacement. Let X be the sum of the two numbers that occur. Derive the probability mass function
of X.
5. The church lottery is going to give away a 3000-dollar car. They sell 10,000 tickets at 1 dollar
apiece. If you buy 1 ticket, what is your expected gain. What is your expected gain if you
buy 100 tickets? Compute the variance of your gain in these two instances.
6. If X is a random variable with probability mass function
p(x) =
(
1/4 if x = 2, 4, 8, 16,
0
otherwise,
compute the following:
(i) E(X);
(ii) E(X 2 );
(iii) E(1/X);
(iv) E(2X/2 );
(v) V ar(X);
(vi) SD(X).
7. Verify that
F (x) =


 0
√
if x < 0,
x if 0 ≤ x ≤ 1,

 1
if x > 1
is a cumulative distribution function and derive the corresponding probability density function.
10
8. A random variable X has the probability density function:
f (x) =
(
10 exp(−10x) if x > 0,
0
otherwise.
What is the cumulative distribution function of X?
9. If X is a random variable with probability density function
f (x) =
(
2(1 − x) if 0 < x < 1,
0
otherwise,
compute the following:
(i) E(X);
(ii) E(X 2 );
(iii) E((X + 10)2 );
(iv) E(1/(1 − X));
(v) V ar(X);
(vi) SD(X).
10. A random variable X has the probability density function:
fX (x) = exp(−x),
x > 0.
Find the probability density function of Y = exp(−X).
3.2
Solutions
1. Check the four conditions (i)-(iv) for F to be a cdf. It is useful to draw F as a function of
x beforehand. From the plot conditions (i) and (ii) would be obvious. Condition (iii) follows
since F (x) = 0 for all x < −3. Condition (iv) follows since F (x) = 1 for all x ≥ 0. For the
probability mass function note that F ‘jumps’ by the amount 1/3 at x = 0, −1, −3. Thus,
p(x) =
(
1/3 if x = 0, −1, −3,
0
otherwise.
2. By the definition of cdf,
F (x) =


