,SOLUTION MANUAL FOR BIOMOLECULAR THERMODYNAMICS FROM THEORY TO APPLICATION 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 (A.1.1)
= pA ×(1−
pB )
= 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 (A.1.2)
= (1− pA
)× pB
= pB −
pApB
,
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 (A.1.1)
= pA ×(1−
pB )
= 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 (A.1.2)
= (1− pA
)× pB
= pB −
pApB
,
