Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 2 Probability
Page 88 Problem 1 Answer
Given: A small manufacturing company will start operating a night shift.
There are 20 machinists employed by the company.
To find how many different crews are possible if a night crew consists of 3 machinists.
We will use a combination formula:
Ck,n=(n/k)=n!/k!(n−k)!
We have n=20 individuals
We have to get a subset of k=3
Hence:
C3,20=(20/3)
C3,20=20!/3!(20−3)!
C3,20=1140.
1140 different crews are possible if a night crew consists of 3 machinists.
Page 88 Problem 2 Answer
Given: A small manufacturing company will start operating a night shift.
There are 20 machinists employed by the company.
To find the number of crews not having the best machinist given that the machinist are ranked 1,2,…,20 in order of competence.
We will use combination formula:
Ck,n=(n/k)
Ck,n =n!/k!(n−k)!
We have to take out the best machinists .
Hence: n=19
We need to crew crew of 3
Hence k=3
Therefore we have:
C3,19=(19/3)
C3,19 =19!/3!(19−3)!
C3,19 =969
The number of crews not having the best machinist given that the machinist are ranked1,2,…,20 in order of competence is 969.
Probability And Statistics For Engineering 8th Edition Supplementary Exercise Solutions Page 88 Problem 3 Answer
Given: A small manufacturing company will start operating a night shift.
There are 20 machinists employed by the company.
To find the number of crews that will have at least 1 of the 10 best machinists.
We will use combination formula:
Ck,n=(n/k)
Ck,n =n!/k!(n−k)!
There are a total of 1140
Hence if we subtract 1140 from the number of to select 3 machinists from of group of 10 bottom machinists we will get the required answer.
Hence we have:
C3,20−C3,10
C3,20−C3,10 =1140−(10/3)
C3,20−C3,10 =1140−120
C3,20−C3,10 =1020
The number of crews that will have at least 1 of the 10 best machinists is 1020.
Page 88 Problem 4 Answer
Given: A small manufacturing company will start operating a night shift.
There are 20 machinists employed by the company.
To find the probability that the best machinist will not work that night given that one of these crews is selected at random to work on a particular night.
We will use combination formula:
Ck,n=(n/k)
=n!/k!(n−k)!
We know that for an event A we have
P(A)= number of favorable outcomes in A number of outcomes in the sample space
Now here A is the event that the best machinist will not work that night
We have found the favorable outcomes in part b) 969 and the total number of outcomes in part a) 1140
Hence P(A)=C3,19/C3,20
P(A)=969/1140
P(A) =0.85
The probability that the best machinist will not work that night is 0.85.
Page 89 Problem 5 Answer
Given: One satellite is scheduled to be launched from Cape Canaveral in Florida, and another launching is scheduled for Vandenberg Air Force Base in California.
Given: A denotes the event that the Vandenberg launch goes off on schedule, and B represent the event that the Cape Canaveral launch goes off on schedule.
Given that A and B are independent events.Given:P(A)>P(B),P(A∪B)=.626, and P(A∩B)=.144
To determine: P(A) and P(B)
We will use the properties: P(A∪B)=P(A)+p(B)-P(A∩B) and
P(A∩B)=P(A).P(B)
Since there are two questions and two unknowns,
Substitute the values in one of the equation,
P(A∩B)=P(A)⋅P(B)
0.144=P(A)⋅P(B)
P(A)=0.144
P(B)
Substitute the values in the other equation,
P(A∪B)=P(A)+P(B)−P(A∩B)
0.626=P(A)+P(B)−0.144
Now replace the value of P(A),
0.626=0.144
P(B)+P(B)−0.144/0.626⋅P(B)=0.144+P(B)2−0.144⋅P(B)
0=0.144+P(B)2−0.77⋅P(B)
After solving the quadratic equation, the solutions are,
P(A) or P(B)=0.45 or 0.32
Since P(A)>P(B),
P(A)=0.45.
P(B)=0.32.
The values of P(A) and P(B), calculated using the given values was found to be 0.45 and 0.32 respectively.
Chapter 2 Supplementary Exercise Probability Solved Examples Page 89 Problem 6 Answer
Given: There are six friends namely, A,B,C,D,E, and F.A has heard a rumor and wants to share it with all the friends.
A selects one of the five at random and tells the rumor to the chosen individual.
Given: individual then selects at random one of the four remaining individuals and repeats the rumor. Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told.
To find the probability that the rumor is repeated in the order B,C,D,E and F.