0







if x ≤ 0,
1/3 if 0 < x ≤ 1,
2/3 if 1 < x ≤ 2,
1
if x ≥ 2.
3. X takes the values 1, 2, 3 or 4 with equal probabilities. Thus,
p(x) =
(
1/4 if x = 1, 2, 3, 4,
0
otherwise.
11
4. The sample space of possible outcomes and the corresponding values for X are given in the
table below.
Outcome
1,2
1,3
1,4
2,1
2,3
2,4
3,1
3,2
3,4
4,1
4,2
4,3
X
3
4
5
3
5
6
4
5
7
5
6
7
Thus, the probability mass function of X is:
x
3
4
5
6
7
p(x)
2/12=1/6
2/12=1/6
4/12 = 1/3
2/12=1/6
2/12=1/6
5. Let X denote the gain. Then X = 3000 − 1 = 2999 with probability 1/10000 and X = 0 − 1 =
−1 with probability 9999/10000. Thus, by definition of expectation, the expected gain is:
E(X) = 2999 ×
9999
7
1
+ (−1) ×
=−
10000
10000
10
and
1
9999
4502
+ (−1)2 ×
=
.
10000
10000
5
Hence, by the definition of the variance, the variance of the gain is:
E(X 2 ) = 29992 ×
V ar(X) = E(X 2 ) − (E(X))2 =
4502
49
89991
−
=
.
5
100
100
If you buy 100 tickets then the expected gain is:
E(X) = 2900 ×
and
E(X 2 ) = 29002 ×
9900
100
+ (−100) ×
= −70
10000
10000
100
9900
+ (−100)2 ×
= 94000.
10000
10000
Hence, the corresponding variance is:
V ar(X) = E(X 2 ) − (E(X))2 = 94000 − 702 = 89100.
12
6.
(i) by the definition of expectation,
1
1
1
1
15
+ 4 × + 8 × + 16 × =
;
4
4
4
4
2
E(X) = 2 ×
(ii) by the definition of expectation,
1
1
1
1
+ 42 × + 82 × + 162 × = 85;
4
4
4
4
E(X 2 ) = 22 ×
(iii) by the definition of expectation,
E(1/X) =
1 1 1 1 1 1
1
1
15
× + × + × +
× = ;
2 4 4 4 8 4 16 4
64
(iv) by the definition of expectation,
1
1
1
139
1
+ 22 × + 24 × + 28 × =
;
4
4
4
4
2
E(2X/2 ) = 2 ×
(v) by the definition of variance,
225
115
=
;
4
4
√
p
(vi) by the definition of standard deviation, SD(X) = 115/4 = 115/2.
V ar(X) = E(X 2 ) − (E(X))2 = 85 −
7. Check the four conditions (i)-(iv). It is useful to draw F as a function of x beforehand. From
the plot conditions (i) and (ii) would be obvious. Condition (iii) follows since F (x) = 0 for all
x < 0. Condition (iv) follows since F (x) = 1 for all x > 1. To obtain the probability density
function simply differentiate:
( √
√
∂ x/∂x = 1/(2 x) if 0 ≤ x ≤ 1,
f (x) =
0
otherwise.
8. If x > 0 by the definition of cdf,
F (x) = 10
Z
x
0
exp(−10y)dy = [− exp(−10y)]x0 = 1 − exp(−10x).
Thus, the cumulative distribution function of X is:
F (x) =
9.
(
1 − exp(−10x) if x > 0,
0
otherwise.
(i) by the definition of expectation,
E(X) = 2
Z
1
x(1 − x)dx = 2
0
Z
0
1
x−x
2
"
x2 x3
dx = 2
−
2
3
#1
0
1 1
=2
−
2 3
=
1 1
=2
−
3 4
1
;
3
(ii) by the definition of expectation,
E X
2
=2
Z
1
0
2
x (1 − x)dx = 2
Z
0
1
2
x −x
13
3
"
x3 x4
dx = 2
−
3
4
#1
0
=
1
;
6
(iii)
E (X + 10)2 = E X 2 + 20X + 100 = E(X 2 )+ 20E(X)+ 100 =
1 20
641
+ + 100 =
;
6 3
6
(iv) by the definition of expectation,
E (1/(1 − X)) = 2
Z
1
0
1−x
dx = 2
1−x
Z
0
1
dx = 2 [x]10 = 2;
(v) by the definition of variance,
1 1
1
− = ;
6 9
18
√
p
(vi) by the definition of standard deviation, SD(X) = 1/18 = 1/(3 2).
V ar(X) = E(X 2 ) − (E(X))2 =
10. We are given that fX (x) = exp(−x) and g(x) = exp(−x). The inverse function of g is
g −1 (y) = − log y and
dg −1 (y)
d log y
1
=−
=− .
dy
dy
y
We get the probability density function of Y as:
1
1
1
fY (y) = fX (− log y) − = exp(log y) = y = 1.
y
y
y
4
Binomial distribution
This is a discrete distribution. It was derived by James Bernoulli in his treatise Ars Conjectandi
published in 1713. The distribution has applications in quality control.
4.1
Definition
Consider an experiment of m repeated trials where the following are valid:
(i) all trials are statistically independent (in the sense that knowing the outcome of any particular
one of them does not change one’s assessment of chance related to any others);
(ii) each trial results in only one of two possible outcomes, labeled as “success” and “failure”;
(iii) and, the probability of success on each trial, denoted by p, remains constant.
Then the random variable X that equals the number of trials that result in a success is said to
have the binomial distribution with parameters m and p.
Examples: number of failures in an exam, number of heads in the coin-tossing experiment
14
Probability mass function The pmf of the random variable X is:
p(x) =
!
m x
p (1 − p)m−x ,
x
x = 0, 1, . . . , m,
(30)
where
m
x
4.2
!
=
m!
.
x!(m − x)!
(31)
Properties
The binomial random variable X has the following properties
(i) the cdf is:
F (x) =
x
X
m
i
i=0
!
pi (1 − p)m−i .
(32)
This can also be written as:
F (x) = 1 −
1
B(x + 1, m − x)
Z
p
0
tx (1 − t)m−x−1 dt = 1 − Ip (x + 1, m − x),
(33)
where
Ip (a, b) =
1
B(a, b)
Z
0
p
ta−1 (1 − t)b−1 dt
(34)
denotes the incomplete beta function.
(ii) the expected value is:
E(X) = mp;
(35)
V ar(X) = mp(1 − p);
(36)
(iii) the variance is:
(iv) the third central moment is:
h
i
E {X − E(X)}3 = mp(1 − p)(1 − 2p);
(37)
(v) the fourth central moment is:
h
i
E {X − E(X)}4 = 3 {mp(1 − p)}2 + mp(1 − p)(1 − 2p);
(38)
(vi) the mean deviation is:
!
m − 1 [mp]+1
E {|X − E(X)|} = m
p
(1 − p)m−[mp] ;
[mp]
15
(39)
(vii) the coefficient of variation is:
CV (X) =
s
1−p
;
mp
(40)
(viii) the skewness measure is:
1 − 2p
γ1 (X) = p
;
mp(1 − p)
(xi) the kurtosis measure is:
γ2 (X) = 3 +
4.3
(41)
1 − 6p(1 − p)
.
mp(1 − p)
(42)
Estimation
Suppose you have a dataset x1 , x2 , . . . , xn from a binomial distribution with the parameter m known
and p unknown. Then an estimate for p is:
p̂ =
Pn
i=1 xi
nm
.
(43)
An exact 95% confidence interval for the true value of p is (pL , pu ), where pL and pU are the
solutions of
Ip L
n
X
i=1
xi , nm −
n
X
!
xi + 1
= 0.025
(44)
n
X
!
= 0.975,
(45)
i=1
and
Ip U
n
X
i=1
xi + 1, nm −
xi
i=1
respectively. For large n, an approximate 95% confidence interval for the true value of p is given
by either:

