Transcript Chapter V

Probability
Experiments, Sample Spaces &
Events
Definitions
1. Experiment
An activity with observable outcomes (results). A special
case of experiments are those in which the considered
outcomes involve chance.
Examples:
1. Tossing a coin with the considered outcomes are the
coin falling head or tail. we will refer to this experiment
by the usual tossing of a coin experiment.
2. Rolling a dice with the considered outcomes are the
numbers of the dots (the integers from 1 to 6) showing
up. we will refer to this experiment by the usual dice
experiment.
2. Sample Space
A simple space is the set of possible outcomes of an experiment.
Examples (1):
1. In the usual experiment of tossing a coin with the considered
outcomes being the coin falling head or tail, the sample space
contains two elements: head and tail. We write:
S = { head , tail }
2. In the usual experiment of rolling a dice with the considered
outcomes are the number of dots (the integers 1 to 6) showing up
(on the top face), the sample space contains 6 elements: the
integers from 1 to 6. We write:
S = {1,2,3,4,5,6}
2. Event
An event is ny subset of the sample space of an
experiment.
Examples(2) :
1. In the usual experiment of tossing a coin with the
considered outcomes are the coin falling head or tail,
where the sample space is S = { head , tail }, each of the
following subsets of S is an event: Φ, S, { tail }, { head }.
2. In the usual experiment of rolling a dice with the
considered outcomes are the number of dots showing
up, where the sample space is S = {1,2,3,4,5,6}, each of
the following subsets of S is an event: Φ, S, {1} , {2} , {3}
,…… {6} , {1,2} , {1,3} ,….. , {5,6} , {1,2,3} , {1,2,4},……..,
{4,5,6} , {1,2,3,4} , {1,2,3,5} ,……,{3,4,5,6} , {1,2,3,4,5}
,…. {1,2,3,4,6} , {2,3,4,5,6}.
Notice
1. Notice that within a given experiment, the sample space
is the universal set containing all of the considered
outcomes for that experiments and the events are the
subsets of this universal set, this includes the empty set
and the sample space itself.
2. The sample space is referred to as the certain event. It is
called certain, because it must occur, since there is no
outcome not belonging to this set.
2. The empty set is referred to as the impossible event. It is
called impossible, because it can not occur, since no
outcome belongs to this set.
Simple Events
Let S = {s1, s2, s3,……,sn} be a sample
space.
A simple event in singleton subset ( a subset
consisting of exactly one element) of S.
Thus, {sk} is a simple event for any positive
integer k ≤ n.
That’s all of the following events are simple
events:
{s1} , {s2} , {s3} , {s4} ,….,{sn}
The algebra of events
1. Since events are subsets of a universal set (the sample space, which
is the universal set here), then all set operations learned in section
IV-01 apply. Thus we can talk about, union and intersection of
events and a complement of an event.
2. A nonintersecting events ( that’s having an empty intersection; that’s,
the intersection of them is equal to Φ) are referred to as mutually
exclusive, because they can not happen at the same time.
For example, the event A = {1,2,3} and B = {5,6}, in the In the usual dice
experiment, are mutually exclusive, since their intersection is the
empty set. If the top face is one of the three numbers (of dots)
belonging to the set A, it certainly, can not be one of the numbers
belonging to the set B, since there is no common element of these
two sets.
Notice that if A is a subset of the complement of B, then no element of A
belongs to B, which means that the intersection of A and B is empty,
and so they are mutually exclusive. This will include the case when
A is equal to BC.
Example (3)
1. In the usual dice experiment, let A, B and C, be the following events:
A = {1,4,6} and B = {1,2,4,5} and C = {3,5}
Find the following:
1. The union and intersection of the events A and B.
2. The complement of the event A.
3. Which two of the sets A, B and C are mutually exclusive.
Solution:
1. AUB = {1,4,6} U {1,2,4,5} = {1,2,4,5,6}
A∩B = {1,4,6} ∩ {1,2,4,5} = {1,4}
2. AC = S - {1,4,6} = {1,2,3,4,5,6} - {1,4,6} = {2,3,5}
3. Since, A∩C = {1,4,6} ∩ {3,5} = Φ, then A and C are mutually exclusive.
Notice that C is a subset of the complement of A
C = {3,5} is a subset of AC = {2,3,5}
Example (4)
In the experiment consisting of tossing a coin three times
and considering the sequence of heads and tails
outcomes. Find the following:
1. The sample space
2. The event E that exactly two heads appear.
3. The event F that at least one tail appears.
4. The event G that two heads appear in a sequence.
5. The event A that three heads appear.
Solution: First let’s study the tree diagram of this
experiment.