We will use permutation to find the favorable outcomes and then find the probability.
The number of ways to arrange all the friends is,P5,5
=5!/(5−5)!
P=5!.
P=120.
But the required condition is a very specific order (B,C,D,E,and ,F),
T=1/120
T=0.0083.
The probability that the rumor is repeated in the order (B,C,D,E,and ,F) is 0.0083.
Page 89 Problem 7 Answer
There are six friends namely,A,B,C,D,E,and F A has heard a rumor and wants to share it with all the friends.
A selects one of the five at random and tells the rumor to the chosen individual.
Given: individual then selects at random one of the four remaining individuals and repeats the rumor.
Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told.
To find the probability that F is the third person at the party to be told the rumor.
We will use permutation to find the favorable outcomes and then find the probability.
The probability of F being the third person,
P4,4=4!/(4−4)!
P=4⋅3⋅1⋅2⋅1
P=24.
The probability of the above situation to occur is,
T=24/120
T=0.2.
The probability that F is the third person at the party to be told the rumor is 0.2.
Page 89 Problem 8 Answer
There are six friends namely, A,B,C,D,E,and F.A has heard a rumor and wants to share it with all the friends.
A selects one of the five at random and tells the rumor to the chosen individual.
Given: individual then selects at random one of the four remaining individuals and repeats the rumor.
Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told.
To find the probability that F is the last person to hear the rumor.
We will use permutation to find the favorable outcomes and then find the probability.
The probability of F being the last person,
P4,4=4!/(4−4)!
P=4⋅3⋅2⋅1⋅1
P=24.
The probability of the situation to occur is,
T=24/120
T=0.2.
The probability that F is the last person to hear the rumor is 0.2.
Probability And Statistics 8th Edition Chapter 2 Supplementary Exercise Walkthrough Page 89 Problem 9 Answer
There are six friends namely, A,B,C,D,E, and F.A has heard a rumor and wants to share it with all the friends, A selects one of the five at random and tells the rumor to the chosen individual.
Given: individual then selects at random one of the four remaining individuals and repeats the rumor.
Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told.
Given: at each stage the person who currently “has” the rumor does not know who has already heard it and selects the next recipient at random from all five possible individuals, To find the probability that F has still not heard the rumor after it has been told ten times at the party.
We will use the definition of probability.
The probability of F to have not heard after 10 times,
F=410
F=1048576.
The total probability of everyone hearing the rumor,T=510
T=9565625.
Now the probability of the event happening is,
P=F/T
P=1058576/9565625
P=0.1074.
The probability that F has still not heard the rumor after it has been told for 10 times is 0.1074.
Page 89 Problem 10 Answer
Given: Each contestant on a quiz show is asked to specify one of six possible categories from which questions will be asked Given: P( contestant requests category i)=1/6
Given: successive contestants choose their categories independently of one another.
To find the probability that exactly one contestant has selected category 1 given that there are three contestants on each show and all three contestants on a particular show select different categories.We will use combination formula n
Cr=(nr)n!/r!(n−r)! in order to find the favorable and total number of outcomes and then find the probability.
We have:
P( contestant requests category i)
i=1/6
=1,2,3,4,5,6
We need 3 of the 6 categories which can be done in C6,3
=(6/3) ways
Hence the number of possible outcomes are:
(6/3)=6!/3!(6−3)!
=20
There are C5,2=(5/2) ways to select the remaining two categories of the five excluding 1
Hence the number of favorable outcomes is:
(5/2)=5!/2!(5−2)!
=10
Therefore the probability that category 1 is selected is:
P( Category 1 is selected )= number of favorable outcomes/ number of possible outcomes
=10/20
=1/2
=0.5
The probability that exactly one contestant has selected category 1 is 0.5.
Page 90 Problem 11 Answer
Given: One percent of all individuals in a certain population are carriers of a particular disease.
A diagnostic test for this disease has a 90% detection rate for carriers and a 5% detection rate for noncarriers
Given: The test is applied independently to two different blood samples from the same randomly selected individual.
To find the probability that both the tests yield the same result.
We will use the complement rule and general multiplication rule: P(A∩B)=P(A)×P(B∣A)
Let: A=Event that the person has the disease
B=Event that the test is positive.
