Lesson4 Probability of an event [Autosaved].pdf

‫جامعة‬
‫سويف‬ ‫بني‬
Probability and Statistics for Engineers

Probability
Chapter 2: Lesson 2

2.4. Probability of an Event: Text book page 48
· To every point (outcome) in the sample
space of an experiment S, we assign a
weight (or probability), ranging from 0 to 1,
such that the sum of all weights
(probabilities) equals 1.
· The weight (or probability) of an outcome
measures its likelihood (chance) of
occurrence.
· To find the probability of an event A, we
sum all probabilities of the sample points
in A. This sum is called the probability of
the event A and is denoted by P(A).

Definition 2.8:
The probability of an event A is the sum of
the weights (probabilities) of all sample
points in A. Therefore,
( ) 1
0 
 A
P
1)
( ) 1
=
S
P
2)
( ) 0
=

P
3)

Example 2.22:
A balanced coin is tossed twice. What is the
probability that at least one head occurs?
Solution:
S = {HH, HT, TH, TT}
A = {at least one head occurs}= {HH, HT, TH}
Since the coin is balanced, the outcomes are
equally likely; i.e., all outcomes have the
same weight or probability.

Outcome Weight
(Probability)
4w =1  w =1/4 = 0.25
P(HH)=P(HT)=P(TH)=P(TT)=0.25
HH
HT
TH
TT
P(HH) = w
P(HT) = w
P(TH) = w
P(TT) = w
sum 4w=1

The probability that at least one head
occurs is:
P(A) = P({at least one head occurs})=P({HH,
HT, TH})
= P(HH) + P(HT) + P(TH)
= 0.25+0.25+0.25
= 0.75

S
in
outcomes
of
no
A
in
outcomes
of
no
N
A
n
S
n
A
n
A
P
.
.
)
(
)
(
)
(
)
( =
=
=
Theorem 2.9:
If an experiment has n(S)=N equally likely
different outcomes, then the probability of
the event A is:

Example 2.25:
A mixture of candies consists of 6 mints, 4
toffees, and 3 chocolates. If a person makes
a random selection of one of these candies,
find the probability of getting:
(a) a mint
(b) a toffee or chocolate.
Solution:
Define the following events:
M = {getting a mint}
T = {getting a toffee}
C = {getting a chocolate}

Experiment: selecting a candy at random
from 13 candies
n(S) = no. of outcomes of the experiment of
selecting a candy.
= no. of different ways of selecting a
candy from 13 candies.
13
1
13
=






=

The outcomes of the experiment are
equally likely because the selection is made
at random.
(a) M = {getting a mint}
n(M) = no. of different ways of selecting
a mint candy from 6 mint candies
6
1
6
=






=
P(M )= P({getting a mint})=
( )
( ) 13
6
=
S
n
M
n

+






=
1
4
7
3
4
1
3
=
+
=






(b) TC = {getting a toffee or chocolate}
n(TC) = no. of different ways of
selecting a toffee or a chocolate
candy
= no. of different ways of selecting
a toffee candy + no. of different
ways of selecting chocolate
candy

= no. of different ways of
selecting a candy from 7 candies
7
1
7
=






=
P(TC )= P({getting a toffee or chocolate})
=
( )
( ) 13
7
=

S
n
C
T
n

Example 2.26:
In a poker hand consisting of 5 cards, find
the probability of holding 2 aces and 3 jacks.

Example 2.26:
In a poker hand consisting of 5 cards, find
the probability of holding 2 aces and 3 jacks.
Solution:
Experiment: selecting 5 cards from 52 cards.
n(S) = no. of outcomes of the experiment of
selecting 5 cards from 52 cards.
2598960
!
47
!
5
!
52
5
52
=

=






=

The outcomes of the experiment are equally
likely because the selection is made at
random.
Define the event A = {holding 2 aces and 3
jacks}
n(A) = no. of ways of selecting 2 aces and 3
jacks
= (no. of ways of selecting 2 aces)  (no.
of ways of selecting 3 jacks)
= (no. of ways of selecting 2 aces from 4
aces)  (no. of ways of selecting 3
jacks from 4 jacks)







=
2
4






3
4

!
2
!
2
!
4

=  24
4
6
!
1
!
3
!
4
=

=

P(A )= P({holding 2 aces and 3 jacks })
( )
( )
000009
.
0
2598960
24
=
=
=
S
n
A
n