 p̂ +
or


1.962
2m
−
r
p̂(1−p̂)1.962
mn
1+

1.962
+
1.964
4(mn)2
p̂ +
,
1.962
2mn
+
p̂ − 1.96
p̂(1−p̂)1.962
mn
1+
mn
s
r
s
p̂(1 − p̂)
, p̂ + 1.96
mn
16
1.962
mn

p̂(1 − p̂) 
.
mn
+
1.964
4(mn)2




(46)
(47)
4.4
Generalizations of the binomial distribution
A doubly truncated binomial distribution is given by
!
m x
p(x) =
p (1 − p)m−x
x
, m−r
X2
!
m j
p (1 − p)m−j ,
j
j=r1
x = r1 , . . . , m − r2 .
(48)
A singly truncated binomial distribution is given by
p(x) =
!
m x
p (1 − p)m−x
x
,
{1 − (1 − p)m } .
(49)
Two mixtures of binomial distributions are given by
p(x) =
!
∞
m X
wj pxj (1 − pj )m−x
x j=1
(50)
(where wj ≥ 0 and w1 + w2 + · · · = 1), and
p(x) =
(where g(p) ≥ 0 and
4.5
Z
!Z
m
x
0
1
g(p)px (1 − p)m−x dp
(51)
1
g(p)dp = 1).
0
Problems
1. In the past, two building contractors, A and B, have competed for 50 contracts. A won 20
and B won 30 of these contracts. The contractors have both been asked to tender for three
new contracts. On the basis of their past performance, what is the probability that
(i) Contractor A will win all the contracts;
(ii) Contractor B will win at least one contract;
(iii) Contractor A will win exactly two contracts?
2. A food-packaging apparatus underfills 10% of the containers. Find the probability that for
any particular 5 containers the number of underfilled will be:
(i) exactly 3;
(ii) exactly 2;
(iii) zero;
(iv) at least 1.
3. An engineer has designed a modified welding robot. The robot will be considered good
enough to manufacture if it misses only 1% of its assigned welds. And it will be judged a
poor performer if it misses 5% of its welds. A test is performed involving 100 welds. The new
design will be accepted if the number of missed welds X is 2 or less and rejected otherwise.
(i) What is the probability that a good design will be rejected?
17
(ii) What is the probability that a poor design will be accepted?
4. A manufacturing process has 100 customer orders to fill. Each order requires one component
that is purchased from a supplier. However, typically, 2% of the components are identified as
defective, and the components can be assumed to be independent.
(i) If the manufacturer stocks 100 components, what is the probability that the 100 orders
can be filled without reordering components?
(ii) If the manufacturer stocks 102 components, what is the probability that the 100 orders
can be filled without reordering components?
(iii) If the manufacturer stocks 105 components, what is the probability that the 100 orders
can be filled without reordering components?
5. An engineer has designed a modified welding robot. The robot will be considered good
enough to manufacture if it misses only 1% of its assigned welds. And it will be judged a
poor performer if it misses 5% of its welds. A test is performed involving 100 welds. The new
design will be accepted if the number of missed welds X is 2 or less and rejected otherwise.
(i) What is the probability that a good design will be rejected?
(ii) What is the probability that a poor design will be accepted?
4.6
Solutions
1. Let X denote the number of contracts won by A. Clearly X is a binomial random variable
with m = 3 and p = 20/50 = 0.4.
(i) by binomial pmf,
P (X = 3) =
!
3
(0.4)3 (0.6)0 = (0.4)3 = 0.064;
3
(ii) by binomial pmf,
P (B wins at least 1 contract) = P (A wins at most 2 contracts)
= P (X ≤ 2)
= 1 − P (X = 3)
= 1 − (0.4)3
= 0.936;
(iii) by binomial pmf,
P (X = 2) =
!
3
3×2
× (0.4)2 × 0.6 = 0.288.
(0.4)2 (1 − 0.4)1 =
1×2
2
2. Let X denote the number of underfilled containers. Clearly X is a binomial random variable
with m = 5 and p = 0.1.
18
(i) by binomial pmf,
!
5
5×4×3
(0.1)3 (0.9)2 = 0.0081;
(0.1)3 (1 − 0.1)2 =
1×2×3
3
P (X = 3) =
(ii) by binomial pmf,
!
5
5×4
(0.1)2 (1 − 0.1)3 =
(0.1)2 (0.9)3 = 0.0729;
2
1×2
P (X = 2) =
(iii) by binomial pmf,
!
5
(0.1)0 (1 − 0.1)5 = (0.9)5 = 0.590;
0
P (X = 0) =
(iv) P (X ≥ 1) = 1 − P (X = 0) = 1 − 0.59 = 0.41.
3. Clearly X is a binomial random variable with the following values for the parameters.
(i) m = 100 and p = 0.01. A good design will be rejected if X > 2. So, by binomial pmf,