Hence:
P(A)=0.01
P(B∣A)=0.9
P(B∣A′)=0.05
Therefore by using complement rule we have:
P(A′)=1−P(A)
=1−0.01
=0.99
P(B′∣A)=1−P(B∣A)
=1−0.9
=0.1 and
P(B′∣A′)=1−P(B∣A′)
=1−0.05
=0.95
Using the multiplication rule we have:
P(A∩B)=P(A)×P(B∣A)
=0.01×0.9
=0.009
P(A∩B′)=P(A)×P(B′∣A)
=0.01×0.1
0.001
P(A′∩B)=P(A′)×P(B∣A′)
=0.99×0.05
=0.0495
P(A′∩B′)=P(A′)×P(B′∣A′)
=0.99×0.95
=0.9405
Hence we have:
P(B)=P(A∩B)+P(A′∩B)
=0.009+0.0495
=0.0585
P(B′)=P(A∩B′)+P(A′∩B′)
=0.001+0.9405
=0.9415
The tests are independent and therefore we get:
P( Both B)=0.0585×0.0585
≈0.0034
P( Both B′)
=0.9415×0.9415
≈0.8864
Hence P( Same result )
=P( Both B)+P( Both B′)
=0.0034+0.8864
=0.8898
The probability that both the tests yield the same result is 0.8898.
Chapter 2 Supplementary Exercise Study Guide Probability And Statistics Page 90 Problem 12 Answer
Given: One percent of all individuals in a certain population are carriers of a particular disease.
A diagnostic test for this disease has a 90% detection rate for carriers and a 5% detection rate for noncarriers.
Given: The test is applied independently to two different blood samples from the same randomly selected individual.
To find the probability that the selected individual is a carrier given that both tests are positive.
We will use conditional probability.
We have:
A=Event that the person has the disease
B=Event that the test is positive.
From part a we have: P( Both B)≈0.0034
Hence P( Both (A∩B))
=P(A∩B)×P(A∩B)
=0.009×0.009
=0.000081
Using the definition of conditional probability we have:P(A∣ Both B)
=P(A∩ Both B)/P( Both B)
=P( Both (A∩B))/P( Both B)
=0.000081/0.0585×0.0585
≈0.02367
=2.367%
The probability that the selected individual is a carrier given that both tests are positive is 0.02367.
Page 90 Problem 13 Answer
Given: System consists of two components.
Given: The probability that the second component functions in a satisfactory manner during its design life=0.9.
The probability that at least one of the two components does so=0.96.
The probability that both components do so=0.75. the first component functions in a satisfactory manner throughout its design life.
To find the probability that the second one also functions in a satisfactory manner throughout its design life.We will use conditional probability P(B∣A)=P(A∩B)/P(A)
Let us consider following events,
A={ the event of the first component functioning}
B={ the event of the second component functioning}
We are given that,
P(B)=0.9
P(A∪B)=0.96
P(A∩B)=0.75
Using Addition Rule,
P(A∪B)=P(A)+P(B)−P(A∩B)
This implies, P(A)=−P(B)+P(A∩B)+P(A∪B)
Thus,P(A)=−0.9+0.75+0.96
P(A)=0.81
Using definition of conditional probability,
P(B∣A)=P(A∩B)/P(A)
P(B∣A)=0.75/0.81
P(B∣A)≈0.9259
The probability that the second component functions in a satisfactory manner throughout its design life given that the first one does is 0.9259.
Page 90 Problem 14 Answer
Given: A certain company sends 40% of its overnight mail parcels via express mail service E1.
of which 2% arrive after the guaranteed delivery time .
Given: a record of an overnight mailing is randomly selected from the company’s file.
To find the probability that the parcel went via E1and was late.
We will use the multiplication property: P(A∩B)=P(A∣B)⋅P(B)
We have: P(E1)=0.4
Now if L denotes the event :late arrived” then:
P(L∣E1)=0.02
Hence by using multiplication rule we have:
P(L∩E1)=P(L∣E1)⋅P(E1)
=(0.02)⋅(0.4)
=0.008
The probability that the parcel went via E1 and was late is 0.008.
Probability Examples From Supplementary Exercise In Engineering And The Sciences Page 90 Problem 15 Answer
Given percentages of the assembly lines that they need rework:
A1=5%
A2=8%
A3=10%
Given percentages of components produced:
A1=50%
A2=30%
A3=20%
Given: a randomly selected component needs rework
To find: the probability that it came from lineA1.
To find: the probability that it came from line A2.
To find: the probability that it came from lineA3.
We will use Bayes’ theorem in order to find the required probability.