2.5 Additive Rules:
Theorem 2.10:
If A and B are any two events, then:
P(AB)= P(A) + P(B) − P(AB)
Corollary 1:
If A and B are mutually exclusive (disjoint)
events, then:
P(AB)= P(A) + P(B)

Corollary 2:
If A1, A2, …, An are n mutually exclusive
(disjoint) events, then:
P(A1 A2 … An)= P(A1) + P(A2) +… + P(An)
( ) 
=
=
=
n
i
i
n
i
i
A
P
A
P 1
1
)
(


Corollary 3:
If A1, A2, …, An is a partition of sample space
S, then
P(A1  A2  …. An) =
P(A1) + P(A2) …+ P(An) = P(S) = 1.

Note: Two event Problems:
Total area= P(S)=1* In Venn diagrams,
consider the probability of an event A as
the area of the region corresponding to the
event A.
* Total area= P(S)=1
Total area= P(S)=1

* Examples:
P(A)= P(AB)+ P(ABC)
P(AB)= P(A) + P(ACB)
P(AB)= P(A) + P(B) − P(AB)
P(ABC)= P(A) − P(AB)
P(ACBC)= 1 − P(AB)
etc.,

Example 2.27:
The probability that Paula passes
Mathematics is 2/3, and the probability that
she passes English is 4/9. If the probability
that she passes both courses is 1/4, what is
the probability that she will:
(a) pass at least one course?
(b) pass Mathematics and fail English?
(c) fail both courses?

Solution:
Define the events:
M={Paula passes Mathematics}
E={Paula passes English}
We know that P(M)=2/3, P(E)=4/9, and
P(ME)=1/4.
(a) Probability of passing at least one
course is:
P(ME)= P(M) + P(E) − P(ME)
36
31
4
1
9
4
3
2
=
−
+
=

(b) Probability of passing Mathematics and
failing English is:
P(MEC)= P(M) − P(ME)
12
5
4
1
3
2
=
−
=
(c) Probability of failing both courses is:
P(MCEC)= 1 − P(ME)
36
5
36
31
1 =
−
=

Theorem 2.12:
If A and AC are complementary events,
then:
P(A) + P(AC) = 1  P(AC) = 1 − P(A)
Proof: Since A U AC = S and the sets A and AC
are disjoint, then
1 = P(S) = P(A U AC) = P(A) + P(AC).

Example 2.31 If the probabilities that an
automobile mechanic will service 3, 4, 5, 6, 7, or
8 or more cars on any given workday are,
respectively, 0.12, 0.19, 0.28, 0.24, 0.10, and
0.07, what is the probability that he will service
at least 5 cars on his next day at work?

Solution: Let E be the event that at least 5 cars
are serviced. Now, P(E) = 1 — P(Ec),
where Ec is the event that fewer than 5 cars are
serviced. Since
P(Ec) = 0.12+ 0.19 = 0.31,
it follows from Theorem 2.12 that
P(E) = 1 - 0.31 = 0.69.

Example 2.32 Suppose the manufacturer
specifications of the length of a certain type of
computer cable are 2000 ± 10 millimeters. In
this industry, it is known that small cable is just
as likely to be defective (not meeting
specifications) as large cable. That is, the
probability of randomly producing a cable with
length exceeding 2010 millimeters

is equal to the probability of producing a cable
with length smaller than 1990 millimeters. The
probability that the production procedure meets
specifications is known to be 0.99.
(a) What is the probability that a cable selected
randomly is too large?
(b) What is the probability that a randomly
selected cable is larger than 1990
millimeters?

Solution: Let M be the event that a cable meets
specifications. Let S and L be the events
that the cable is too small and too large,
respectively. Then
(a) P(M) = 0.99 and P(S) = P(L) = = 0.005.
(b) Denoting by X the length of a randomly
selected cable, we have
2
99
.
0
1−

P(1990 < X < 2010) = P(M) = 0.99.
Since P(X > 2010) = P(L) = 0.005
then
P(X > 1990) = P(M) + P(L) = 0.995.
This also can be solved by using Theorem 2.12:
P(X > 1990) + P(X < 1990) = 1.
Thus, P(X > 1990) = 1 – P(S)
= 1 - 0.005 = 0.995.

