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SECTION 1 - BASIC PROBABILITY CONCEPTS
37
SECTION 1 - BASIC PROBABILITY CONCEPTS
PROBABILITY SPACES AND EVENTS
Sample point and sample space: A sample point is the simple outcome of a random
experiment. The probability space (also called sample space) is the collection of all possible sample
points related to a specified experiment. When the experiment is performed, one of the sample points will
be the outcome. The probability space is the "full set" of possible outcomes of the experiment.
Mutually exclusive outcomes: Outcomes are mutually exclusive if they cannot occur
simultaneously. They are also referred to as disjoint outcomes.
Exhaustive outcomes: Outcomes are exhaustive if they combine to be the entire probability
space, or equivalently, if at least one of the outcomes must occur whenever the experiment is performed.
Event: Any collection of sample points, or any subset of the probability space is referred to
as an event. We say that "event E has occurred" if the experimental outcome was one of the sample
points in E.
Union of events E and F : E ∪ F denotes the union of events E and F , and consists of
all sample points that are in either E or F .
8
3œ"
the events E" ß E# ß ÞÞÞß E8 , and consists of all sample points that are in at least one of the E3 's. This
Union of events E" ß E# ß ÞÞÞß E8 : E" ∪ E# ∪ â ∪ E8 œ ∪ E3 denotes the union of
definition can be extended to the union of infinitely many events.
8
3œ"
intersection of the events E" ß E# ß ÞÞÞß E8 , and consists of all sample points that are
Intersection of events E" ß E# ß ÞÞÞß E8 : E" ∩ E# ∩ â ∩ E8 œ ∩ E3 denotes the
simultaneously in all of the E3 's. (E ∩ F is also denoted E † F or EF ).
© ACTEX 2015
SOA Exam P - Probability
38
SECTION 1 - BASIC PROBABILITY CONCEPTS
Mutually exclusive events E" ß E# ß ÞÞÞß E8 : Two events are mutually exclusive if they
have no sample points in common, or equivalently, if they have empty intersection. Events
E" ß E# ß ÞÞÞß E8 are mutually exclusive if E3 ∩ E4 œ g for all 3 Á 4 , where g denotes the empty set with
no sample points. Mutually exclusive events cannot occur simultaneously.
Exhaustive events F" ß F# ß ÞÞÞß F8 : If F" ∪ F# ∪ â ∪ F8 œ W , the entire probability space, then
the events F" ß F# ß ÞÞÞß F8 are referred to as exhaustive events.
Complement of event E: The complement of event E consists of all sample points in

the probability space that are not in E. The complement is denoted Eß µ Eß Ew or E- and is equal to
ÖB À B Â E×. When the underlying random experiment is performed, to say that the complement of E has
occurred is the same as saying that E has not occurred.
Subevent (or subset) E of event F : If event F contains all the sample points in event
E, then E is a subevent of F , denoted E § F . The occurrence of event E implies that event F has
occurred.
8
3œ"
Partition of event E: Events G" ß G# ß ÞÞÞß G8 form a partition of event E if E œ ∪ G3
and the G3 's are mutually exclusive.
DeMorgan's Laws:
(i)
ÐE ∪ FÑw œ Ew ∩ F w , to say that E ∪ F has not occurred is to say that E has not occurred
and F has not occurred ; this rule generalizes to any number of events;
w
8
8
 ∪ E3  œ ÐE" ∪ E# ∪ â ∪ E8 Ñw œ E"w ∩ E#w ∩ â ∩ E8w œ ∩ E3w
3œ"
3œ"
(ii)
ÐE ∩ FÑw œ Ew ∪ F w , to say that E ∩ F has not occurred is to say that either E has not
occurred or F has not occurred (or both have not occurred) ; this rule generalizes to any
w
8
8
number of events,  ∩ E3  œ ÐE" ∩ E# ∩ â ∩ E8 Ñw œ E"w ∪ E#w ∪ â ∪ E8w œ ∪ E3w
3œ"
3œ"
Indicator function for event E: The function ME ÐBÑ œ 
" if B−E
! if BÂE
is the indicator
function for event E, where B denotes a sample point. ME ÐBÑ is 1 if event E has occurred.
Basic set theory was reviewed in Section 0 of these notes. The Venn diagrams presented there apply in
the sample space and event context presented here.
© ACTEX 2015
SOA Exam P - Probability
SECTION 1 - BASIC PROBABILITY CONCEPTS
39
Example 1-1:
Suppose that an "experiment" consists of tossing a six-faced die. The probability space of outcomes
consists of the set Ö"ß #ß $ß %ß &ß '×, each number being a sample point representing the number of spots
that can turn up when the die is tossed. The outcomes " and # (or more formally, Ö"× and Ö#×) are an
example of mutually exclusive outcomes, since they cannot occur simultaneously on one toss of the die.
The collection of all the outcomes (sample points) 1 to 6 are exhaustive for the experiment of tossing a
die since one of those outcomes must occur. The collection Ö#ß %ß '× represents the event of tossing an
even number when tossing a die. We define the following events.
E œ Ö"ß #ß $× œ "a number less than % is tossed" ,
F œ Ö#ß %ß '× œ "an even number is tossed" ,
G œ Ö%× œ "a 4 is tossed" ,
H œ Ö#× œ "a 2 is tossed" .
Then
(i)
E ∪ F œ Ö"ß #ß $ß %ß '× ,
(ii)
E ∩ F œ Ö#× ,
(iii) E and G are mutually exclusive since E ∩ G œ g (when the die is tossed it is not
possible for both E and G to occur),
(iv)
H§F ,
(v)
Ew œ Ö%ß &ß '× (complement of E) ,
(vi)
F w œ Ö"ß $ß &×,
(vii) E ∪ F œ Ö"ß #ß $ß %ß '× , so that ÐE ∪ FÑw œ Ö&× œ Ew ∩ F w (this illustrates one of
DeMorgan's Laws).
© ACTEX 2015

