Topic 2-2 Basic probability rules
Meaning of probability
(i) Experimental: if an experiment is repeated N times (e.g. throwing a die), and one records the
frequency f that a certain event E occurs (e.g. “a prime number comes up”), then as N gets very
large, the ratio
f/N
will approach a fixed number, say 0.33333…. This number is called the probability of the event,
denoted P(E).
(ii) Predicted: If an experiment has N different and equally likely outcomes (e.g. the six faces on a
fair die; N = 6), while NE of them lead to the occurrence of event E (e.g. “a prime number
occurs” (3 and 5 are prime, so NE = 2), then we define probability, P(E) as
NE
N
Methods (i) and (ii) should always give the same answer.
Basic formulas
Most basic formulas are obvious by examining Venn diagrams: draw the sample space S (all
possible outcomes) as a rectangle, and events with areas that reflect the number of outcomes in them.
In Fig. 1, there are n1 outcomes in event “A but excluding B”, n2 outcomes in event “both A and B”,
n3 outcomes in “B but excluding A”, and n4 outcomes in event “neither A nor B”.
Sample space S with N = n1 + n2 + n3 + n4 outcomes
Events:
n4
AB
n1
n2 n 3
Fig. 1
Sample
space S and
various
events
AB
(AUB)'
B
A
Consider any event E, such as one drawn in Fig. 1. By definition of probability, we must have (a)
P(E)  0 for any event E, and (b) P(S) = 1. These then lead to 0  P(E)  1, so a probability negative
or bigger than 1 indicates error in calculations. Also, it is clear from Fig. 1 that
P(AB) = P(A) + P(B) – P(AB)
(1)
since (n1 + n2 + n3) = (n1 + n2) + (n2 + n3) – n2. Furthermore, if A and B are mutually exclusive (m.e.),
i.e. one’s occurrence precludes the other, then A and B must have no overlap on a Venn diagram, and
hence n2 = P(AB) = 0, so
P(AB) = P(A) + P(B) if A and B are m.e.
(2)
Another fact easily seen by considering any event E in Fig. 1 is
P(E) = 1 - P(E’)
(3)
(3) is useful when one seeks the probability of some complicated event E, whose compliment
(denoted by E’, E , or E*) is simple.
Conditional probability and statistical independence
It is clear from Fig. 1 that to find the probability of A given B occurs, one should shrink the sample
space to B, thus
n2
P(A|B) =
,
n 2  n3
Dividing top and bottom by N, we get the formula for conditional probability,
P ( A B) 
P( A  B)
P( B)
(4)
Similarly, P(B|A) = n2/(n1+n2), hence
P( B A) 
P( A  B)
P( A)
(5)
Rearranging (4) and (5), we have the “multiplication rule”
P(AB) = P(A|B)P(B)
(6a)
P(AB) = P(B|A)P(A)
(6b)
(6) is useful to find probability of intersections when conditional probabilities are given.
If the probability of A does not change whether B occurs or not, i.e. P(A|B) = P(A), as indicated in
Fig. 2, then we say they are statistically independent (s.i.). But from (4) this means
P(AB) = P(A)P(B) if A and B are s.i.
Plugging (7) into (5), we see that P(B|A) = P(B) and P(A|B) = P(A) actually imply each other.
B
Fig. 2
Statistical
independence
B'
B
A'
B'
A'
A
A
A and B are s.i.
A and B are not s.i.; P(A|B) > P(A)
(7)
Solved problems
Problem 2-2-1
A contractor has two subcontractors for his excavation work. Experience shows that in 60% of the
time, subcontractor A was available to do a job; whereas subcontractor B was available 80% of the
time. Also, the contractor is able to get at least one of these two subcontractors 90% of the time.
(a)
What is the probability that both subcontractors will be available to do the next job?
(b) If the contractor learned that subcontractor A is not available for the job, what is the probability
that the other subcontractor will be available?
(c) Suppose A denotes the event that subcontractor A is available, and B denotes that
subcontractor B is available.
(i) Are A and B statistically independent?
(ii) Are A and B mutually exhaustive?
(iii) Are A and B collectively exhaustive?
Solution:
(a) The event “both subcontractors will be available” = AB (the  sign is omitted), hence
since P(AB) = P(A) + P(B) - P(AB)
 P(AB) = P(A) + P(B) - P(AB)
= 0.6 + 0.8 - 0.9 = 0.5
(b) P(B is availableA is not available) = P(B A )
=
P( BA )
P( A )
while it is clear from the following Venn diagram that P( BA ) = P(B) - P(AB).
