Biomolecular Thermodynamics, From Theory to Applic
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Solutions Manual for Biomolecular Thermodynamics, From Theory
to Application, 1st Edition by Douglas Barrick (All Chapters) A+
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:
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:
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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):
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.
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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.
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:
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Because Event 1 and Event 2 are independent, the “and” combination of these two outcomes is
given by the intersection, and the probability of the
intersection is given by the product of the two separate probabilities, leading to the expressions
for probabilities for the gray cross-hatched crescent.
(a) These are two independent elementary events each with an outcome probability of 0.5.
We are asked for the probability of the sequence H1 T2, which requires multiplication of the
elementary probabilities:
We can arrange this probability, along with the probability for the other three possible sequences,
in a table:
Toss 1
Toss 2 H (0.5) T (0.5)
H (0.5) H1H2 T1H2
(0.25) (0.25)
T (0.5) H1T2 T1T2
(0.25) (0.25)
The probability of getting a head on the first toss or a tail on the second toss, but not both, is
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