SOA Exam P - Probability
40
SECTION 1 - BASIC PROBABILITY CONCEPTS
Some rules concerning operations on events:
(i)
E ∩ ÐF" ∪ F# ∪ â ∪ F8 Ñ œ ÐE ∩ F" Ñ ∪ ÐE ∩ F# Ñ ∪ â ∪ ÐE ∩ F8 Ñ and
E ∪ ÐF" ∩ F# ∩ â ∩ F8 Ñ œ ÐE ∪ F" Ñ ∩ ÐE ∪ F# Ñ ∩ â ∩ ÐE ∪ F8 Ñ for any events
Eß F" ß F# ß ÞÞÞß F8
(ii)
8
If F" ß F# ß ÞÞÞß F8 are exhaustive events  ∪ F3 œ W , the entire probability space,
3œ"
then for any event E,
E œ E ∩ ÐF" ∪ F# ∪ â ∪ F8 Ñ œ ÐE ∩ F" Ñ ∪ ÐE ∩ F# Ñ ∪ â ∪ ÐE ∩ F8 Ñ .
If F" ß F# ß ÞÞÞß F8 are exhaustive and mutually exclusive events, then they form a
partition of the probability space. For example, the events F" œ Ö"ß #× , F# œ Ö$ß %× and
F$ œ Ö&ß '× form a partition of the probability space for the outcomes of tossing a single die.
The general idea of a partition is illustrated in the diagram below. As a special case of a
partition, if F is any event, then F and F w form a partition of the probability space. We then
get the following identity for any two events E and F :
E œ E ∩ ÐF ∪ F w Ñ œ ÐE ∩ FÑ ∪ ÐE ∩ F w Ñ ; note also that E ∩ F and E ∩ F w form a
partition of event E.
B2
B1
A  B
A  B1
A B
AB2
AB3
A  B4
B4
B3
A
B
(iii) For any event E, E ∪ Ew œ W , the entire probability space, and E ∩ Ew œ g
(iv)
E ∩ F w œ ÖB À B − E and B  F× is sometimes denoted E  F , and consists of
all sample points that are in event E but not in event F
(v)
If E § F then E ∪ F œ F and E ∩ F œ E .
© ACTEX 2015
SOA Exam P - Probability
SECTION 1 - BASIC PROBABILITY CONCEPTS
41
PROBABILITY
Probability function for a discrete probability space: A discrete probability space (or sample
space) is a set of a finite or countable infinite number of sample points. T Ò+3 Ó or :3 denotes the
probability that sample point (or outcomeÑ +3 occurs. There is some underlying "random experiment"
whose outcome will be one of the +3 's in the probability space. Each time the experiment is performed,
one of the +3 's will occur. The probability function T must satisfy the following two conditions:
(i) ! Ÿ T Ò+3 Ó Ÿ " for each +3 in the sample space, and
(ii) T Ò+" Ó  T Ò+# Ó  â œ D T Ò+3 Ó œ " (total probability for a probability space is always 1).
all 3
This definition applies to both finite and infinite probability spaces.
Tossing an ordinary die is an example of an experiment with a finite probability space Ö"ß #ß $ß %ß &ß '× .
An example of an experiment with an infinite probability space is the tossing of a coin until the first head
appears. The toss number of the first head could be any positive integer, 1, or 2, or 3, .... The probability
space for this experiment is the infinite set of positive integers Ö"ß #ß $ß ÞÞÞ× since the first head could occur
on any toss starting with the first toss. The notion of discrete random variable covered later is closely
related to the notion of discrete probability space and probability function.
Uniform probability function: If a probability space has a finite number of sample points, say 5
points, +" ß +# ß ÞÞÞß +5 , then the probability function is said to be uniform if each sample point has the
same probability of occurring ; T Ò+3 Ó œ 5" for each 3 œ "ß #ß ÞÞÞß 5 . Tossing a fair die would be an
example of this, with 5 œ '.
Probability of event E: An event E consists of a subset of sample points in the probability space. In
the case of a discrete probability space, the probability of E is
T ÒEÓ œ
D T Ò+3 Ó , the sum of T Ò+3 Ó over all sample points in event E.
+3 − E
Example 1-2: In tossing a "fair" die, it is assumed that each of the six faces has the same chance of
turning up. If this is true, then the probability function T Ð4Ñ œ
"
'
"
'
of
for 4 œ "ß #ß $ß %ß &ß ' is a uniform
probability function on the sample space Ö"ß #ß $ß %ß &ß '×.
The event "an even number is tossed" is E œ Ö#ß %ß '×, and has probability
T ÒEÓ œ "'  "'  "' œ "2 .
© ACTEX 2015