H
H
(H,H,H)
T
(H,H,T)
H
H
(H,T,H)
T
T
(H,T,T)
Third Toss
Second Toss
(T,H,H)
H
First Toss
H
(T,H,T)
T
T
H
(T,T,H)
T
(T,T,T)
T
Solution
1. The sample space S
= {(H,H,H), (H,H,T), (H,T,H) , (H,T,T), , (T,H,H), , (T,H,T), ,
(T,T,H), , (T,T,T }
Convention: For the sake of brevity, we will use the
notation HHH for (H,H,H), HHT for (H,H,T)……etc. Thus:
S = { HHH, HHT, HTH, HTT,THH, THT, TTH, TTT}
2. E (the event that exactly 2 heads appear)
= { HHT,HTH,THH}
3. F (the event that at least one tail appears)
= {HHT, HTH, HTT,THH, THT, TTH, TTT}
4. The event G that two heads appear in sequence
= { HHT,THH,HHH}
5. The event A that three heads appear = {HHH}
Example (5)
In the experiment consisting of tossing a pair of
dice and considering the numbers of dots falling
uppermost on each dice . Find the following:
1. The sample space
2. The events En that the sum of the numbers of
dots falling uppermost is n, for n = 2,3,4, ….,12
First study the table on the next slide
Sum of
uppermost numbers
Event
2
3
{ (1 ,1) }
4
5
{ (1,3) , (2 , 2) , (3 ,1) }
6
7
{ (1 , 5 ) , (2 , 4) , (3 , 3) , (4 , 2) , (5 ,1) }
8
{ (2 , 6) , (3 , 5) , (4 , 4) , (5 ,1) , (6 , 2) }
9
10
11
12
{ (1,2) , (2 ,1) }
{ (1, 4) , (2 , 3) , (3 , 2) , (4 ,1) }
{(1, 6) , (2 , 5) , (3,4) , (4 , 3) , (5 , 2) , (6 ,1)}
{ (3 , 6) , (4 , 5) , (5 , 4) , (6 , 3) }
{ (4 , 6) , (5 , 5) , (6 , 4) }
{ (5 , 6) , (6 , 5) }
{(6,6)}
Solution
1. The sample space S
= { (k,m) : where k and m are the numbers of dots appearing
uppermost on the first and the second dice respectively}
= {(1,1), (1,2), ….(1,6), (2,1), (2,2),…..,(2,6),…..(6,1),
(6,2),…..(6,6) }
2.
E2 = {(1,1)}
E3 = {(1,2), (2,1)}
E4 = {(1,3), (2,2), (3,1)}, Find , E5 and E6
E7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
E10 = {(4,6), (5,5), (4,6)}, Find E8 , E9
E11 = {(5,6), (6,5)}
E12 = {(6,6)}
Example (6)
In the experiment consisting of guessing the
number of tickets that will be sold for a play . If
there are 40 seats in the theater. Find the
following:
1. The sample space
2. The event E that fewer than 30 tickets will be
sold.
3. The event F that at least half of tickets will be
sold.
Solution
The number of the tickets that can be sold could run from 0
to 40
1. The sample space S = { m : m the number of tickets that
could be sold}
= {0,1,2,3,4,5……….,40 }
2. E (the event that fewer than 30 tickets will be sold)
= { 0,1,2,3,…….29}
3. E (the event that the theater will be at least half full, i.e.
at least half of the tickets will be sold)
= { 20,21,22,23,…….40}
Example (7)
Infinite Sample Space
In the experiment consisting of guessing the
distance that an animal travels far away from its
original home.
1. The sample space
2. The event E that the animal travels no more
than 1000 KM away from home.
3. The event F that the animal travels more than
1000 KM away from home.
4. The event G that the animal travels between
500 KM and 1000 Km (inclusive) away from
home.
Solution
1. The distance covered by the animal away from home is a
nonnegative number
The sample space S
= {x: x = the distance the animal travels away from home} =
{ x: x ≥ 0 } = [0,∞) This is an infinite set
2. E (The event E that the animal travels no more than
1000 KM away from home)
= {x: 0 ≤ x ≤ 1000 } = [0 , 1000]
3. F (the event that the animal travels more than 1000 KM
away from home)
= {x: x > 1000 } = (1000,∞)
4. G (the event that the animal travels between 500 KM
and 1000 Km (inclusive) away from home)
= {x: 500 ≤ x ≤ 1000 } = [500 , 1000]
Probability
Relative Frequency & Empirical Probability
of an Event & Probability Distribution
1. In an experiment, repeatable under independent and
similar condition, if in n trails, an event E occurs m times,
then the ratio m/n is called the relative frequency of the
event E.