The probability of an event B occurring
when it is known that some event A has
occurred is called a conditional
probability and is denoted by P(B|A). The
symbol P (B | A) is usually read "the
probability that B occurs given that A
occurs"
or simply "the probability of B, given A."

Definition 2.9: The conditional probability
of B, given A, denoted by P(B|A) is defined
by
𝑷(𝑩|𝑨) =
)
𝐏(𝐀∩𝐁
)
𝐏(𝐀
provided P(A) > 0.

Example:
Suppose that our sample space S is
the population of adults in a small
town who have completed the
requirements for a college degree.
We shall categorize them according
to gender and employment status.
The data are given in the following
table.

Employed Unemployed Total
Male 460 40 500
Female 140 260 400
Total 600 300 900

One of these individuals is to be selected
at random for a tour throughout the
country to publicize the advantages of
establishing new industries in the town.
We shall be concerned with the
following events:
M: a man is chosen,
E: the one chosen is employed.
What is 𝐏 𝐌|𝐄 ?

Solution:
Let )
𝐧(𝐀 denote the number of elements in any
set A, then we write
𝐏(𝐌|𝐄) =
)
𝐧(𝐌 ∩ 𝑬
)
𝐧(𝑬
=
)
𝒏(𝑴 ∩ 𝑬
)
𝒏(𝑺
)
𝒏(𝑬
)
𝒏(𝑺
=
)
𝑷(𝑴 ∩ 𝑬
)
𝑷(𝑬
where )
𝐏(𝐌 ∩ 𝑬 and )
𝐏(𝑬 are found from the
original sample space S. To verify this result,
note that 𝐏(𝐄) =
𝟔𝟎𝟎
𝟗𝟎𝟎
=
𝟐
𝟑
, 𝐏(𝐌 ∩ 𝐄) =
𝟒𝟔𝟎
𝟗𝟎𝟎
=
𝟐𝟑
𝟒𝟓
Hence 𝐏(𝐌|𝐄) =
ൗ
𝟐𝟑
𝟒𝟓
ൗ
𝟐
𝟑
=
𝟐𝟑
𝟑𝟎

The probability that a regularly scheduled
flight departs on time is 𝐏 𝐃 = 𝟎. 𝟖𝟑;
the probability that it arrives on time is 𝐏 𝑨
= 𝟎. 𝟖2; and the probability that it
departs and arrives on time is 𝐏 𝐃 ∩ 𝐀
= 𝟎. 𝟕𝟖. Find the probability that a plane
(a)arrives on time given that it departed on
time,
(b)departed on time given that it has arrived
on time.

(a)The probability that a plane arrives
on time given that it departed on
time is
𝑷 𝑨|𝑫 =
𝑷 𝑨 ∩ 𝑫
𝑷 𝑫
=
𝟎. 𝟕𝟖
𝟎. 𝟖𝟑
= 𝟎. 𝟗𝟒
(b) The probability that a plane
departed on time given that it has
arrived on time is
𝐏 𝐃|𝐀 =
𝐏 𝐀 ∩ 𝐃
𝐏 𝐀
=
𝟎. 𝟕𝟖
𝟎. 𝟖𝟐
= 𝟎. 𝟗𝟓

Independent Events :
Although conditional probability allows for an
alteration of the probability of an event in the light
of additional material, it also enables us to
understand better the very important concept of
independence or, in the present context,
independent events. Consider the situation where
we have events 𝑨 and B and 𝑷 𝑨|𝑩 = 𝑷 𝑨
That is, the occurrence of B had no impact on the
odds of occurrence of A.

Definition 2.10:
Two events A and B are independent if
and only if
𝐏 𝐁|𝐀 = 𝐏 𝐁 or 𝐏 𝐀|𝐁 = 𝐏 𝐀 ,
provided the existences of the
conditional probabilities. Otherwise A
and B are dependent.

Example:
Consider an experiment in which 2 cards are
drawn in succession from an ordinary deck,
with replacement. The events are defined as
A: the first card is an ace,
B: the second card is a spade.
Find 𝐏 𝐁|𝐀 , 𝐏 𝐀|𝐁 .