SOA Exam P - Probability
42
SECTION 1 - BASIC PROBABILITY CONCEPTS
Continuous probability space: An experiment can result in an outcome which can be any real
number in some interval. For example, imagine a simple game in which a pointer is spun randomly and
ends up pointing at some spot on a circle. The angle from the vertical (measured clockwise) is between !
and #1. The probability space is the interval Ð!ß #1Ó , the set of possible angles that can occur. We
regard this as a continuous probability space. In the case of a continuous probability space (an interval),
we describe probability by assigning probability to subintervals rather than individual points. If the spin
is "fair", so that all points on the circle are equally likely to occur, then intuition suggests that the
probability assigned to an interval would be the fraction that the interval is of the full circle. For instance,
the probability that the pointer ends up between "3 O'clock" and "9 O'clock" (between 1Î# or 90° and
$1Î# or 270° from the vertical) would be .5, since that subinterval is one-half of the full circle. The
notion of a continuous random variable, covered later in this study guide, is related to a continuous
probability space.
Some rules concerning probability:
(i)
T ÒWÓ œ " if W is the entire probability space (when the underlying experiment is
performed, some outcome must occur with probability 1; for instance W œ Ö"ß #ß $ß %ß &ß '×
for the die toss example).
(ii)
T ÒgÓ œ ! (the probability of no face turning up when we toss a die is 0).
(iii) If events E" ß E# ß ÞÞÞß E8 are mutually exclusive (also called disjoint) then
8
8
T Ò ∪ E3 Ó œ T ÒE" ∪ E# ∪ â ∪ E8 Ó œ T ÒE" Ó  T ÒE# Ó  â  T ÒE8 Ó œ  T ÒE3 Ó .
3œ"
3œ"
This extends to infinitely many mutually exclusive events. This rule is similar to the rule
discussed in Section 0 of this study guide, where it was noted that the number of elements in the
union of mutually disjoint sets is the sum of the numbers of elements in each set. When we have
mutually exclusive events and we add the event probabilities, there is no double counting.
(iv)
For any event E, ! Ÿ T ÒEÓ Ÿ "Þ
(v)
If E § F then T ÒEÓ Ÿ T ÒFÓ .
© ACTEX 2015
SOA Exam P - Probability
SECTION 1 - BASIC PROBABILITY CONCEPTS
(vi)
43
For any events E, F and G , T ÒE ∪ FÓ œ T ÒEÓ  T ÒFÓ  T ÒE ∩ FÓ .
This relationship can be explained as follows. We can formulate E ∪ F as the union of three
mutually exclusive events as follows: E ∪ F œ ÐE ∩ F w Ñ ∪ ÐE ∩ FÑ ∪ ÐF ∩ Ew Ñ.
This is expressed in the following Venn diagram.
A  B A  B B  A
Since these are mutually exclusive events, it follows that
T ÒE ∪ FÓ œ T ÒE ∩ F w Ó  T ÒE ∩ FÓ  T ÒF ∩ Ew Ó .
From the Venn diagram we see that E œ ÐE ∩ F w Ñ ∪ ÐE ∩ FÑ , so that
T ÒEÓ œ T ÒE ∩ F w Ó  T ÒE ∩ FÓ , and we also see that T ÒF ∩ Ew Ó œ T ÒFÓ  T ÒE ∩ FÓ .
It then follows that
T ÒE ∪ FÓ œ ÐT ÒE ∩ F w Ó  T ÒE ∩ FÓÑ  T ÒF ∩ Ew Ó œ T ÒEÓ  T ÒFÓ  T ÒE ∩ FÓ Þ
We subtract T ÒE ∩ FÓ because T ÒEÓ  T ÒFÓ counts T ÒE ∩ FÓ twice. T ÒE ∪ FÓ is the
probability that at least one of the two events Eß F occurs. This was reviewed in Section 0,
where a similar rule was described for counting the number of elements in E ∪ F .
For the union of three sets we have
T ÒE ∪ F ∪ GÓ œ T ÒEÓ  T ÒFÓ  T ÒGÓ  T ÒE ∩ FÓ  T ÒE ∩ GÓ  T ÒF ∩ GÓ  T ÒE ∩ F ∩ GÓ
(vii)
For any event E, T ÒEw Ó œ "  T ÒEÓ .
(viii) For any events E and F , T ÒEÓ œ T ÒE ∩ FÓ  T ÒE ∩ F w Ó
(this was mentioned in (vi), it is illustrated in the Venn diagram above).
(ix)
8
For exhaustive events F" ß F# ß ÞÞÞß F8 , T Ò ∪ F3 Ó œ " .
3œ"
If F" ß F# ß ÞÞÞß F8 are exhaustive and mutually exclusive, they form a partition of the
entire probability space, and for any event E,
T ÒEÓ œ T ÒE ∩ F" Ó  T ÒE ∩ F# Ó  â  T ÒE ∩ F8 Ó œ  T ÒE ∩ F3 Ó .
8
3œ"
(x)
If T is a uniform probability function on a probability space with 5 points, and if event
E consists of 7 of those points, then T ÒEÓ œ 7
5 .
© ACTEX 2015
SOA Exam P - Probability
44
(xi)
SECTION 1 - BASIC PROBABILITY CONCEPTS
The words "percentage" and "proportion" are used as alternatives to "probability".
As an example, if we are told that the percentage or proportion of a group of people that are of a
certain type is 20%, this is generally interpreted to mean that a randomly chosen person from
the group has a 20% probability of being of that type. This is the "long-run frequency"
interpretation of probability. As another example, suppose that we are tossing a fair die. In the
long-run frequency interpretation of probability, to say that the probability of tossing a 1 is "' is
the same as saying that if we repeatedly toss the die, the proportion of tosses that are 1's will
approach "' .
(xii)
8
8
for any events E" ß E# ß ÞÞÞß E8 , T Ò ∪ E3 Ó Ÿ  T ÒE3 Ó , with equality holding if and only if the
3œ"
3œ"
events are mutually exclusive
Example 1-3: Suppose that T ÒE ∩ FÓ œ Þ# ß T ÒEÓ œ Þ' ß and T ÒFÓ œ Þ& .
Find T ÒEw ∪ F w Ó ß T ÒEw ∩ F w Ó , T ÒEw ∩ FÓ and T ÒEw ∪ FÓ.
Solution: Using probability rules we get the following.
T ÒEw ∪ F w Ó œ T ÒÐE ∩ FÑw Ó œ "  T ÒE ∩ FÓ œ Þ) Þ
T ÒE ∪ FÓ œ T ÒEÓ  T ÒFÓ  T ÒE ∩ FÓ œ Þ'  Þ&  Þ# œ Þ*
p T ÒEw ∩ F w Ó œ T ÒÐE ∪ FÑw Ó œ "  T ÒE ∪ FÓ œ "  Þ* œ Þ" Þ
T ÒFÓ œ T ÒF ∩ EÓ  T ÒF ∩ Ew Ó p T ÒEw ∩ FÓ œ Þ&  Þ# œ Þ$ Þ
T ÒEw ∪ FÓ œ T ÒEw Ó  T ÒFÓ  T ÒEw ∩ FÓ œ Þ%  Þ&  Þ$ œ Þ' .
The following Venn diagrams illustrate the various combinations of intersections, unions and
complements of the events E and F .
A
B
P[ A]  .6
© ACTEX 2015
A
B
P[B]  .5
A
B
P[ A B]  .2
SOA Exam P - Probability
SECTION 1 - BASIC PROBABILITY CONCEPTS
A
45
A
B
P[ A B]  .4
B
P[ A B]  .3
From the following Venn diagrams we see that T ÒE ∩ F w Ó œ T ÒEÓ  T ÒE ∩ FÓ œ Þ'  Þ# œ Þ%
and T ÒEw ∩ FÓ œ T ÒFÓ  T ÒE ∩ FÓ œ Þ&  Þ# œ Þ$Þ