2. If the relative frequency of an event E approaches some
value as n becomes larger and larger, then this value is
called the empirical probability of the event E, and
denoted by p(E).
An event’s probability is a measure of the proportion of the
time that the event will occur.
The meaning of the phrase “the relative frequency of E
approaches some value”, can be understood by studying
the table on the next slide.
The relative frequency of a coin’s toss
results in showing tail – approaching 0.5 = ½
n (number of
tosses)
10
m (number of
tails)
4
m/n ( relative
frequency
0.4000
100
58
0.5800
1000
492
0.4920
10000
5034
0.5034
20000
10024
0.5012
40000
20032
0.5008
Probability Function
Let S = {sk : k = 1, 2, 3, ….,n}
= {s1 , s2 , s3 , s4 ,………., sn }
be a sample space
1. The function p assigning probability to an event E is
called a probability function.
The probability p({sk}) of the simple event {sk} is written in
short as p(sk) , for any k = 1, 2, 3, ….,n.
2. The property function has the following rules:
a. 0 ≤ p(sk) ≤ 1 ; k = 1, 2, 3, ….,n
b. p(s1) + p(s2) +……. + p(sn) = 1
C. p( {si} U {sj} ) = p(si) + p(sj) ; i not equal j
Probability distribution table
The table assigning probabilities to each of the simple events is called
a Probability distribution for the experiment.
Simple Event Probability
{s1}
P({s1})
Or in short p(s1)
{s2}
P({s2})
Or in short p(s2)
----
------
-------------------
{sn}
P({sn})
Or in short p(sn)
General Rules of Probability
1. The property function has the following rules:
a. p(Φ) = 0 ≤ p(E) ≤ 1 = p(S) ; for any event E in the sample
space S
2. p(EUF) = p(E) + p(F) – p(E∩F)
Special case: If E & F are mutually exclusive, then:
p(EUF) = p(E) + p(F) (Why?)
3. p(EC) = 1 – p(E)
Probability in a Uniform (Equiprobable)
Sample Space
1. A sample space is said to be uniform (equiprobable) if
simple events are equally likely to occur.
2. If S is uniform (equiprobable) and n(S) = k, then:
p({s}) = 1 / k; s in S.
Example (8)
For the experiment of the usual rolling of the dice:
1. Exhibit the probability distribution table
2. Compute the probability that the dice shows an odd
number of dots.
3. Compute the probability that the dice shows less than 5
dots.
Solution:
1. The sample space S = {1,2,3,4,5,6} consists of six
outcomes. Thus each simple event is to be assigned a
probability of 1/6. See the probability distribution table on
the next slide
Simple Event
Probability
{1}
1/6
{2}
1/6
{3}
1/6
{4}
1/6
{5}
1/6
{6}
1/6
2. The event E that the dice shows an odd number
of dots
= {1 , 3 , 5 }
→ p(E) = p({1}) + p({3}) + p({5})
=1/6 +1/6 + 1/6 = 3(1/6) = ½
3. The event F that the dice shows less than 5 dots
= { 1, 2 , 3 , 4 }
→ p(E) = p({1}) + p({2}) + p({3}) + p({4})
=1/6 +1/6 + 1/6 + 1/6 = 4(1/6) = 2/3
Example (9)
A pair of dice is rolled. Calculate the following probabilities:
1. The probability that the dice show the same number.
2. The probability that the sum of the numbers shown by
the two dice is 10.
3. The probability that the sum of the numbers shown by
the two dice is 12.
4. The probability that the number shown on one dice is
exactly five times that shown on the other.
Solution:
See Example (5) for the sample space S of this
experiment.
The sample spaces S contains (consists of) 36 outcomes,
hence each simple event is to be assigned the
probability 1/36.
1. The event E that the two dice show the same number is:
E = { ( 1, 1 ) , ( 2, 2 ) , ( 3, 3 ) , ( 4, 4 ) , ( 5, 5 ) , ( 6, 6 ) }
This event contains 6 element outcomes). Since (E is the
union of 6 simple events), then
p(E) =
p({( 1, 1 ) }) + p({( 2, 2 ) }) + p({( 1, 1 ) }) + …….p({( 6, 6 ) })
= 1/36 + 1/36+ ……+1/36 ( 6 term)
= 6(1/36) = 1/6
2. The probability that the sum of the numbers shown by
the two dice is 10.