Let us consider the following event,
R={Component needs rework}
It is given that,
P(A1)=0.5
P(A2)=0.3
P(A3)=0.2
P(R∣A1)=0.05;
P(R∣A2)=0.08
P(R∣A3)=0.1
Now, we need to find the probability such that if a random selected component needs rework, it is from the lineA1
i.e.,P(A1∣R)
Using Bayes’ theorem, we get,
P(A1∣R)=P(A1∩R)/P(R)=P(A1)⋅P(R∣A1)
P(A1)⋅P(R∣A1)+P(A2)⋅P(R∣A2)+P(A2)⋅P(R∣A2)
=0.5⋅0.05/0.5⋅0.05+0.3⋅0.08+0.2⋅0.1
=0.025/0.025+0.024+0.02
=0.025/0.069
P(A1∣R)=0.377
Now, we need to find the probability such that if a random selected component needs rework, it is from the lineA2
i.e.,P(A2∣R).
Using Bayes’ Theorem, we get
P(A2∣R)=P(A2∩R)/P(R)
=P(A2)⋅P(R∣A2)
P(A1)⋅P(R∣A1)+P(A2)⋅P(R∣A2)+P(A2)⋅P(R∣A2)
=0.3⋅0.08/0.5⋅0.05+0.3⋅0.08+0.2⋅0.1
=0.024/0.025+0.024+0.02
=0.024/0.069
P(A2∣R)=0.348
Now, we need to find the probability such that if a random selected component needs rework, it is from the line A3
i.e.,P(A3∣R).
Using Bayes’ Theorem, we get
P(A3∣R)=P(A3∩R)/P(R)
=P(A3)⋅P(R∣A3)
P(A1)⋅P(R∣A1)+P(A2)⋅P(R∣A2)+P(A2)⋅P(R∣A2)
=0.2⋅0.1/0.5⋅0.05+0.3⋅0.08+0.2⋅0.1
=0.02/0.025+0.024+0.02
=0.02/0.069
P(A3∣R)=0.29 the probability that the component came from lineA1 is 0.377 the probability that the component came from line A2 is 0.348 the probability that the component came from line A3 is 0.29
Page 90 Problem 16 Answer
Given: A subject is allowed a sequence of glimpses to detect a target where Gi={ the target is detected on the ith glimpse } with h pi=P(Gi)
Given: Gi′s are independent events.
To write an expression for the probability that the target has been detected by the end of the nth glimpse.
We will use the multiplication property of mutually independent events.
An event Gi is defined as follows,
Gi= the target is detected on theith glimpse
Also, the probability of the event Gi is denoted by pi, then the probability that the target is not observed on ithglimpse is denoted by(1−pi).
The objective is to find the probability that the target has been detected by the end of thenth glimpse.
As it is given thatGi′s are independent, therefore their complementary events are also independent.
Now, the probability that the target has been detected by the end of then th glimpse is equal to the combined probability that the target is not detected by first′n−1′ trials.
Therefore,
P(Gn) = P(target that is not detected in previous ′n′ trials)
P(Gn) = 1 − P(G1∪G2∪……∪Gn)
P(Gn) = 1 − P(G1′∪G2′∪……∪Gn′)
P(Gn) = 1 − [P(G1′)×P(G2′)×……P(Gn′)]
P(Gn) = 1 − [(1−p1)×(1−p2)×….…×(1−pn)]
The expression for the probability that the target has been detected by the end of then th glimpse is P(Gn)=1−[(1−p1)×(1−p2)×….…×(1−pn)]=1−i=1∏n(1−pi)
Page 91 Problem 17 Answer
Given : A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles and the events are describes as:
A= New York flight is full
B =The Atlanta flight is full
C =The Los Angeles flight is full
Given: All the three events are independent events.
Given probabilities: P(A)=0.9
P(B)=0.7
P(C)=0.8
To find the probability that all three flights are full.To find the probability that at least one flight is not full.
We will use the complement rule and multiplication property for independent events in order to find the required probabilities.
Now, we will find the probability that all three flights are full.
This means that all the events are occurring together, since all the events are independent, so by using the multiplication rule of probability
Let probability that all the fights are full be P(A∩B∩C)
As P(A1∩A2∩A3∩………An)=P(A1)⋅P(A2)⋅P(A3)⋅…………P(An)
Therefore, P(A∩B∩C)=P(A)⋅P(B)⋅P(C)
P(A∩B∩C)=0.9⋅0.7⋅0.8
P(A∩B∩C)=0.504
Now, we will calculate the probability if at least one flight is not full.