P[ A]
.6

P[ A  B]
.2



P[B]
.5
A
P[ A  B]
.4



P[ A  B]
.2
B
A
B
P[ A  B]
.3


The following Venn diagrams shows how to find T ÒE ∪ FÓ .
P[ A  B]
.4
© ACTEX 2015





P[ A  B]
.2


P[ A  B]
.3


A
B
P[ A  B]
.9
SOA Exam P - Probability
46
SECTION 1 - BASIC PROBABILITY CONCEPTS
The relationship T ÒE ∪ FÓ œ T ÒEÓ  T ÒFÓ  T ÒE ∩ FÓ is explained in the following Venn


P[ A  B]




P[ A]

P[B]

P[ A  B]
 .6  .5  .2  .9
The components of the events and their probabilities are summarized in the following diagram.
( A B)  A B
.1
A B
A B
.4
.2
A B
.3
We can represent a variety of events in Venn diagram form and find their probabilities from the
component events described in the previous diagram. For instance, the complement of E is the combined
shaded region in the following Venn diagram, and the probability is T ÒEw Ó œ Þ$  Þ" œ Þ% Þ We can get
this probability also from T ÒEw Ó œ "  T ÒEÓ œ "  Þ' œ Þ%Þ
A
© ACTEX 2015
SOA Exam P - Probability
SECTION 1 - BASIC PROBABILITY CONCEPTS
47
Another efficient way of summarizing the probability information for events E and F is in the form of a
table.
T ÒEÓ œ Þ' (given)
T ÒEw Ó
T ÒFÓ œ Þ& (given)
T ÒE ∩ FÓ œ Þ# (given)
T ÒEw ∩ FÓ
T ÒF w Ó
T ÒE ∩ F w Ó
T ÒEw ∩ F w Ó
Complementary event probabilities can be found from T ÒEw Ó œ "  T ÒEÓ œ Þ% and
T ÒF w Ó œ "  T ÒFÓ œ Þ& . Also, across each row or along each column, the "intersection probabilities"
add up to the single event probability at the end or top:
T ÒFÓ œ T ÒE ∩ FÓ  T ÒEw ∩ FÓ p Þ& œ Þ#  T ÒEw ∩ FÓ p T ÒEw ∩ FÓ œ Þ$ ,
T ÒEÓ œ T ÒE ∩ FÓ  T ÒE ∩ F w Ó p Þ' œ Þ#  T ÒE ∩ F w Ó p T ÒE ∩ Fw Ó œ Þ% , and
T ÒEw Ó œ T ÒEw ∩ FÓ  T ÒEw ∩ F w Ó p Þ% œ Þ$  T ÒEw ∩ F w Ó p T ÒEw ∩ Fw Ó œ Þ" or
T ÒF w Ó œ T ÒE ∩ F w Ó  T ÒEw ∩ F w Ó p Þ& œ Þ%  T ÒEw ∩ Fw Ó p T ÒEw ∩ Fw Ó œ Þ" .
These calculations can be summarized in the next table.
T ÒFÓ œ Þ&
É
T ÒEÓ œ Þ' (given)
Ë
T ÒE ∩ FÓ œ Þ#
Ê

T ÒEw Ó œ "  T ÒEÓ œ Þ%
Ë
T ÒEw ∩ FÓ œ Þ$

T ÒF w Ó œ Þ&
É
T ÒE ∩ F w Ó œ Þ%


T ÒEw ∩ F w Ó œ Þ"

Example 1-4:
A survey is made to determine the number of households having electric appliances in a certain city. It is
found that 75% have radios (V ), 65% have electric irons (M ), 55% have electric toasters (X ), 50% have
(MV ), 40% have (VX ), 30% have (MX ), and 20% have all three. Find the probability that a household has
at least one of these appliances.
Solution:
T ÒV ∪ M ∪ X Ó œ T ÒVÓ  T ÒMÓ  T ÒX Ó
 T ÒV ∩ MÓ  T ÒV ∩ X Ó  T ÒM ∩ X Ó  T ÒV ∩ M ∩ X Ó
œ Þ(&  Þ'&  Þ&&  Þ&  Þ%  Þ$  Þ# œ Þ*& .
© ACTEX 2015
SOA Exam P - Probability
48
SECTION 1 - BASIC PROBABILITY CONCEPTS
The following diagram deconstructs the three events.
I
.05
.30
.10
.20
R
.05
.20
T
.05
The entries in this diagram were calculated from the "inside out". For instance, since T ÐV ∩ MÑ œ Þ&
(given) , and since T ÐV ∩ M ∩ X Ñ œ Þ# (also given) , it follows that
T ÐV ∩ M ∩ X w Ñ œ Þ$ , since
Þ& œ T ÐV ∩ MÑ œ T ÐV ∩ M ∩ X Ñ  T ÐV ∩ M ∩ X w Ñ œ Þ#  T ÐV ∩ M ∩ X w Ñ
(this uses the rule T ÐEÑ œ T ÐE ∩ FÑ  T ÐE ∩ Fw Ñ , where E œ V ∩ M and F œ X ).
This is illustrated in the following diagram.
I
.05
R
R  I T 
.10
.30
.20
R  I T
.05
.20
T
.05
The value ".05" that is inside the diagram for event V refers to T ÐV ∩ M w ∩ X w Ñ (the proportion who have
radios but neither irons nor toasters). This can be found in the following way.
First we find T ÐV ∩ M w Ñ :
Þ(& œ T ÐVÑ œ T ÐV ∩ MÑ  T ÐV ∩ M w Ñ œ Þ&  T ÐV ∩ M w Ñ p T ÐV ∩ M w Ñ œ Þ#&Þ
© ACTEX 2015
SOA Exam P - Probability
SECTION 1 - BASIC PROBABILITY CONCEPTS
49
T ÐV ∩ M w Ñ is the proportion with radios but not irons; this is the ".05" inside V combined with the ".2" in
the lower triangle inside V ∩ X . Then we find T ÐV ∩ M w ∩ X Ñ :
Þ% œ T ÐV ∩ X Ñ œ T ÐV ∩ M ∩ X Ñ  T ÐV ∩ M w ∩ X Ñ
œ Þ#  T ÐV ∩ M w ∩ X Ñ p T ÐV ∩ M w ∩ X Ñ œ Þ# Þ
Finally we find T ÐV ∩ M w ∩ X w Ñ :
Þ#& œ T ÐV ∩ M w Ñ œ T ÐV ∩ M w ∩ X Ñ  T ÐV ∩ M w ∩ X w Ñ
œ Þ#  T ÐV ∩ M w ∩ X w Ñ p T ÐV ∩ M w ∩ X w Ñ œ Þ!& Þ
The other probabilities in the diagram can be found in a similar way. Notice that
T ÐV ∪ M ∪ X Ñ is the sum of the probabilities of all the disjoint pieces inside the three events,
T ÐV ∪ M ∪ X Ñ œ Þ!&  Þ!&  Þ!&  Þ"  Þ#  Þ$  Þ# œ Þ*& .
We can also use the rule
T ÐV ∪ M ∪ X Ñ œ T ÐVÑ  T ÐMÑ  T ÐX Ñ  T ÐV ∩ MÑ  T ÐV ∩ X Ñ  T ÐM ∩ X Ñ  T ÐV ∩ M ∩ X Ñ
œ Þ(&  Þ'&  Þ&&  Þ&  Þ%Þ  $  Þ# œ Þ*& .