Solution :
Since the first card is replaced, our
sample space for both the first and
second draws consists of 52 cards,
containing 4 aces and 13 spades.
Hence
𝐏 𝐁|𝑨 =
𝟏𝟑
𝟓𝟐
=
𝟏
𝟒
, 𝐏 𝑩 =
𝟏𝟑
𝟓𝟐
then 𝐏 𝐁|𝐀 = 𝐏 𝐁 =
𝟏
𝟒
and 𝐏 𝐀|𝑩 =
𝟒
𝟓𝟐
=
𝟏
𝟏𝟑
, 𝐏 𝑨 =
𝟒
𝟓𝟐
then 𝐏 𝐀|𝑩 = 𝐏 𝐁 =
𝟏
𝟏𝟑

Multiplicative Rules:
Multiplying the formula of
Definition 2.9 by P(A), we obtain
the following important
multiplicative rule, which enables
us to calculate the probability that
two events will both occur.

Theorem 2.13:
If in an experiment the events A
and B can both occur, then
𝐏 𝐀 ∩ 𝐁 = 𝐏 𝐀 𝐏 𝐁|𝐀
provided P(A) > 0.

Example:
Suppose that we have a fuse box
containing 20 fuses, of which 5 are
defective. If 2 fuses are selected at
random and removed from the box
in succession without replacing the
first, what is the probability that
both fuses are defective?

Solution:
Let A be the event that the first fuse is
defective and B the event that the second
fuse is defective; then we interpret 𝐏(
)
𝐀
∩ 𝐁 as the event that A occurs, and then
B occurs after A has occurred. The
probability of first removing a defective
fuse is
𝟏
𝟒
; then the probability of
removing a second defective fuse from
the remaining 4 is
𝟒
𝟏𝟗
. Hence
𝐏 𝐀 ∩ 𝑩 =
𝟏
𝟒
𝟒
𝟏𝟗
=
𝟏
𝟏𝟗

Theorem 2.14:
Two events A and B are independent if and only
if
𝐏 𝐀 ∩ 𝐁 = 𝐏 𝐀 𝐏 𝐁
Therefore, to obtain the probability that two
independent events will both occur, we simply
find the product of their individual probabilities.

Example:
An electrical system consists of four
components as illustrated in Figure 1 below.
The system works if components A and B work
and either of the components C or D work. The
reliability (probability of working) of each
component is also shown in Figure 1. Find the
probability that
(a)the entire system works, and
(b) the component C does not work, given that
the entire system works. Assume that four
components work independently.

Figure 1: An electrical system
0.9
0.8
0.9
0.8

Solution:
In this configuration of the system, A, B, and
the subsystem C and D constitute
a serial circuit system, whereas the subsystem
C and D itself is a parallel circuit system.
(a) Clearly the probability that the entire
system works can be calculated as the
following:

𝐏 𝐀 ∩ 𝐁 ∩ 𝐂 ∪ 𝐃 = 𝐏 𝐀 𝐏 𝐁 𝐏 𝐂 ∪ 𝐃
= 𝐏 𝐀 𝐏 𝐁 𝟏 − 𝐏 𝐂′ ∩ 𝐃′ = 𝐏 𝐀 𝐏 𝐁 𝟏 − 𝐏 𝐂′ 𝐏 𝐃′
= (𝟎. 𝟗)(𝟎. 𝟗) )
𝟏 − (𝟏 − (𝟎. 𝟖))(𝟏 − (𝟎. 𝟖) = 𝟎. 𝟕𝟕𝟕𝟔
The equalities above hold because of the
independence among the four components.

Conditional
Probability - Bayes’
Rule

Theorem 2.15:
If, in an experiment, the events 𝐀𝟏, 𝐀𝟐, … , 𝐀𝐤
can occur, then
If the events 𝐀𝟏, 𝐀𝟐, … , 𝐀𝐤 are independent, then
𝐏 𝐀𝟏 ∩ ⋯ ∩ 𝐀𝐤 = 𝐏 𝐀𝟏 𝐏 𝐀𝟐 … 𝐏 𝐀𝐤

Example:
Three cards are drawn in succession,
without replacement, from an ordinary
deck of playing cards. Find the
probability that the event 𝐏(
)
𝐀𝟏 ∩ 𝐀𝟐
∩ 𝐀𝟑 occurs, where
𝐀𝟏 is the event that the first card is a red
ace, 𝐀𝟐 is the event that the second card
is a 10 or a jack, and 𝐀𝟑 is the event that
the third card is greater than 3 but less
than 7.