We will calculate it using the definition of the complement of an event as the event that at least one flight is not full is a complement to that event that all the three flights are full (let the probability that all the flights are full be P(A) )
P(Z) = 1 − P(A)
P(Z) = 1 − 0.504
P(Z) = 0.496
The probability that all three flights are full is 0.0504 .
The probability that at least one flight is not full is0.496.
Notes For Supplementary Exercise In Chapter 2 Probability Page 91 Problem 18 Answer
Given : A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles and the events are describes as:
A= New York flight is full
B = The Atlanta flight is full
C =The Los Angeles flight is full
Given: All three events are independent events.Given probabilities:
P(A)=0.9
P(C)=0.8
P(B)=0.7
To find the probability that only the New York flight is full.To find the probability that exactly one of the three flights is full.
We will use the complement rule and multiplication property for independent events in order to find the required probabilities.
First, we will find the probability that only the New York flight is full. It means that event A will occur but at the same time events, A and B will not occur.
The probability that only the New York flight is full-
By using the multiplication rule of probability –
P(A∩Bˉ∩Cˉ)=P(A)⋅P(Bˉ)⋅P(Cˉ)
we have ,P(Aˉ) = 1−P(A)
P(A∩Bˉ∩Cˉ)=0.9⋅ (1 − 0.7)⋅(1−0.8)
P(A∩Bˉ∩Cˉ)=0.9⋅0.3⋅0.2
P(A∩Bˉ∩Cˉ)=0.054
Now, we need to find the probability that exactly one of the three flights is full.
There are three possible ways either the New York flight is full, or The Atlanta flight is full or The Los Angeles flight is full.
The probability that exactly one of the three flights is full –
P(exactly one of the three flights is full) = P(A∩Bˉ∩Cˉ)+ P(Aˉ∩B∩Cˉ)+ P(Aˉ∩Bˉ∩C)
P(exactly one of the three flights is full) = P(A)P(Bˉ)P(Cˉ)+ P(Aˉ)P(B)P(Cˉ) + P(A)ˉP(B)ˉP(C)
Now, we will use the multiplication rule of probability,
P(exactly one of the three flights is full)= (0.9)⋅(1−0.7)⋅(1−0.8?) + (1−0.9)⋅(0.7)⋅(1−0.8) + (1−0.9)⋅(1−0.7)⋅(0.8)
P(exactly one of the three flights is full)= 0.9⋅0.3⋅0.2 + 0.1⋅0.7⋅0.2 + 0.1⋅0.3⋅0.8
P(exactly one of the three flights is full)= 0.092
The probability that only the New York flight is full is0.054 .
The probability that exactly one of the three flights is full is0.092 .
Page 91 Problem 19 Answer
Given: a box is containing four slips and each having same dimension
Given events:A1 = {win prize 1}
A2 = {win prize 2}
A3 = {win prize 3}
To show that A1 and A2 are independent,A2 and A3 are independent and A1 and A3 are also independent.
To showP(A1∩A2∩A3) ≠ P(A1)⋅P(A2)⋅P(A3), so the three are not mutually independent.
We will use the multiplication property is not hold to prove that they are not mutually independent.
P(A1)=win prize 1(A1)+win prize 1,2,3(A4)
Therefore,P(A1) = 1/2, P(A2) = 1/2 , P(A3) = 1/2
P(A1∩A2) = A4
A1+A2+A3+A4
P(A1∩A2) = 1/4
As, P(A1)⋅P(A2) = 1/2⋅1/2
P(A1)⋅P(A2)= 1/4 which is equal to P(A1∩A2)
This shows that A1 and A2 are independent events.
Similarly, we will solve
P(A1∩A3) = P(A1)⋅P(A3)
P(A1∩A3) = 1/2 ⋅ 1/2
P(A1∩A3) = 1/4 which is equal to P(A1 ∩A3)
This shows that A1 and A3 are independent events
Now, for P(A2∩A3) = P(A2)⋅P(A3)
P(A2∩A3) = 1/2 ⋅ 1/2
P(A2∩A3) = 1/4 which is equal to P(A2∩A3)
This shows that A2 and A3 are independent events
P(A1∩A2∩A3) = A4
A1+A2 +A3+A4
P(A1∩A2∩A3) = 1/4
And P(A1)⋅P(A2)⋅P(A3) = 1/2 ⋅ 1/2 ⋅ 1/2
P(A1)⋅P(A2)⋅P(A3) = 1/8
Therefore, it shows that P(A1∩A2∩A3)≠ P(A1)⋅P(A2)⋅P(A3)
Hence Proved.