Either way, this implies that 5% of the households have none of the three appliances.
It is possible that information is given in terms of numbers of units in each category rather than
proportion of probability of each category that was given in Example 1-4.
Example 1-5:
In a survey of 120 students, the following data was obtained.
60 took English, 56 took Math, 42 took Chemistry, 34 took English and Math, 20 took Math and
Chemistry, 16 took English and Chemistry, 6 took all three subjects.
Find the number of students who took
(i) none of the subjects,
(ii) Math, but not English or Chemistry,
(iii) English and Math but not Chemistry.
Solution:
Since I ∩ Q has 34 and I ∩ Q ∩ G has 6 , it follows that I ∩ Q ∩ G w has 28 .
The other entries are calculated in the same way (very much like the previous example).
(i) The total number of students taking any of the three subjects is I ∪ Q ∪ G , and is
"'  #)  '  "!  )  "%  "# œ *% . The remaining #' (out of 120) students are not taking
any of the three subjects (this could be described as the set I w ∩ Q w ∩ G w ).
(ii) Q ∩ I w ∩ G w has 8 students .
(iii) I ∩ Q ∩ G w has 28 students .
© ACTEX 2015
SOA Exam P - Probability
50
SECTION 1 - BASIC PROBABILITY CONCEPTS
Example 1-5 continued
The following diagram illustrates how the numbers of students can be deconstructed.
E
16
28
10
6
M
8
14
C
12