Solution: First we define the events
𝐀𝟏: the first card is a red ace,
𝐀𝟐: the. second card is a 10 or a jack,
𝐀𝟑: the third card is greater than 3 but less than
7. Now
𝐏 𝐀𝟏 =
𝟐
𝟓𝟐
, 𝐏 𝐀𝟐|𝐀𝟏 =
𝟖
𝟓𝟏
, 𝐏 𝐀𝟑|𝐀𝟏 ∩ 𝐀𝟐
=
𝟏𝟐
𝟓𝟎
and hence, by Theorem 2.15,
𝐏 𝐀𝟏 ∩ 𝐀𝟐 ∩ 𝐀𝟑 = 𝐏 𝐀𝟏 𝐏 𝐀𝟐|𝐀𝟏 𝐏 𝐀𝟑|𝐀𝟏 ∩ 𝐀𝟐
=
𝟐
𝟓𝟐
𝟖
𝟓𝟏
𝟏𝟐
𝟓𝟎
=
𝟖
𝟓𝟓𝟐𝟓

Bayes’Rule:
Consider the following figure
Figure 2:Venn diagram for the events A, E,
E E’
A
A

Referring to Figure 2, we can write A as
the union of the two mutually exclusive
events 𝐄 ∩ 𝐀 and 𝐄′
∩ 𝐀. Hence
𝐀 = 𝐄 ∩ 𝐀 ∪ 𝐄′
∩ 𝐀
and by additive rule and multiplicative
rule, we can write
𝐏(𝐀) = 𝐏 𝐄 ∩ 𝐀 ∪ 𝐄′
∩ 𝐀
= 𝐏 𝐄 ∩ 𝐀 + 𝐏 𝐄′
∩ 𝐀
= 𝐏 𝐄 𝐏 𝐀|𝐄 + 𝐏 𝐄′
𝐏 𝐀|𝐄′

A generalization of the foregoing illustration
to the case where the sample space is
partitioned into k subsets is covered by the
following theorem, sometimes called the
theorem of total probability or the rule of
elimination.

Theorem 2.16:
If, the events 𝑩𝟏, 𝑩𝟐, … , 𝑩𝐤 constitute a
partition of the sample space S such that
𝐏 𝐁𝐢 ≠ 𝟎 for 𝐢 = 𝟏, 𝟐, … , 𝐤
then for any event A of S,
𝐏 𝐀 = σ𝐢=𝟏
𝐤
𝐏 𝐁𝐢 ∩ 𝐀 = σ𝐢=𝟏
𝐤
𝐏 𝐁𝐢 𝐏 𝐀|𝐁𝐢 .

B2
B1
Bk
Figure 3: Partitioning the sample space S
A

Proof:
Consider the Venn diagram of the figure 3.
The events A is seen to be the union of the
mutually exclusive events
𝐁𝟏 ∩ 𝐀, 𝐁𝟐 ∩ 𝐀, … , 𝐁𝐤 ∩ 𝐀
that is
𝑨 = 𝑩𝟏 ∩ 𝑨 ∪ 𝑩𝟐 ∩ 𝑨 ∪ ⋯ ∪ 𝑩𝒌 ∩ 𝑨
Using theorem 2.10 (additive rule) and
theorem 2.13 (multiplicative rule) we have

𝐏 𝐀 = 𝐏 𝐁𝟏 ∩ 𝐀 ∪ 𝐁𝟐 ∩ 𝐀 ∪ ⋯ ∪ 𝐁𝐤 ∩ 𝐀
= 𝐏 𝐁𝟏 ∩ 𝐀 + 𝐏 𝐁𝟐 ∩ 𝐀 + ⋯ + 𝐏 𝐁𝐤 ∩ 𝐀
= ෍
𝐢=𝟏
𝐤
𝐏 𝐁𝐢 ∩ 𝐀 = ෍
𝐢=𝟏
𝟑
𝐏 𝐁𝐢 𝐏 𝐀|𝐁𝐢

Example:
In certain assembly plant In a certain assembly
plant, three machines, 𝐁𝟏, 𝐁𝟐, 𝐁𝟑, make 30%,
45%, and 25%, respectively, of the products. It
is known from past experience that 2%, 3%,
and 2% of the products made by each machine,
respectively, are defective. Now, suppose that a
finished product is randomly selected. What is
the probability that it is defective?