A AB
B
A
A
Hence
P( BA ) P( B)  P( AB)
=
P( A )
1  P( A)
= (0.8 - 0.5)/(1 - 0.6)
= 0.3/0.4 = 0.75
(c)
(i) If A and B are s.i., we must have P(BA) = P(B) = 0.8. However, using Bayes’ rule,
P(BA) = P(AB)/P(A)
= 0.5/0.6 = 0.8333
So A and B are not s.i. (A’s being available boosts the chances that B will be available)
(ii) From (a), P(AB) is nonzero, hence AB  , i.e. A and B are not m.e.
(iii) Given: P(AB) = 0.9  AB does not generate the whole sample space (otherwise the
probability would be 1), i.e. A and B are not collectively exhaustive.
Problem 2-2-2
An underground site is being considered for the storage of hazardous waste. Within the next 100
years, there is a 1% chance that the hazardous material could leak outside of the storage containment.
Two adjacent towns, A and B, rely on ground water for their water supply. The water to each town
would be contaminated if there is a leakage in the waste storage and if there exists a continuous seam
of sand between the storage containment and the given town. Observe that the presence of a
continuous seam of sand would allow the contaminant to move freely and quickly access the region.
SA ?
A
L?
waste storage site X
SB ?
B
Suppose there is a 2% chance of a continuous seam of sand from the site to A, and the probability of
a continuous seam of sand to B is slightly higher and equals 3%. However, if indeed a continuous
seam of sand exists between X and A, the probability of continuous seam of sand between X and B is
increased to 20%. You may assume the event of leakage from the storage is independent of the
presence of seams of sand. Consider the period over the next 100 years.
(a) What is the probability that water in town A will be contaminated?
(b) What is the probability that at least one of the two towns’ water will be contaminated?
Solution:
(a) Let L, SA, SB denote the respective events “leakage at site”, “seam of sand from X to A”, “seam of
sand from X to B”.
Given probabilities:
P(L) = 0.01, P(SA) = 0.02, P(SB) = 0.03, P(SBSA) = 0.2,
Also given: independence between leakage and seams of sand, i.e.
P(SBL) = P(SB), P(LSB) = P(L), P(SAL) = P(SA), P(L SA) = P(L)
The event “water in town A will be contaminated” = LSA, whose probability is
P(LSA) = P(L SA) P(SA)
= P(L)P(SA)
= 0.010.02 = 0.0002.
(b) The desired event is (L SA)  (L SB), whose probability is
P(L SA) + P(L SB) - P(L SA L SB)
= P(L)P(SA) + P(L)P(SB) - P(L SA SB)
= P(L) [P(SA) + P(SB) - P(SA SB)]
= P(L) [P(SA) + P(SB) - P(SBSA) P(SA)]
= 0.01 (0.02 + 0.03 - 0.20.02) = 0.00046
Problem 2-2-3
Towns A, B and C lie along a river which may be subject to overflow (flooding). The probabilities of
flooding in each year are 0.2, 0.3 and 0.1 for towns A, B and C, respectively. The events of flooding
in each of the towns A, B and C are not statistically independent. If town C is flooded in a given year,
the probability that town B will also be flooded that year is increased to 0.6; if both towns B and C
are flooded in a given year, the probability that town A will also be flooded that year is increased to
0.8. However, if town C experiences no flood in a given year, the probability that both towns A and B
will also suffer no floods in that year is 0.9. In a given year, if all three towns are flooded, it is
regarded as a disaster year. Suppose the flooding events between any two years are statistically
independent. Answer the following:
(a) What is the probability that a given year is a disaster year?
(b) If town B is flooded in a given year, what is the probability that town C is also flooded?
(c) What is the probability that at least one town is flooded in a given year?
C
B
A
Solution:
Let A,B,C denote the respective events that the named towns are flooded. Given probabilities: P(A) =
0.2, P(B) = 0.3, P(C) = 0.1, P(B | C) = 0.6, P(A | BC) = 0.8, P( AB | C ) = 0.9, where an overbar
denotes compliment of an event.
(a) P(disaster year) = P(ABC)
= P(A | BC)P(BC)
= P(A | BC)P(B | C)P(C)
= 0.80.60.1 = 0.048
(b) P(C | B) = P(BC) / P(B)
= P(B | C)P(C) / P(B)
= 0.60.10.3 = 0.2
(c) The event of interest is ABC. Since this is the union of many items, we can work with its
compliment instead, allowing us to apply De Morgan’s rule and rewrite as
P(ABC) = 1 – P( A  B  C )
= 1 – P( A BC )
= 1 – P( AB | C )P( C )
= 1 – 0.9(1 – 0.1)
= 1 – 0.81 = 0.19
by De Morgan’s rule,
Problem 2-2-4
Successful completion of a construction project depends on the supply of materials and labor as well
as the weather condition. Consider a given project which will be successfully completed if either one
of the following conditions prevail:
(i)
Good weather and at least labor or materials are adequately available.