We have proved that P(A1∩A2∩A3) ≠ P(A1)⋅P(A2)⋅P(A3) and also we have shown that the three events are pairwise independent.
Page 91 Problem 20 Answer
Given: A1,A2, and A3are independent events.To show that: P(A1∣A2∩A3)=P(A1)
We will use Bayes’ Theorem in order to show the required.
We have to consider that A1,A2 and A3 are independent events
By using Bayes Theorem, we will simplify for P(A1∣A2∩A3).
P(A1∣A2∩A3) = P(A1∩A2∩A3)
P(A2∩A3)
As all the events are independent, then it follows the multiplication rule,
P(A1∩A2∩A3) = P(A1)P(A2)P(A3)
P(A2∩A3) = P(A2)P(A3)
Therefore, P(A1∣A2∩A3) = P(A1∩A2∩A3)
P(A2∩A3)
P(A1∣A2∩A3) = P(A1)P(A2)P(A3)
P(A2)P(A3)
P(A1∣A2∩A3) = P(A1)
Hence proved.
We have shown that if A1,A2 , and A3 are independent events then P(A1∣A2∩A3)=P(A1).
Study Materials For Chapter 2 Supplementary Exercise In Probability Page 91 Problem 21 Answer
Given: a box is containing four slips and each having same dimension
Given events:A1 = {win prize 1}
A2 = {win prize 2}
A3 = {win prize 3}
To show that A1 and A2 are independent,A2 and A3 are independent and A1 and A3 are also independent.To show P(A1∩A2∩A3) ≠ P(A1)⋅P(A2)⋅P(A3), so the three are not mutually independent.
We will use the multiplication property is not hold to prove that they are not mutually independent.
P(A1)=win prize 1(A1)+win prize 1,2,3(A4)
Therefore,P(A1) = 1/2, P(A2) = 1/2 , P(A3) = 1/2
P(A1∩A2) = A4
A1+A2+A3+A4
P(A1∩A2) = 1/4
As, P(A1)⋅P(A2) = 1/2⋅1/2
P(A1)⋅P(A2)= 1/4 which is equal to P(A1∩A2)
This shows that A1 and A2 are independent events.
Similarly, we will solve
P(A1∩A3) = P(A1)⋅P(A3)
P(A1∩A3) = 1/2 ⋅ 1/2
P(A1∩A3) = 1/4 which is equal to P(A1 ∩A3)
This shows that A1 and A3 are independent events
Now, for P(A2∩A3) = P(A2)⋅P(A3)
P(A2∩A3) = 1/2 ⋅ 1/2
P(A2∩A3) = 1/4 which is equal to P(A2∩A3)
This shows that A2 and A3 are independent events
P(A1∩A2∩A3) = A4
A1+A2 +A3+A4
P(A1∩A2∩A3) = 1/4 And P(A1)⋅P(A2)⋅P(A3) = 1/2 ⋅ 1/2 ⋅ 1/2
P(A1)⋅P(A2)⋅P(A3) = 1/8
Therefore, it shows that P(A1∩A2∩A3)≠ P(A1)⋅P(A2)⋅P(A3)
Hence Proved.
We have proved that P(A1∩A2∩A3) ≠ P(A1)⋅P(A2)⋅P(A3) and also we have shown that the three events are pairwise independent.
Page 91 Problem 22 Answer
Given: A1,A2, and A3 are independent events.To show that: P(A1∣A2∩A3)=P(A1)
We will use Bayes’ Theorem in order to show the required.
We have to consider that A1,A2and A3 are independent events
By using Bayes Theorem, we will simplify for P(A1∣A2∩A3).
P(A1∣A2∩A3) = P(A1∩A2∩A3)
P(A2∩A3)
As all the events are independent, then it follows the multiplication rule ,
P(A1∩A2∩A3) = P(A1)P(A2)P(A3)
P(A2∩A3) = P(A2)P(A3)
Therefore, P(A1∣A2∩A3) = P(A1∩A2∩A3)
P(A2∩A3)
P(A1∣A2∩A3) = P(A1)P(A2)P(A3)
P(A2)P(A3)
P(A1∣A2∩A3) = P(A1)
Hence proved.
We have shown that if A1,A2, and A3 are independent events then P(A1∣A2∩A3)=P(A1).