Solution:
Consider the following events:
A: the product, is defective,
𝐁𝟏: the product is made by machine 𝐁𝟏,
𝐁𝟐: the product is made by machine 𝐁𝟐,
𝐁𝟑: product is made by machine 𝐁𝟑.
Applying the rule of elimination, we can
write
𝐏 𝐀
= 𝐏 𝐁𝟏 𝐏 𝐀|𝐁𝟏 + 𝐏 𝐁𝟐 𝐏 𝐀|𝐁𝟐
+ 𝐏 𝐁𝟑 𝐏 𝐀|𝐁𝟑

B1 P(A|B1)=0.02
P(B1)=0.3 A
P(B2)=0.45 P(A|B2)=0.03
B2 A
P(B3)=0.25
P(A|B3)=0.02
B3 A
Figure 4:Tree diagram

Referring to the tree diagram of Figure 4, we
find that the three branches give
the probabilities
𝐏 𝐁𝟏 𝐏 𝐀|𝐁𝟏 = (0.3) (0.02) = 0.006.
𝐏 𝐁𝟐 𝐏 𝐀|𝐁𝟐 = (0.45)(0.03) = 0.0135,
𝐏 𝐁𝟑 𝐏 𝐀|𝐁𝟑 = (0.25)(0.02) = 0.005,
and hence
𝐏 𝐀 = 0.006 + 0.0135 + 0.005 = 0.0245.

Theorem 2.17: (Bayes' Rule)
If the events 𝑩𝟏, 𝑩𝟐, … , 𝑩𝐤 constitute a
partition of the sample space S such that
𝐏 𝐁𝐢 ≠ 𝟎 for 𝐢 = 𝟏, 𝟐, … , 𝐤, then for any
event A in S such that 𝐏 𝐀 ≠ 𝟎,
𝐏 𝐁𝐫|𝐀 =
𝐏 𝐁𝐫 ∩ 𝐀
σ𝐢=𝟏
𝐤
𝐏 𝐁𝐢 ∩ 𝐀
=
𝐏 𝐁𝐫 𝐏 𝐀|𝐁𝐫
σ𝐢=𝟏
𝐤
for 𝐫 = 𝟏, 𝟐, … , 𝐤

Proof:
By the: definition of conditional
probability,
𝐏 𝐀
and then using Theorem 2.16 in the
denominator, we have
σ𝐢=𝟏
𝐤
𝐏 𝐁𝐢 ∩ 𝐀

Applying theorem 2.13 to both numerator and
denominator, we obtain
𝐏 𝐁𝐫 𝐏 𝐀|𝐁𝐫
σ𝐢=𝟏
𝐤
which completes the proof.

Example :
With reference to the last example above, if a
product was chosen randomly and found to be
defective, what is the probability that it was
made by machine 𝐁𝟑 ?
Solution : Using Bayes' rule to write
𝐏 𝐁𝟑|𝐀
=
𝐏 𝐁𝟑 𝐏 𝐀|𝐁𝟑
𝐏 𝐁𝟏 𝐏 𝐀|𝐁𝟏 + 𝐏 𝐁𝟐 𝐏 𝐀|𝐁𝟐 + 𝐏 𝐁𝟑 𝐏 𝐀|𝐁𝟑

and then substituting the: probabilities calculated
in last example, we have
𝐏 𝐁𝟑|𝐀 =
𝟎. 𝟎𝟎𝟓
𝟎. 𝟎𝟎𝟔 + 𝟎. 𝟎𝟏𝟑𝟓 + 𝟎. 𝟎𝟎𝟓
=
𝟎. 𝟎𝟎𝟓
𝟎. 𝟎𝟐𝟒𝟓
=
𝟏𝟎
𝟒𝟗
In view of the fact that a defective product was
selected, this result suggests that it probably was
not made by machine 𝐁𝟑.

Lesson4 Probability of an event [Autosaved].pdf

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Lesson4 Probability of an event [Autosaved].pdf