The event F that the sum of the numbers shown by the two
dice is 10 is:
F = { ( 4, 6 ) , ( 5, 5 ) , ( 6, 4 ) }
This event contains 3 element. Since (F is the union of 3
simple events), then
p(F) = p({( 4, 6 ) }) + p({( 5, 5 ) }) + p({( 6, 4 ) })
= 1/36 + 1/36 + 1/36
= 3(1/36) = 1/12
3. The probability that the sum of the numbers shown by
the two dice is 12.
The event G that the sum of the numbers shown by the two
dice is 12 is the singleton set:
G = { ( 6, 6 ) }
This event contains only one element.
p(G) = p({( 6, 6 ) }) = 1/36
4. The probability that the number shown on one dice is
exactly five times that shown on the other.
The event H that the number shown on one dice is exactly
5 times that shown on the other is the set:
H = { ( 1, 5) , ( 5, 1 )}
This event contains 2 elements (outcomes).
p(H) = p({ ( 1, 5 )}) + p({ ( 5, 1 )})
= 1/36 + 1/36
= 2(1/36) = 1/18
Example (10)
The next slide shows the data gathered from an
experiment of 200 test runs of the distance a
prototype electric car is able to cover, using a
fully charged battery of a certain brand.
1. Find the sample space S of this experiment.
2. Find the empirical property distribution of this
experiment.
3. Compute the probability that the such a car will
be able to cover more than 150 KM.
4. Compute the probability that the such a car will
be able to cover no more than 150 KM
Distant covered in KM, x
Frequency of occurrence
x ε (0 , 50]
x is less than 50
x ε (50 ,100]
4
10
x is greater than 50 but less or equal 100
x ε (100 , 150]
30
x is greater than 100 but less or equal 150
x ε (150 , 200]
100
x is greater than 150 but less or equal 200
x ε (200 , 250]
40
x is greater than 200 but less or equal 250
x ε (250 , ∞)
x is greater than 250
16
Solution
1. The sample space S of this experiment
= { s1 , s2 , s3 , s4 , s5 , s6 }
S1 denotes the outcome that the car was able to cover a
distance no more than 50 KM
S2 denotes the outcome that the car was able to cover a
distance greater than 50 KM, but less or equal 100 KM.
S3 denotes the outcome that the car was able to cover a
distance greater than 100 KM, but less or equal 150 KM
S4 denotes the outcome that the car was able to cover a
distance greater than 150 KM, but less or equal 200 KM
S5 denotes the outcome that the car was able to cover a
distance greater than 200 KM, but less or equal 250 KM
S6 denotes the outcome that the car was able to cover a
distance more than 250 KM
2. Find the empirical probability distribution of this
experiment.
p(s1) = relative frequency of s1
= number of trails in which s1 occurs / total number of trails
= 4 / 200 = 2 / 100 = 0.02
p(s2) = 10 / 200 = 5 / 100 = 0.05
p(s3) = 30 / 200 = 15 / 100 = 0.15
p(s4) = 100 / 200 = 50 / 100 = 0.50
p(s5) = 40 / 200 = 20 / 100 = 0.20
p(s6) = 16 / 200 = 8 / 100 = 0.08
This results in the probability distribution table of the next
slide
Distant covered in KM, x
Frequency of occurrence
{s1}
0.02
The distance is less than 50
{s2}
0.05
The distance is greater than 50 but less or
equal 100
{s3}
0.15
The distance is greater than 100 but less
or equal 150
{s4}
0.50
The distance is greater than 150 but less
or equal 200
{s5}
0.20
The distance is greater than 200 but less
or equal 250
{s6}
The distance is greater than 250
0.08
3. The event E that such a car will be able to cover more
than 150 KM is:
E = {s4 , s5 , s6 }
→ p(E) = p(s4) + p(s5) + p(s6)
= 0.50 + o.20 + 0.08 = 0.78
4. The event F that such a car will be able to cover no more
than 150 KM is:
F = {s1 , s2 , s3 }
→ p(E) = p(s1) + p(s2) + p(s3)
= 0.02 + o.05 + 0.15 = 0.22
Notice that F = EC and that p(F) = 1 – p(E) = 1 0.78 = 0.22
Example (11)
The next slide shows the probability distribution
with a final exam scores of a Math course. If we
select at random a student who has done the
exam, what is the probability that her score will
be:
