Solution Manual –
Biomolecular Thermodynamics
By (Barrick)
1st Edition
,CHAPTER 1
1.1 Using The Same Venn Diagram For Illustration, We Want The Probability
Of Outcomes From The Two Events That Lead To The Cross-Hatched Area
Shown Below:
A1 A1 N B2 B2
This Represents Getting A In Event 1 And Not B In Event 2, Plus Not Getting A
In Event 1 But Getting B In Event 2 (These Two Are The Common “Or But Not
Both” Combination Calculated In Problem 1.2) Plus Getting A In Event 1 And B
In Event 2.
1.2 First The Formula Will Be Derived Using Equations, And Then Venn Diagrams
Will Be Compared With The Steps In The Equation. In Terms Of Formulas And
Probabilities, There Are Two Ways That The Desired Pair Of Outcomes Can
Come About. One Way Is That We Could Get A On The First Event And Not B
On The
Second ( A1 ∩ (∼B2 )). The Probability Of This Is Taken As The Simple Product,
Since Events 1 And 2 Are Independent:
Pa1 ∩ (∼B2 ) = Pa × P∼B
= Pa ×(1− Pb ) (A.1.1)
= Pa − Papb
The Second Way Is That We Could Not Get A On The First Event And We Could Get
B On The Second ((∼ A1) ∩ B2 ) , With Probability
P(∼A1) ∩ B2 = P∼A × Pb
= (1− Pa )× Pb (A.1.2)
= Pb − Papb
,
, 2 SOLUTION MANUAL
Since Either One Will Work, We Want The Or Combination. Because The Two
Ways Are Mutually Exclusive (Having Both Would Mean Both A And ∼A In The
First Outcome, And With Equal Impossibility, Both B And ∼B), This Or
Combination Is Equal To The Union { A1 ∩ (∼B2 )} ∪ {(∼ A1) ∩ B2}, And Its Probability
Is Simply The Sum Of The Probability Of The Two Separate Ways Above
(Equations A.1.1 And A.1.2):
P{A1 ∩ (∼B2 )} ∪ {(~A1) ∩ B2} = Pa1 ∩ (∼B2) + P(∼A1) ∩ B2
= Pa − Papb + Pb − Papb
= Pa + Pb − 2papb
The Connection To Venn Diagrams Is Shown Below. In This Exercise We Will
Work Backward From The Combination Of Outcomes We Seek To The Individual
Outcomes. The Probability We Are After Is For The Cross-Hatched Area Below.
{ A1 ∩ (∼B2 )} ∪ {(∼ A1) ∩ B2 }
A1 B2
As Indicated, The Circles Correspond To Getting The Outcome A In Event 1
(Left) And Outcome B In Event 2. Even Though The Events Are Identical, The
Venn Diagram Is Constructed So That There Is Some Overlap Between These
Two (Which We Don’t Want To Include In Our “Or But Not Both” Combination.
As Described Above, The Two Cross-Hatched Areas Above Don’t Overlap, Thus
The Probability Of Their Union Is The Simple Sum Of The Two Separate Areas
Given Below.
A1 N ~B2
~ A1 N B2
Pa × P~B
p ~A × Pb
= Pa (1 – Pb)
= (1 – PA )PB
A1 N ~B2 ~ A1 N B2
Adding These Two Probabilities Gives The Full “Or But Not Both” Expression
Above. The Only Thing Remaining Is To Show That The Probability Of Each Of
The Crescents Is Equal To The Product Of The Probabilities As Shown In The
Top Diagram. This Will Only Be Done For One Of The Two Crescents, Since
The Other Follows In An Exactly Analogous Way. Focusing On The Gray
Crescent Above, It
Represents The A Outcomes Of Event 1 And Not The B Outcomes In Event 2.