the corresponding probability is:
P (X > 2) = 1 − P (X ≤ 2)
= 1 − P (X = 0) − P (X = 1) − P (X = 2)
!
!
100
100
= 1−
(0.01)0 (0.99)100 −
(0.01)1 (0.99)99
0
1
!
100
−
(0.01)2 (0.99)98
2
= 1 − (0.99)100 − 100 × 0.01 × (0.99)99 − 99 × 50 × (0.01)2 × (0.99)98
= 0.0793732;
(ii) m = 100 and p = 0.05. A poor design will be accepted if X ≤ 2. So, by binomial pmf,
the corresponding probability is:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
=
!
!
100
100
(0.05)0 (0.95)100 +
(0.05)1 (0.95)99
0
1
!
100
+
(0.05)2 (0.95)98
2
= (0.95)100 + 100 × 0.05 × (0.95)99 + 99 × 50 × (0.05)2 × (0.95)98
= 0.118263.
4. Let X denote the number of defective components. Clearly X is a binomial random variable
with the following values for the parameters.
(i) m = 100 and p = 0.02. Orders can be filled without reordering components if there are
no defective components. So, by binomial pmf, the corresponding probability is:
P (X = 0) =
!
100
(0.02)0 (0.98)100 = (0.98)100 = 0.1326196;
0
19
(ii) m = 102 and p = 0.02. Orders can be filled without reordering components if there are
at most two defective components. So, by binomial pmf, the corresponding probability
is:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
=
!
!
102
102
(0.02)0 (0.98)102 +
(0.02)1 (0.98)101
0
1
!
102
+
(0.02)2 (0.98)100
2
= (0.98)102 + 102 × 0.02 × (0.98)101 + 101 × 51 × (0.02)2 × (0.98)100
= 0.6657502;
(iii) m = 105 and p = 0.02. Orders can be filled without reordering components if there are
at most five defective components. So, by binomial pmf, the corresponding probability
is:
P (X ≤ 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+P (X = 4) + P (X = 5)
=
!
!
105
105
(0.02)0 (0.98)105 +
(0.02)1 (0.98)104
0
1
!
!
!
!
105
105
+
(0.02)2 (0.98)103
(0.02)3 (0.98)102
2
3
105
105
+
(0.02)4 (0.98)101
(0.02)5 (0.98)100
4
5
= (0.98)105 + 105 × 0.02 × (0.98)104 + 105 × 52 × (0.02)2 × (0.98)103
+35 × 52 × 103 × (0.02)3 × (0.98)102
+35 × 13 × 103 × 102 × (0.02)4 × (0.98)101
+7 × 13 × 103 × 102 × 101 × (0.02)5 × (0.98)100
= 0.9807593.
5. Clearly X is a binomial random variable with the following values for the parameters.
(i) m = 100 and p = 0.01. A good design will be rejected if X > 2. So, by binomial pmf,
the corresponding probability is:
P (X > 2) = 1 − P (X ≤ 2)
= 1 − P (X = 0) − P (X = 1) − P (X = 2)
!
!
100
100
= 1−
(0.01)0 (0.99)100 −
(0.01)1 (0.99)99
0
1
!
100
−
(0.01)2 (0.99)98
2
= 1 − (0.99)100 − 100 × 0.01 × (0.99)99 − 99 × 50 × (0.01)2 × (0.99)98
= 0.0793732;
20
(ii) m = 100 and p = 0.05. A poor design will be accepted if X ≤ 2. So, by binomial pmf,
the corresponding probability is:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
!
!
100
100
(0.05)0 (0.95)100 +
(0.05)1 (0.95)99
0
1
=
!
100
+
(0.05)2 (0.95)98
2
= (0.95)100 + 100 × 0.05 × (0.95)99 + 99 × 50 × (0.05)2 × (0.95)98
= 0.118263.
4.7
Figures
Binomial PMF
•
2
•
4
•
•
6
•
•
8
•
10
•
•
•
•
•
2
4
6
•
n=10,p=0.7
n=10,p=0.9
•
•
0
•
•
2
•
•
4
6
8
0.0
•
•
10
•
•
•
2
•
•
•
4
•
6
x
21
•
•
0
x
•
10
•
0.2
0.2
•
PMF
0.4
x
•
•
8
x
•
0.0
PMF
•
0
0.4
0
•
•
•
0.0
PMF
•
0.2
0.4
n=10,p=0.4
•
0.2
•
0.0
PMF
0.4
n=10,p=0.1
•
8
10
5
Poisson distribution
The Poisson distribution (another discrete distribution) is associated with counts of number of
occurrences of a relatively rare event over a specified interval of time, volume, length, area, etc.
In 1837 Poisson published the derivation of this distribution which bears his name. An important
application of the distribution is in queueing theory.
Examples: number of breakdowns of a generator in a week, the number of accidents on a
construction site per month, the number of plating faults per square meter of metal sheet, etc.
5.1
Definition
Given a continuous interval (in time, length, etc), assume discrete events occur randomly throughout the interval. If the interval can be partitioned into subintervals of small enough length such
that
(i) the probability of more than one occurrence in a subinterval is zero;
(ii) the probability of one occurrence in a subinterval is the same for all subintervals and proportional to the length of the subinterval;
