Solution Manual
Biomolecular Thermodynamics 1/E Douglas Barrick
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 ∩ 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 − pA pB
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 − pA pB
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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):
p{ A1 ∩ (∼B2 )} ∪ {(~ A1) ∩ B2 } = pA1 ∩ (∼B2 ) + p(∼ A1) ∩ B2
= pA − pA pB + pB − pA pB
= pA + pB − 2 pA pB
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 ∩ ~B2
~ A1 ∩ B2
pA × p~B
p~A × pB
= pA (1 – pB)
= (1 – pA)pB
A1 ∩ ~B2 ~ A1 ∩ 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
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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) T
hese 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:
1 1 1
pH1H2 = H1 ∩ T2 = pH1 × pT2 = × =
2 2 4
e can arrange this probability, along with the probability for the other
W
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)
Note: Probabilities are given in parentheses.
he probability of getting a head on the first toss or a tail on the second
T
toss, but not both, is
pH1 or H2 = pH1 + pH2 − 2( pH1 × pH2 )
1 1 1 1
= + − 2 ×
2 2 2 2
1
=
2
In the table above, this combination corresponds to the sum of the two off-
diagonal elements (the H1T2 and the T1H2 boxes).
(b) T
his is the "and" combination for independent events, so we multiply the
elementary probability pH for each of N tosses:
pH1H2H3…HN = pH1 × pH2 × pH3 ×
× pHN
N
1
=
2
his is both a permutation and a composition (there is only one
T
permutation for all-heads). And note that since both outcomes have
equal probability (0.5), this gives the probability of any permutation of any
number NH of heads with any number N − NH of tails.
1.3 Two different approaches will be given for this problem. One is an
approximation that is very close to being correct. The second is exact. By
comparing the results, the reasonableness of the first approximation can be
examined.
Whichever approach we use to solve this problem, we begin by representing
the probability that you know a randomly selected person from the population.
This is pk = 2000/300,000,000 = 2/300,000 = 6.67 × 10−6. To avoid dealing
with "or" combinations, we can greatly simplify the problem by calculating the
probability that you do NOT know anyone on the plane, and then recognize
that one minus this probability represents all the ways you could know at
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least one person (that is, you know only one, or you know only two, or…
you know all of them). Working with the probability you do not know a given
passenger (pnk) converts a complicated "or" problem to a simple "and"
problem. The probability you do not know an individual selected randomly
from the population is pnk = p∼k = 1 − pk = 0.9999933 (nk for “not know”).
The approximate to find the answer is to use the binomial distribution, and
calculate the probability that you do not know randomly selected people.
This is
p(200; 200) = W (200; 200) × pnk 200 × ( pk )( 200−200 )
200 ! 2000 200
= ×A 1− × ( pk )( 0 )
200 !(200 − 200)! 300, 000, 000
2000 200
= 1−
300, 000, 000
= 0.9986675507...
Thus, the chances you know somebody on the plane is about
1 − p(200;200) = 0.0013324492779… , a little better than one in one thousand
chance (note, the rounding error in these calculations can be a big deal,
especially when comparing to the exact result below). A bit of thought about
the binomial distribution underscores the value of doing the calculation on
people you don’t know, rather than on the ones you do. There is only one
composition the binomial distribution in which each person (besides you) is
not known (200 nk’s), and only one way to arrange the 200 nk’s (hence the
factorial term goes to one). If we had calculated probabilities for knowing
people, we would have had calculate for 200 different compositions (1k, 2k,
3k,…200k) and add them all, and the numbers in the compositions in the
middle of the distribution would involve calculating huge factorials, which is
not straightforward.†
The reason the binomial approach is approximate is that it assumes that for
each person, the probability is independent. Although this seems reasonable
(and because of the large total population, the extent to which it is wrong is
fairly small), it is not quite right. The probability of knowing the first one on the
plane is still pk = 200/300,000,000. But if you do not know the first person on
the plane, the probability you know the second is 2000/299,999,999; you still
know 2000 people outside the plane, but there are no longer 300,000,000 left.
And if you do not know the first two people on the plane, the probability of
knowing the third is 2000/299,999,998,… and if you didn’t know the first 199
people on the plane, the probability you don’t know the last person on (200th)
is 2000/299,999,801. In general, this relationship is
2000
pk , i =
300, 000, 000 − i + 1
where pk,i is the probability that you know the ith person on the plane, given
you did not know any of the previous passengers aboard. And the probability
you don’t know the ith person, given you didn’t know any of the previous
passengers (pk,i), is one minus this:
2000
pnk , i = 1−
300, 000, 000 − i + 1
The total probability you don’t know anyone on the plane is the product of
these dependent probabilities from the first person to the 200th, that is,
† This can be done, but it requires Stirling’s approximation, to be introduced in Chapter 4.
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