1. More than 40.
2. Less than or equal 50
3. greater than 40 but less or equal 70
4. greater than 40 but less or equal 60
Score
Probability
{s1}
0.01
The score is greater than 70
{s2}
0.07
The score is greater than 60 but less or
equal 70
{s3}
0.19
The score is greater than 50 but less or
equal 60
{s4}
0.23
The score is greater than 40 but less or
equal 50
{s5}
0.31
The score is greater than 30 but less or
equal 40
{s6}
The score is less than 30
0.19
Solution
1. The sample space S of this experiment
= { s1 , s2 , s3 , s4 , s5 , s6 }
S1 denotes the outcome that the score is greater than 70
S2 denotes the outcome that the score is greater than 60,
but less or equal 70
S3 denotes the outcome that the score is greater than 50,
but less or equal 60
S4 denotes the outcome that the score is greater than 40,
but less or equal 50
S5 denotes the outcome that the score is greater than 30,
but less or equal 40
S6 denotes the outcome that the score is less or equal 30
1. The event E that the score is greater than 40
E = { s1 , s2 , s3 , s4 }
→ p(E) =p(s1) + p(s2) + p(s3) + p(s4)
= 0.01 + 0.07 + 0.19 +0.23 = 0.5 = ½
2. The event F that the score is less than or equal 50
F = { s4 , s5 , s6 }
→ p(F) =p(s4) + p(s5) + p(s6)
= 0.23 + 0.31 + 0.19 = 0.73
3. The event G that the score is greater than 40 but less than or equal 70
F = { s4 , s3 , s2 }
→ p(F) = p(s4) + p(s5) + p(s6)
= 0.23 + 0.19 + 0.07 = 0.49
4. The event H that the score is greater than 40 but less or equal 60
H = {s3 , s4 , }
→ p(H) = p(s3) + p(s4)
= 0.19 +0.23 = 0.42
Example (12)
A card is drawn from a deck of 52 playing cards. What’s the probability
that it is a king or a diamond.
Solution:
The sample space consists of 52 outcomes.
Thus, there are 52 simple events, each with a property equal to 1/52.
The event E that the outcome is a king consists of 4 outcomes
→ p(E) = 1/52 + 1/52 + 1/52 + 1/52 = 4 (1/52) = 4/52 = 1/13
The event F that the outcome is a diamond consists of 13 outcomes
→ p(F) = 1/52 + 1/52 + …… + 1/52 = 13 (1/52) = 13/52 = ¼
Notes:
1. The event H that the outcome is a king or a diamond = EUF
We have:
p(EUF) = p(E) + p(F) – p(E∩F) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13
2. The event G that the outcome is a king and a diamond consists of
one outcome → p(G) = 1/52. Notice that G = E∩F.
Example (13)
The data shows that 3% of a brand of TV sold experience video problems, 1%
audio problems and 0.1% both type of problems before the expiration of the
warranty. What’s the probability that a purchased such TV
1. will experience video or audio problems before the expiry of the warranty.
1. will not experience video or audio problems before the expiry of the warranty
Solution:
1. Let E denote the event that the TV will experience video problems and F the
event that the TV will experience audio problems. Then E∩F is the event
that the TV will experience both video and audio problems, while EUF is the
event that the TV will experience video or audio problems.
→ p(E) = 0.03 , p(F) = 0.01
and
p(E∩F) = 0.001
We have;
p(EUF) = p(E) = p(F) – p(E∩F) = 0.03 + 0.01 – 0.001
= 0.040 -0.001 = 0.039
2. Let G denote the event that the TV will not experience video or audio
problems, then G = (EUF)C , and so p(G) = 1 - p(EUF) = 1 – 0.039= 0.961
Example (14)
Let E & F be mutually exclusive events and that p(E) = 0.1 and p(F) =
0.6. Compute:
1. p(E∩F)
2. p(EUF)
3. p(EC)
4. p(EC∩FC)
5. p(ECUFC)
Solution:
1. p(E∩F) = 0, since E∩F = Φ
2. p(EUF) = p(E) + p(F) = 0.1 + 0.6 = 0.7
3. p(EC) = 1 – P(E) = 1 – 0.1 = 0.9
4. EC∩FC = (EUF)C → p(EC∩FC)= p(EUF)C = 1 - p(EUF) = 1 – 0.7 = 0.3
5. ECUFC = (E ∩ F)C =(Φ) C = S → p(EC∩FC)= p(S) = 1
Another way: ECUFC = (E ∩ F)C =(Φ) C
→ p(EC∩FC) = p (Φ) C= 1 - p (Φ) = 1 – 0 =1
Example (16)
Let p(E) = 0.2 , p(F) = 0.1 and p(E∩F) =
0.05. Compute:
1. p(EUF) 2. p(EC∩FC)
Solution:
1. p(EUF) = p(E) + p(F) - p(E∩F) = 0.2 + 0.1
- 0.05 = 0.25
2. EC∩FC = (EUF)C → p(EC∩FC)= p(EUF)C
= 1 - p(EUF) = 1 – 0.25 = 0.75