(iii) and, the occurrence of an event in one subinterval has no effect on the occurrence or nonoccurrence in another non-overlapping subinterval,
If the mean number of occurrences in the interval is λ, the random variable X that equals the
number of occurrences in the interval is said to have the Poisson distribution with parameter λ.
Probability mass function The pmf of the random variable X is:
p(x) =
5.2
λx exp(−λ)
,
x!
x = 0, 1, 2, . . . .
(52)
Properties
The poisson random variable X has the following properties
1. the expected value is:
E(X) = λ;
(53)
V ar(X) = λ;
(54)
2. the variance is:
22
5.3
Problems
1. The number of cracks requiring repair in a section of motor way is assumed to follow a Poisson
distribution with a mean of two cracks per mile:
(i) What is the probability that there are no cracks that require repair in 5 miles of motor
way?
(ii) What is the probability that at least one crack requires repair in one-half mile of motor
way?
(iii) What is the probability that there are exactly three cracks that require repair in 2 miles
of motor way?
2. One-hundred-year records of earthquakes in a region show that there were 12 earthquakes of
intensity 6 or more. Calculate the following:
(i) the probability that such earthquakes will occur in the next 3 years;
(ii) the probability that no such earthquake will occur in the next 10 years.
3. Suppose that in one year the number of industrial accidents follows a Poisson distribution
with mean 3. What is the probability that in a given year there will be at least 1 accident?
4. Cars arrive at a particular highway patrol safety checkpoint with the mean rate λ = 100 per
hour. Each car encountered during a 1-hour period is tested for mechanical defects, and the
drivers are informed of needed repairs. Suppose that 10% of all cars in the state need such
repairs.
(i) What is the probability that at least 4 cars will be found in need of repairs?
(ii) To test the effectiveness of the warnings, the highway patrol will trace the defective cars
to determine if the necessary repairs were eventually made. Supposing that only half of
the originally warned operators ever fix their cars, what is the probability that at least
2 cars will have been repaired?
5. The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson
distribution with a mean of 0.05 per square foot of plastic panel. Assume an automobile
interior contains 10 square feet of plastic panel.
(i) What is the probability that there are no surface flaws in an auto’s interior?
(ii) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars
has any surface flaws?
(iii) If 10 cars are sold to a rental company, what is the probability that at most one car has
any surface flaws?
6. Cars arrive at a particular highway patrol safety checkpoint with the mean rate λ = 100 per
hour. Each car encountered during a 1-hour period is tested for mechanical defects, and the
drivers are informed of needed repairs. Suppose that 10% of all cars in the state need such
repairs.
(i) What is the probability that at least 4 cars will be found in need of repairs?
23
(ii) To test the effectiveness of the warnings, the highway patrol will trace the defective cars
to determine if the necessary repairs were eventually made. Supposing that only half of
the originally warned operators ever fix their cars, what is the probability that at least
2 cars will have been repaired?
7. The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson
distribution with a mean of 0.05 per square foot of plastic panel. Assume an automobile
interior contains 10 square feet of plastic panel.
(i) What is the probability that there are no surface flaws in an auto’s interior?
(ii) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars
has any surface flaws?
(iii) If 10 cars are sold to a rental company, what is the probability that at most one car has
any surface flaws?
5.4
Solutions
1. Let X denote the number of cracks in the motor way. X is a poisson random variable with
the following values for the parameter λ.
(i) λ = 5 × 2 = 10; by Poisson pmf, the required probability is:
P (X = 0) =
100 exp(−10)
= exp(−10) = 4.54 × 10−5 ;