(ii)
Bad weather but both labor and materials are adequately available.
Define
G = Good weather
G’ = Bad weather
L = Adequate labor supply
M = Adequate materials supply
C = Successful completion
Suppose P(L) = 0.7; P(G) = 0.6. L is independent of both M and G. If weather is good, adequate
supply of materials is guaranteed whereas the probability of adequate supply of material is only 50%
if bad weather prevails.
(a)
Formulate the event of successful completion in terms of G, L and M.
(b)
Determine the probability of successful completion. (ans. 0.74)
(c)
If the project were successfully completed, what is the probability that labor supply had been
inadequate? (ans. 0.243)
Solution:
(a)
C = [(LM) G)] [(LM) G’]
(b)
Since (LM)G is contained in G, it is mutually exclusive to (LM)G’ which is contained in G’.
Hence P(C) is simply the sum of two terms,
P(C) = P[(LM) G)] + P[(LM) G’]
= P(LGMG) + P(L)P(MG’)
= P(LG) + P(MG) – P(LMG) + P(L)P(M|G’)P(G’)
= P(L)P(G) + P(M|G)P(G) – P(L)P(M|G)P(G)+ P(L)P(M|G’)P(G’)
= 0.70.6 + 10.6 – 0.710.6 + 0.70.50.4 = 0.74
(c)
P(L’|C) = P(L’C)/P(C) = P(L’{[(LM) G)] [(LM) G’]}) / P(C)
= P[L’(LM) G)] / P(C)
since (L’L)MG’ is an impossible event
= P(L’MG) / P(C)
= P(L’)P(MG) / P(C)
= P(L’)P(M|G)P(G) / P(C)
= 0.310.6 / 0.74  0.243
Exercises
Exercise 2-2-1
The accidents at rail-highway grade-crossings reported for XY province over the last 10 years are
summarized and classified as follows:
Time of
Occurrence
Day (D)
Night (N)
Type of Accident
(R) Run into Train
(S) Struck by Train
30
60
20
20
Suppose there are 1000 rail-highway grade-crossings in XY province.
(a)
What is the probability that an accident will occur at a given crossing next year? (ans. 0.013)
(b)
If an accident is reported to have occurred in daytime, what is the probability that it is a “struck
by train” accident? (ans. 2/3)
(c)
Suppose that 50% of the “run into train” accidents are fatal; and 80% of the “struck by train”
accidents are fatal. What is the probability that the next accident will be fatal? (ans. 0.685)
(d)
Suppose D = event that the next accident occurs in daytime
R = event that the next accident is a “run into train” accident
(i)
(ii)
Are D and R mutually exclusive? Justify. (ans. no)
Are D and R statistically independent? Justify. (ans. no)
Exercise 2-2-2
A team of two engineers, A and B respectively, are assigned to check a set of computations. Each
person works simultaneously but separately and independently. The probability of engineer A
spotting a given error is 0.8, whereas that for B is 0.9.
(a)
Suppose there is only one error in the computation. What is the probability that this error will
be spotted by this team? (ans. 0.98)
(b)
If the error in part (a) were indeed identified, what is the probability that is has been spotted by
A alone? (ans. 0.082)
(c)
Suppose there is an alternate team consisting of 3 engineers C1, C2 and C3, each of whom
works separately and independently and has a probability of 0.75 of spotting a given error.
Would this team of 3 engineers be selected instead, if the objective is to maximize the chance
of spotting the error? Please justify. (ans. yes)
(d)
In part (a), suppose there were two errors in the computations instead. What is the probability
that both errors will be spotted by this team of two engineers? Assume the event of spotting
between any two errors are statistically independent. (ans. 0.960)
Exercise 2-2-3
From previous records on winter weather for a town, the probability that it snows on a given day is
0.2. Records also reveal that low temperature (say less than 0F) occurs on 5% of the days, whereas
windy days occur 10% of the time. If temperature is low, the probability of snow on that day is
increased to 30%. Let S, C, and W denote the events of snow, low temperature and wind,
respectively, on a given day. The event W may be assumed to be statistically independent of S and C.