Each Of These Outcomes Is Shown Below:
Event 1 Event 2
A1 ~B
P~B = 1 – Pb
pA
A1 ~B2
Biomolecular Thermodynamics
By (Barrick)
1st Edition
,CHAPTER 1
1.1 Using The Same Venn Diagram For Illustration, We Want The Probability
Of Outcomes From The Two Events That Lead To The Cross-Hatched Area
Shown Below:
A1 A1 N B2 B2
This Represents Getting A In Event 1 And Not B In Event 2, Plus Not Getting A
In Event 1 But Getting B In Event 2 (These Two Are The Common “Or But Not
Both” Combination Calculated In Problem 1.2) Plus Getting A In Event 1 And B
In Event 2.
1.2 First The Formula Will Be Derived Using Equations, And Then Venn Diagrams
Will Be Compared With The Steps In The Equation. In Terms Of Formulas And
Probabilities, There Are Two Ways That The Desired Pair Of Outcomes Can
Come About. One Way Is That We Could Get A On The First Event And Not B
On The
Second ( A1 ∩ (∼B2 )). The Probability Of This Is Taken As The Simple Product,
Since Events 1 And 2 Are Independent:
Pa1 ∩ (∼B2 ) = Pa × P∼B
= Pa ×(1− Pb ) (A.1.1)
= Pa − Papb
The Second Way Is That We Could Not Get A On The First Event And We Could Get
B On The Second ((∼ A1) ∩ B2 ) , With Probability
P(∼A1) ∩ B2 = P∼A × Pb
= (1− Pa )× Pb (A.1.2)
= Pb − Papb
,
, 2 SOLUTION MANUAL
Since Either One Will Work, We Want The Or Combination. Because The Two
Ways Are Mutually Exclusive (Having Both Would Mean Both A And ∼A In The
First Outcome, And With Equal Impossibility, Both B And ∼B), This Or
Combination Is Equal To The Union { A1 ∩ (∼B2 )} ∪ {(∼ A1) ∩ B2}, And Its Probability
Is Simply The Sum Of The Probability Of The Two Separate Ways Above
(Equations A.1.1 And A.1.2):
P{A1 ∩ (∼B2 )} ∪ {(~A1) ∩ B2} = Pa1 ∩ (∼B2) + P(∼A1) ∩ B2
= Pa − Papb + Pb − Papb
= Pa + Pb − 2papb
The Connection To Venn Diagrams Is Shown Below. In This Exercise We Will
Work Backward From The Combination Of Outcomes We Seek To The Individual
Outcomes. The Probability We Are After Is For The Cross-Hatched Area Below.
{ A1 ∩ (∼B2 )} ∪ {(∼ A1) ∩ B2 }
A1 B2
As Indicated, The Circles Correspond To Getting The Outcome A In Event 1
(Left) And Outcome B In Event 2. Even Though The Events Are Identical, The
Venn Diagram Is Constructed So That There Is Some Overlap Between These
Two (Which We Don’t Want To Include In Our “Or But Not Both” Combination.
As Described Above, The Two Cross-Hatched Areas Above Don’t Overlap, Thus
The Probability Of Their Union Is The Simple Sum Of The Two Separate Areas
Given Below.
A1 N ~B2
~ A1 N B2
Pa × P~B
p ~A × Pb
= Pa (1 – Pb)
= (1 – PA )PB
A1 N ~B2 ~ A1 N B2
Adding These Two Probabilities Gives The Full “Or But Not Both” Expression
Above. The Only Thing Remaining Is To Show That The Probability Of Each Of
The Crescents Is Equal To The Product Of The Probabilities As Shown In The
Top Diagram. This Will Only Be Done For One Of The Two Crescents, Since
The Other Follows In An Exactly Analogous Way. Focusing On The Gray
Crescent Above, It
Represents The A Outcomes Of Event 1 And Not The B Outcomes In Event 2.
Each Of These Outcomes Is Shown Below:
Event 1 Event 2
A1 ~B
P~B = 1 – Pb
pA
A1 ~B2