© ACTEX 2015
SOA Exam P - Probability
PROBLEM SET 1
51
PROBLEM SET 1
Basic Probability Concepts
1.
A survey of 1000 people determines that 80% like walking and 60% like biking, and all like at least
one of the two activities. What is the probability that a randomly chosen person in this survey likes
biking but not walking?
A) 0
2.
B) .1
C) .2
D) .3
E) .4
(SOA) Among a large group of patients recovering from shoulder injuries, it is found that 22%
visit both a physical therapist and a chiropractor, whereas 12% visit neither of these. The
probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a
physical therapist. Determine the probability that a randomly chosen member of this group visits a
physical therapist.
A) 0.26
3.
B) 0.38
C) 0.40
D) 0.48
E) 0.62
(SOA) An insurer offers a health plan to the employees of a large company. As part of this plan,
the individual employees may choose exactly two of the supplementary coverages A, B, and C, or
they may choose no supplementary coverage. The proportions of the company’s employees that
"
&
choose coverages A, B, and C are "
% , $ , and "# , respectively. Determine the probability that a
randomly chosen employee will choose no supplementary coverage.
A) 0
© ACTEX 2015
B)
%(
"%%
C)
"
#
D)
*(
"%%
E)
(
*
SOA Exam P - Probability
52
4.
PROBLEM SET 1
(SOA) An auto insurance company has 10,000 policyholders.
Each policyholder is classified as
(i)
young or old;
(ii)
male or female; and
(iii) married or single.
Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. The policyholders
can also be classified as 1320 young males, 3010 married males, and 1400 young married persons.
Finally, 600 of the policyholders are young married males. How many of the company’s
policyholders are young, female, and single?
A) 280
5.
B) 423
C) 486
D) 880
E) 896
(SOA) The probability that a visit to a primary care physician’s (PCP) office results in neither lab
work nor referral to a specialist is 35% . Of those coming to a PCP’s office, 30% are referred to
specialists and 40% require lab work. Determine the probability that a visit to a PCP’s office
results in both lab work and referral to a specialist.
A) 0.05
6.
B) 0.12
D) 0.25
E) 0.35
(SOA) You are given T ÒE∪FÓ œ !Þ( and T ÒE∪F w Ó œ !Þ*.
A) 0.2
7.
C) 0.18
B) 0.3
C) 0.4
D) 0.6
Determine T ÒEÓÞ
E) 0.8
(SOA) A survey of a group’s viewing habits over the last year revealed the following information:
(i) 28% watched gymnastics
(ii) 29% watched baseball
(iii)
19% watched soccer
(iv) 14% watched gymnastics and baseball
(v) 12% watched baseball and soccer
(vi) 10% watched gymnastics and soccer
(vii)
8% watched all three sports.
Calculate the percentage of the group that watched none of the three sports during the last year.
A) 24
© ACTEX 2015
B) 36
C) 41
D) 52
E) 60
SOA Exam P - Probability
PROBLEM SET 1
8.
53
(SOA) Under an insurance policy, a maximum of five claims may be filed per year by a
policyholder. Let :8 be the probability that a policyholder files 8 claims during a given year, where
8 œ !ß "ß #ß $ß %ß & . An actuary makes the following observations:
i) :8 :8" for 8 œ !ß "ß #ß $ß % .
ii) The difference between :8 and :8" is the same for 8 œ !ß "ß #ß $ß % .
iii) Exactly 40% of policyholders file fewer than two claims during a given year.
Calculate the probability that a random policyholder will file more than three claims
during a given year.
A) 0.14
9.
B) 0.16
C) 0.27
D) 0.29
E) 0.33
(SOA) The probability that a member of a certain class of homeowners with liability and property
coverage will file a liability claim is 0.04, and the probability that a member of this class will file a
property claim is 0.10. The probability that member of this class will file a liability claim but not a
property claim is 0.01.
Calculate the probability that a randomly selected member of this class of homeowners will not file
a claim of either type.
A) 0.850
10.
B) 0.860
C) 0.864
D) 0.870
E) 0.890
(SOA) A mattress store sells only, king, queen and twin-size mattresses. Sales records at the store
indicate that one-fourth as many queen-size mattresses are sold as king and twin-size mattresses
combined. Records also indicate that three times as many king-size mattresses are sold as twin-size
mattresses.
Calculate the probability that the next mattress sold is either king or queen-size
A) 0.12
© ACTEX 2015
B) 0.13
C) 0.080
D) 0.85
E) 0.95
SOA Exam P - Probability
54
PROBLEM SET 1
PROBLEM SET 1 SOLUTIONS
1.
Let E œ "like walking" and F œ "like biking" . We use the interpretation that "percentage" and
"proportion" are taken to mean "probability".
We are given T ÐEÑ œ Þ) ß T ÐFÑ œ Þ' and T ÐE ∪ FÑ œ " .
From the diagram below we can see that since E ∪ F œ E ∪ ÐF ∩ Ew Ñ we have
T ÐE ∪ FÑ œ T ÐEÑ  T ÐEw ∩ FÑ p T ÐEw ∩ FÑ œ Þ# is the proportion of people who like biking
but (and) not walking . In a similar way we get T ÐE ∩ F w Ñ œ Þ% .
A
.4
B
.4
A  B
.2
An algebraic approach is the following. Using the rule T ÐE ∪ FÑ œ T ÐEÑ  T ÐFÑT ÐE ∩ FÑ ,
we get " œ Þ)  Þ'  T ÐE ∩ FÑ p T ÐE ∩ FÑ œ Þ% . Then, using the rule
T ÐFÑ œ T ÐF ∩ EÑ  T ÐF ∩ Ew Ñ , we get T ÐF ∩ Ew Ñ œ Þ'  Þ% œ Þ# .
2.
Answer: C
G - chiropractor visit ; X - therapist visit.
w
We are given T ÐG ∩ X Ñ œ Þ## , T ÐG ∩ X w Ñ œ Þ"# , T ÐGÑ œ T ÐX Ñ  Þ"% .
w
)) œ "  T ÐG ∩ X w Ñ œ T ÐG ∪ X Ñ œ T ÐGÑ  T ÐX Ñ  T ÐG ∩ X Ñ
œ T ÐX Ñ  Þ"%  T ÐX Ñ  Þ## p T ÐX Ñ œ Þ%) .
3.
Answer: D
Since someone who chooses coverage must choose exactly two supplementary coverages, in order
for someone to choose coverage A, they must choose either A-and-B or A-and-C. Thus, the
proportion of "% of individuals that choose A is
T ÒE ∩ FÓ  T ÒE ∩ GÓ œ "% (where this refers to the probability that someone chosen at random in
the company chooses coverage A). In a similar way we get
Then, ÐT ÒE ∩ FÓ  T ÒE ∩ GÓÑ  ÐT ÒF ∩ EÓ  T ÒF ∩ GÓÑ  ÐT ÒG ∩ EÓ  T ÒG ∩ FÓÑ
&
œ #ÐT ÒE ∩ FÓ  T ÒE ∩ GÓ  T ÒF ∩ GÓÑ œ "%  "$  "#
œ".
"
It follows that T ÒE ∩ FÓ  T ÒE ∩ GÓ  T ÒF ∩ GÓ œ # .
© ACTEX 2015
SOA Exam P - Probability
PROBLEM SET 1
55
3. continued
This is the probability that a randomly chosen individual chooses some form of coverage, since if
someone who chooses coverage chooses exactly two of A,B, and C. Therefore, the probability that
a randomly chosen individual does not choose any coverage is the probability of the
complementary event, which is also "# .
4.
Answer: C
We identify the following subsets of the set of 10,000 policyholders:
] œ young, with size 3000 (so that ] w œ old has size 7000) ,
Q œ male, with size 4600 (so that Q w œ female has size 5400), and
w
G œ married, with size 7000 (so that G œ single has size 3000).
We are also given that ] ∩ Q has size 1320 , Q ∩ G has size 3010 ,
] ∩ G has size 1400 , and ] ∩ Q ∩ G has size 600 .
w
We wish to find the size of the subset ] ∩ Q w ∩ G .
We use the following rules of set theory:
(i) if two finite sets are disjoint (have no elements in common, also referred to as empty
intersection), then the total number of elements in the union of the two sets is the sum of
the numbers of elements in each of the sets;