0!
(ii) λ = 1/2 × 2 = 1; by Poisson pmf, the required probability is:
P (X ≥ 1) = 1 − P (X = 0) = 1 −
10 exp(−1)
= 1 − exp(−1) = 0.632;
0!
(iii) λ = 2 × 2 = 4; by Poisson pmf, the required probability is:
P (X = 3) =
64 exp(−4)
43 exp(−4)
=
= 0.195.
3!
6
2. Let X denote the number of earthquakes in the region. X is a poisson random variable with
the following values for the parameter λ.
(i) λ = (12/100) × 3 = 0.36; by Poisson pmf, the required probability is:
P (X ≥ 1) = 1 − P (X = 0) = 1 −
(0.36)0 exp(−0.36)
= 1 − exp(−0.36) = 0.302;
0!
(ii) λ = (12/100) × 10 = 1.2; by Poisson pmf, the required probability is:
P (X = 0) =
(1.2)0 exp(−1.2)
= exp(−1.2) = 0.301.
0!
3. Let X denote the number of industrial accidents per year. X is a poisson random variable
with λ = 3. By Poisson pmf, the required probability is:
P (X ≥ 1) = 1 − P (X = 0) = 1 −
24
30 exp(−3)
= 1 − exp(−3) = 0.95.
0!
4.
(i) Let X denote the number of cars in need of repairs. Clearly X is a poisson random
variable with λ = 100 × 0.1 = 10. By Poisson pmf, the probability that at least 4 cars
will be found in need of repair is:
P (X ≥ 4) = 1 − P (X < 4)
= 1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3)
100 exp(−10) 101 exp(−10) 102 exp(−10) 103 exp(−10)
−
−
−
= 1−
0! 1!
2!
3!
500
= 1 − 61 +
exp(−10)
3
= 0.9896639;
(ii) Let X denote the number of cars fixed. Clearly X is a poisson random variable with
λ = 100 × 0.1 × (1/2) = 5. By Poisson pmf, the probability that at least 2 cars will have
been repaired is:
P (X ≥ 2) = 1 − P (X < 2)
= 1 − P (X = 0) − P (X = 1)
50 exp(−5) 51 exp(−5)
−
= 1−
0!
1!
= 1 − 6 exp(−5)
= 0.9595723.
5. Let X denote the number of surface flaws. X is a poisson random variable with the following
parameter values.
(i) λ = 10 × 0.05 = 0.5. So, by Poisson pmf, the probability that there are no flaws in an
auto’s interior is:
P (X = 0) =
(0.5)0 exp(−0.5)
= exp(−0.5) = 0.6065307;
0!
(ii) λ = 10 × 10 × 0.05 = 5. So, by Poisson pmf, the probability that none of the 10 cars has
any flaws is:
50 exp(−5)
= exp(−5) = 0.006737947;
P (X = 0) =
0!
(iii) λ = 10 × 10 × 0.05 = 5. At most one car has any surface flaws means that either there
are no flaws or there is exactly one flaw in one of the 10 cars. So, by Poisson pmf, the
corresponding probability is:
P (X = 0) + P (X = 1) =
6.
50 exp(−5) 51 exp(−5)
+
= 6 exp(−5) = 0.04042768.
0!
1!
(i) Let X denote the number of cars in need of repairs. Clearly X is a poisson random
variable with λ = 100 × 0.1 = 10. By Poisson pmf, the probability that at least 4 cars
will be found in need of repair is:
P (X ≥ 4) = 1 − P (X < 4)
= 1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3)
25
100 exp(−10) 101 exp(−10) 102 exp(−10) 103 exp(−10)
−
−
−
0! 1!
2!
3!
500
exp(−10)
= 1 − 61 +
3
= 0.9896639;
= 1−
(ii) Let X denote the number of cars fixed. Clearly X is a poisson random variable with
λ = 100 × 0.1 × (1/2) = 5. By Poisson pmf, the probability that at least 2 cars will have
been repaired is:
P (X ≥ 2) = 1 − P (X < 2)
= 1 − P (X = 0) − P (X = 1)
50 exp(−5) 51 exp(−5)
−
= 1−
0!
1!
= 1 − 6 exp(−5)
= 0.9595723.
7. Let X denote the number of surface flaws. X is a poisson random variable with the following
parameter values.
(i) λ = 10 × 0.05 = 0.5. So, by Poisson pmf, the probability that there are no flaws in an
auto’s interior is:
P (X = 0) =
(0.5)0 exp(−0.5)
= exp(−0.5) = 0.6065307;
0!
(ii) λ = 10 × 10 × 0.05 = 5. So, by Poisson pmf, the probability that none of the 10 cars has
any flaws is:
50 exp(−5)
P (X = 0) =
= exp(−5) = 0.006737947;
0!
(iii) λ = 10 × 10 × 0.05 = 5. At most one car has any surface flaws means that either there
are no flaws or there is exactly one flaw in one of the 10 cars. So, by Poisson pmf, the
corresponding probability is:
P (X = 0) + P (X = 1) =
5.5
50 exp(−5) 51 exp(−5)
+
= 6 exp(−5) = 0.04042768.
0!
1!
Generalizations of the Poisson distribution
A positive Poisson (also called conditional Poisson) distribution is given by
p(x) =
λx
1
,
exp(λ) − 1 x!
x = 1, 2, . . . .
(55)
A left truncated Poisson distribution is given by
p(x) =
λx exp(−λ)
x!
,
1 − exp(−λ)
rX
1 −1
j=0
26
λj
j!