(a)
What is the probability of having a “doomsday” (i.e., snow, low temperature and wind all
occurring on the same day)? (ans. 0.0015)
(b)
Suppose a construction project cannot progress if it is windy or cold, but it is not affected by
the snow. What is the probability that construction work will be stopped on a given day? (ans.
0.145)
(c)
On a day without snow, what is the probability that construction work cannot proceed? (ans.
0.139)
Exercise 2-2-4
Lead and bacteria are the two common sources of contamination in a water distribution system.
Suppose 4% of the water distribution systems are contaminated by lead, and only 2% of the water
distribution systems are contaminated by bacteria. Assume the events of lead and bacterial
contamination are statistically independent.
(a)
Determine the probability that a water distribution system selected at random for inspection is
contaminated. (ans. 0.0592)
(b)
If indeed a system is contaminated, what is the probability that it is caused by lead only? (ans.
0.662)
Exercise 2-2-5
On a given day, casting of concrete at a construction site depends on the availability of material and
the weather. The material can be ordered from either pre-mix concrete supplier A or B. However, it is
not always certain that these sources will be available. Moreover, casting cannot be performed on a
rainy day. Suppose on a given day, the following events are defined:
R = it rains
A = pre-mixed is available from supplier A
B = pre-mixed is available from supplier B
with probabilities: P(R) = 0.2, P(A) = 0.9, and P(B) = 0.6
The availability of either sources of materials is statistically independent of the weather. However, if
pre-mix concrete is not available from supplier B on a given day, the likelihood that supplier A can
supply it is reduced to 0.5.
(a)
Determine the probability that casting of concrete cannot be performed on a given day. (ans.
0.36)
(b)
If indeed pre-mix concrete were not available from supplier B, what is the probability that
casting of concrete can still be performed that day? (ans. 0.4)
Exercise 2-2-6
A landfill containment system is shown in the following figure. A thick layer of clay (which is a
highly impermeable material) was placed between the landfill and the surrounding soil stratum to
prevent the contaminants leaking from the landfill into the soil stratum resulting from rainfall
infiltration. A layer of synthetic material called geomembrane was also placed above the clay
material to provide additional protection against leakage of contaminants. Nevertheless, the quality of
workmanship during construction may not be completely satisfactory. First, the clay might have been
compacted poorly. Second, the geomembrane might have holes punctured by sharp stones, that were
not detected during inspection. Moreover, extremely heavy rainfall could happen during the operation
of the landfill, which could induce excessive pore pressure on the geomembrane/clay layers.
Rainfall
Clay
Landfill
Geomembrane
Soil stratum
The engineer believes that leakage will happen “during extremely heavy rainfall, and either the clay
was not well compacted or there were holes in the geomembrane” (Event I). Leakage could also
occur “under ordinary rainfalls (i.e. without extremely heavy rainfall), but only when the clay was not
well compacted and the geomembrane contained holes” (Event II).
Let
W = event of well compacted clay; and this occurs with 90% probability
H = event of geomembrane containing holes; and this is 30% likely
E = event of extremely heavy rainfall; and the likelihood is only 20%
The quality of construction has no effect on the future amount of rainfall. However, if the
geomembrane contained holes, the probability of a well-compacted clay is reduced to 60%.
(a)
Express Event I in terms of the symbols defined above. Repeat for Event II.
(b)
Determine the probability of Event I. Repeat for Event II. (ans. 0.056, 0.096)
(c1) Are the W and H mutually exclusive? Are they statistically independent? (ans. no, no)
(c2) Are the events I and II mutually exclusive? Are they collectively exhaustive? (ans. yes, no)
(d)
Determine the probability of leakage for this landfill containment system. (ans. 0.152)
Exercise 2-2-7
The damage in a structure after an earthquake can be classified as none (N), light (L) or heavy (H).
For a new undamaged structure, the probability that it will suffer light and heavy damages after an
earthquake is 0.2 and 0.05 respectively. However, if a structure was already lightly damaged, its
probability of getting heavy damage during the next earthquake is increased to 0.5.
(a)
For a new structure, what is the probability that it will be heavily damaged after 2 earthquakes?
Assume that no repair was performed after the first earthquake. (ans. 0.188)
(b)
If a structure is indeed heavily damaged after 2 earthquakes, what is the probability that the
structure was either undamaged or lightly damaged before the second earthquake? (ans. 0.733)
(c)
If the structure is restored to undamaged condition after each earthquake, what is the
probability that the structure will ever experience heavy damage during three earthquakes?
(ans. 0.143)