(ii) for any sets E and F , E œ ÐE ∩ FÑ ∪ ÐE ∩ F w Ñ , and E ∩ F and E ∩ F are disjoint.
Applying rule (ii), we have ] œ Ð] ∩ Q Ñ ∪ Ð] ∩ Q w Ñ . Applying rule (i), it follows that the size
of ] ∩ Q w must be 3000  1320 œ 1680 .
We now apply rule (ii) to ] ∩ G to get ] ∩ G œ Ð] ∩ G ∩ Q Ñ ∪ Ð] ∩ G ∩ Q w Ñ .
Applyng rule (i), it follows that ] ∩ G ∩ Q w has size 1400  600 œ 800.
w
Now applying rule (ii) to ] ∩ Q w we get ] ∩ Q w œ Ð] ∩ Q w ∩ GÑ ∪ Ð] ∩ Q w ∩ G Ñ .
w
Applying rule (i), it follows that ] ∩ Q w ∩ G has size 1680  800 œ 880 .
Within the "Young" category, which we are told is 3000, we can summarize the calculations in the
following table. This is a more straightforward solution.
Married
1400 (given)
Single
1600 œ 3000  1400
Male
1320 (given)
600 (given)
720 œ 1320  600
Female
1680 œ
3000  1320
800 œ 1400  600
880 œ 1600  720
© ACTEX 2015
Answer: D
SOA Exam P - Probability
56
5.
PROBLEM SET 1
We identify events as follows:
P À lab work needed
V À referral to a specialist needed
We are given T ÒPw ∩ V w Ó œ Þ$& ß T ÒVÓ œ Þ$ ß T ÒPÓ œ Þ% . It follows that
T ÒP ∪ VÓ œ "  T ÒPw ∩ V w Ó œ Þ'& , and then since
T ÒP ∪ VÓ œ T ÒPÓ  T ÒVÓ  T ÒP ∩ VÓ, we get T ÒP ∩ VÓ œ Þ$  Þ%  Þ'& œ Þ!& .
L  R
.35
L
.05
.35
.25
R
These calculations can be summarized in the following table
6.
P ß Þ%
given
Pw ß Þ'
Þ' œ "  Þ%
V ß Þ$
given
P∩V
Þ!& œ Þ%  Þ$&
Pw ∩ V
Þ#& œ Þ$  Þ!&
V w , Þ(
Þ( œ "  Þ$
P ∩ Vw
Þ$& œ Þ(  Þ$&
Pw ∩ V w , Þ$&
given
Answer: A
T ÒE ∪ FÓ œ T ÒEÓ  T ÒFÓ  T ÒE ∩ FÓ ß T ÒE ∪ Fw Ó œ T ÒEÓ  T ÒFw Ó  T ÒE ∩ Fw Ó.
We use the relationship T ÒEÓ œ T ÒE ∩ FÓ  T ÒE ∩ F w Ó . Then
T ÒE ∪ FÓ  T ÒE ∪ F w Ó œ T ÒEÓ  T ÒFÓ  T ÒE ∩ FÓ  T ÒEÓ  T ÒFw Ó  T ÒE ∩ Fw Ó
œ #T ÒEÓ  "  T ÒEÓ œ T ÒEÓ  " (since T ÒFÓ  T ÒF w Ó œ ") .
Therefore, Þ(  Þ* œ T ÒEÓ  " so that T ÒEÓ œ Þ' .
© ACTEX 2015
SOA Exam P - Probability
PROBLEM SET 1
57
6. continued
An alternative solution is based on the following Venn diagrams.
P[ A  B]  .90
P[ A  B]  .70
P[( A B)]  P[ A B]  .10
In the third diagram, the shaded area is the complement of that in the second diagram
(using De Morgan's Law, we have ÐE ∪ F w Ñw œ Ew ∩ F ww œ Ew ∩ F ) . Then it can be
seen from diagrams 1 and 3 that E œ ÐE ∪ FÑ  ÐEw ∩ FÑ , so that
T ÒEÓ œ T ÒE ∪ FÓ  T ÒEw ∩ FÓ œ Þ(  Þ" œ Þ' .
7.
Answer: D
We identify the following events:
K - watched gymnastics , F - watched baseball ,
W - watched soccer .
We wish to find T ÒKw ∩ F w ∩ W w Ó . By DeMorgan's rules we have
T ÒKw ∩ F w ∩ W w Ó œ "  T ÒK ∪ F ∪ WÓ .
We use the relationship
T ÒK ∪ F ∪ WÓ œ T ÒKÓ  T ÒFÓ  T ÒWÓ
 T ÒK ∩ FÓ  T ÒK ∩ WÓ  T ÒF ∩ WÓ  T ÒK ∩ F ∩ WÓ .
We are given T ÒKÓ œ Þ#) ß T ÒFÓ œ Þ#* ß T ÒWÓ œ Þ"* ß
T ÒK ∩ FÓ œ Þ"% ß T ÒK ∩ WÓ œ Þ"! ß T ÒF ∩ WÓ œ Þ"# ß T ÒK ∩ F ∩ WÓ œ Þ!) .
Then T ÒK ∪ F ∪ WÓ œ Þ%) and T ÒKw ∩ F w ∩ W w Ó œ "  Þ%) œ Þ&# . Answer: D
© ACTEX 2015
SOA Exam P - Probability
58
8.
PROBLEM SET 1
The probability in question is :%  :& . We know that :!  :"  :#  :$  :%  :& œ ".
We are given that :!  :" œ :"  :# œ :#  :$ œ :$  :% œ :%  :& ,
and :!  :" œ Þ% . If we let > be equal to the common difference :8  :8" , then
:" œ :!  > ß :# œ :!  #> ß :$ œ :!  $> ß :% œ :!  %> and :& œ :!  &> .
Then :!  :"  :#  :$  :%  :& œ ':!  "&> œ " . When we combine this equation
&
"
with :!  :" œ #:!  > œ Þ% , we can solve the two equations to get :! œ #%
, and > œ  '!
.
"(
"&
$#
Then :% œ :!  %> œ "#! , and :& œ :!  &> œ "#! , so that :%  :& œ "#! œ Þ#( . Answer: C
9.
We define the following events: P œ file a liability claim , T œ file a property claim.
We are given T ÐPÑ œ !Þ!% ß T ÐT Ñ œ !Þ"! ß T ÐP ∩ T w Ñ œ !Þ!" . We wish to find T ÐPw ∩ T w Ñ .
T ÐT w Ñ œ "  T ÐT Ñ! œ *! and !Þ*! œ T ÐT w Ñ œ T ÐP ∩ T w Ñ  T ÐPw ∩ T w Ñ œ !Þ!"  T ÐPw ∩ T w ÑÞ
It follows that T ÐPw ∩ T w Ñ œ !Þ)*. Answer: E
10.
We define X to be the event that the next mattress sold is twin-size, and similarly we define O and
U as the events that the next mattress sold is king-size and queen-size, respectively. We define
T ÐX Ñ œ - . Then T ÐOÑ œ $- and T ÐUÑ œ "% ‚ ÒT ÐOÑ  T ÐX ÑÓ œ - .
Since T ÐX Ñ  T ÐUÑ  T ÐOÑ œ ", we have &- œ " , so that - œ !Þ#.
Then T ÐO ∪ UÑ œ %- œ !Þ)Þ Answer: C
© ACTEX 2015
SOA Exam P - Probability
SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS
233
SECTION 8 - JOINT, MARGINAL,
AND CONDITIONAL DISTRIBUTIONS
Joint distribution of random variables \ and ]
A random variable \ is a numerical outcome that results from some random experiment, such as the
number that turns up when tossing a die. It is possible that an experiment may result in two or more
numerical outcomes. A simple example would be the numbers that turn up when tossing two dice. \
