TEST BANK FOR Molecular Driving Forces Statistical Thermodynamics in 2nd Ed By Bromberg, Dill

Chapter 1
Principles of Probability
1. Combining independent probabilities.
You have applied to three medical schools: University of California at San Francisco (UCSF),
Duluth School of Mines (DSM), and Harvard (H). You guess that the probabilities you’ll
be accepted are: p(UCSF) = 0.10, p(DSM) = 0.30, and p(H) = 0.50. Assume that the
acceptance events are independent.
(a) What is the probability that you get in somewhere (at least one acceptance)?
(b) What is the probability that you will be accepted by both Harvard and Duluth?
(a) The simplest way to solve this problem is to recall that when probabilities are
independent, and you want the probability of events A and B, you can multiply them.
When events are mutually exclusive and you want the probability of events A or B,
you can add the probabilities. Therefore we try to structure the problem into an and
and or problem. We want the probability of getting into H or DSM or UCSF. But
this doesn’t help because these events are not mutually exclusive (mutually exclusive
means that if one happens, the other cannot happen). So we try again. The probability
of acceptance somewhere, P(a), is P(a) = 1 − P(r), where P(r) is the probability that
you’re rejected everywhere. (You’re either accepted somewhere or you’re not.) But this
probability can be put in the above terms. P(r) = the probability that you’re rejected
at H and at DSM and at UCSF. These events are independent, so we have the answer.
The probability of rejection at H is p(rH) = 1 − 0.5 = 0.5. Rejection at DSM is
p(rDSM) = 1 − 0.3 = 0.7. Rejection at UCSF is p(rUCSF) = 1 − 0.1 = 0.9. Therefore
P(r) = (0.5)(0.7)(0.9) = 0.315. Therefore the probability of at least one acceptance
= P(a) = 1 − P(r) = 0.685.
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(b) The simple answer is that this is the intersection of two independent events:
p(aH)p(aDSM) = (0.50)(0.30)
= 0.15.
A more mechanical approach to either part (a) or this part is to write out all the
possible circumstances. Rejection and acceptance at H are mutually exclusive. Their
probabilities add to one. The same for the other two schools. Therefore all possible
circumstances are taken into account by adding the mutually exclusive events together,
and multiplying independent events:
[p(aH) + p(rH)][p(aDSM) + p(rDSM)][p(aUCSF) + p(rUCSF)] = 1,
or equivalently,
= p(aH)p(aDSM)p(aUCSF) + p(aH)p(aDSM)p(rUCSF)
+p(aH)p(rDSM)p(aUCSF) + · · ·
where the first term is the probability of acceptance at all 3, the second term represents
acceptance at H and DSM but rejection at UCSF, the third term represents acceptance
at H and UCSF but rejection at DSM, etc. Each of these events is mutually exclusive
with respect to each other; therefore they are all added. Each individual term
represents independent events of, for example, aH and aDSM and aUCSF. Therefore it
is simple to read off the answer in this problem: we want aH and aDSM, but notice we
don’t care about UCSF. This probability is
p(aH)p(aDSM) = p(aH)p(aDSM)[p(aUCSF) + p(rUCSF)]
= (0.50)(0.30)
= 0.15.
Note that we could have solved part (a) the same way; it would have required adding
up all the appropriate possible mutually exclusive events. You can check that it gives
the same answer as above (but notice how much more tedious it is).
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2. Probabilities of sequences.
Assume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequence
of nine monomers.
(a) What is the probability of finding the sequence AAATCGAGT through random
chance?
(b) What is the probability of finding the sequence AAAAAAAAA through random
chance?
(c) What is the probability of finding any sequence that has four A’s, two T’s, two G’s,
and one C, such as that in (a)?
(a) Each base occurs with probability 1/4. The probability of an A in position 1 is 1/4, of
A in position 2 is 1/4, of A in position 3 is 1/4, of T in position 4 is 1/4, and so on.
There are 9 bases. The probability of this specific sequence is (1/4)9 = 3.8 × 10−6.
(b) Same answer as (a) above.
(c) Each specific sequence has the probability given above, but in this case there are many
possible sequences which satisfy the requirement that we have 4 A’s, 2 T’s, 2 G’s, and 1
C. How many are there? We start as we have done before, by assuming all nine objects
are distinguishable. There are 9! arrangements of nine distinguishable objects in a
linear sequence. (The first one can be in any of nine places, the second in any of the
remaining eight places, and so on.) But we can’t distinguish the four A’s, so we have
overcounted by a factor of 4!, and must divide this out. We can’t distinguish the two
T’s, so we have overcounted by 2!, and must also divide this out. And so on. So the
probability of having this composition is
"
9!
4!2!2!1!
#
1
4
9
= 0.014.