,
x = r1 , r1 + 1, . . . .
(56)
A right truncated Poisson distribution is given by
λx
p(x) =
x!
,

r2
X
λj
j=0
j!

,
x = 0, . . . , r2 .
(57)
x = r1 , r1 + 1, . . . , r2 .
(58)
A doubly truncated Poisson distribution is given by
p(x) =
λx
x!
,

r2
X
λj
j=r1
j!
A modified Poisson distribution is given by
p(x) =
(

,
λx exp(−λ)/x!,
if x = 0, 1, . . . , K − 1,
j
j=K λ exp(−λ)/j!, if x = K.
P∞
(59)
A mixture of two positive Poisson distributions is given by
p(x) =
p
λx1
λx2
1−p
+
,
exp(λ1 ) − 1 x!
exp(λ2 ) − 1 x!
27
x = 1, 2, . . . .
(60)
5.6
Figures
Poisson PMF
Lambda=1
Lambda=2
0
10
0.2
15
•
•
• •
• • • • • • • • •
0
5
10
x
Lambda=5
Lambda=10
• • •
•
•
• •
• • • • • •
5
10
15
• • •
• •
• •
•
• •
•
• • • • •
0
x
5
10
x
28
15
0.2
x
PMF
•
• •
• • • • • • • • • • • •
5
0.2
0.0
PMF
0
•
0.0
•
• •
0.0
PMF
0.2
•
0.0
PMF
• •
15