could be the number that turns up on the first die and ] could be the number on the second die. Another
example could be the following experiment. A coin is tossed and if the outcome is head then toss one die,
and if the outcome is tails then toss two dice. We could set \ œ " for a head and \ œ # for a tail and
] œ total on the dice thrown. In both of the examples just described, we have a pair of random variables
\ and ] , that result from the experiment. \ and ] might be unrelated or independent of one another (as
in the example of the toss of two independent dice), or they might be related to each other (as in the coindice example).
We describe the probability distribution of two or more random variables together as a joint distribution.
As in the case of a single discrete random variable, we still describe probabilities for each possible pair of
outcomes for a pair of discrete random variables. In the case of a pair of random variables \ and ] , there
would be probabilities of the form T ÒÐ\ œ BÑ ∩ Ð] œ CÑÓ for each pair ÐBß CÑ of possible outcomes. For
a pair of continuous random variables \ and ] , there would be a density function to describe density
over a two dimensional region.
A joint distribution of two random variables has a probability function or probability density function
0 ÐBß CÑ that is a function of two variables (sometimes denoted 0\ß] ÐBß CÑ). It is defined over a twodimensional region. For joint distributions of continuous random variables \ and ] , the region of
probability (the probability space) is usually a rectangle or triangle in the B-C plane.
If \ and ] are discrete random variables, then 0 ÐBß CÑ œ T ÒÐ\ œ BÑ ∩ Ð] œ CÑÓ is the joint probability
function, and it must satisfy
(i)
(ii)
! Ÿ 0 ÐBß CÑ Ÿ " and
0 ÐBß CÑ œ " .
B C
If \ and ] are continuous random variables, then 0 ÐBß CÑ must satisfy
(i)
(ii)
0 ÐBß CÑ ! and
∞
∞
0 ÐBß CÑ .C .B œ ".
© ACTEX 2015
SOA Exam P - Probability
234
SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS
In the two dice example described above, if the two dice are tossed independently of one another then
"
0 ÐBß CÑ œ T ÒÐ\ œ BÑ ∩ Ð] œ CÑÓ œ T Ò\ œ BÓ ‚ T Ò] œ CÓ œ "' ‚ "' œ $'
for each pair with
B œ "ß #ß $ß %ß &ß ' and C œ "ß #ß $ß %ß &ß ' . The coin-die toss example above is more complicated because
the number of dice tossed depends on whether the toss is head or tails. If the coin toss is a head then
\ œ " and ] œ "ß #ß $ß %ß &ß ' so
"
0 Ð"ß CÑ œ T ÒÐ\ œ "Ñ ∩ Ð] œ CÑÓ œ "# ‚ "' œ "#
for C œ "ß #ß $ß %ß &ß ' .
If the coin toss is tail then \ œ # and ] œ #ß $ß ÞÞÞß "# with
"
"
0 Ð#ß #Ñ œ T ÒÐ\ œ #Ñ ∩ Ð] œ #ÑÓ œ "# ‚ $'
œ (#
,
"
#
"
0 Ð#ß $Ñ œ T ÒÐ\ œ #Ñ ∩ Ð] œ $ÑÓ œ # ‚ $' œ $' , etc.
It is possible to have a joint distribution in which one variable is discrete and one is continuous, or either
has a mixed distribution. The joint distribution of two random variables can be extended to a joint
distribution of any number of random variables.
If E is a subset of two-dimensional space, then T ÒÐ\ß ] Ñ − EÓ is the summation (discrete case) or
double integral (continuous case) of 0 ÐBß CÑ over the region E.
Example 8-1:
\ and ] are discrete random variables which are jointly distributed
with the probability function 0 ÐBß CÑ defined in the following table:
\
"
]
!
"
"
!
"
"
")
"
*
"
'
"
*
"
'
"
'
"
*
!
"
*
From this table we see, for example, that
T Ò\ œ !ß ] œ  "Ó œ 0 Ð!ß  "Ñ œ "* .
Find (i) T Ò\  ] œ "Ó , (ii) T Ò\ œ !Ó and (iii) T Ò\  ] Ó .
Solution:
(i)
We identify the ÐBß CÑ-points for which \  ] œ ", and the probability is the sum of 0 ÐBß CÑ over
those points. The only Bß C combinations that sum to 1 are the points Ð!ß "Ñ and Ð"ß !Ñ .
(ii)
&
Therefore, T Ò\  ] œ "Ó œ 0 Ð!ß "Ñ  0 Ð"ß !Ñ œ "*  "' œ ")
Þ
We identify the ÐBß CÑ-points for which \ œ ! . These are Ð!ß  "Ñ and Ð!ß "Ñ (we omit Ð!ß !Ñ
since there is no probability at that point). T Ò\ œ !Ó œ 0 Ð!ß  "Ñ  0 Ð!ß "Ñ œ "*  "* œ #*
(iii) The ÐBß CÑ-points satisfying \  ] are Ð  "ß !Ñ ß Ð  "ß "Ñ and Ð!ß "Ñ .
"
&
Then T Ò\  ] Ó œ 0 Ð  "ß !Ñ  0 Ð  "ß "Ñ  0 Ð!ß "Ñ œ "*  ")
 *" œ ")
Þ
© ACTEX 2015

SOA Exam P - Probability
SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS
235
Example 8-2:
Suppose that 0 ÐBß CÑ œ OÐB#  C# Ñ is the density function for the joint distribution of the continuous
random variables \ and ] defined over the unit square bounded by the points Ð!ß !Ñ ß Ð"ß !Ñ ß Ð"ß "Ñ and
Ð!ß "Ñ, find O . Find T Ò\  ] "Ó .
Solution:
In order for 0 ÐBß CÑ to be a properly defined joint density, the (double) integral of the density function
over the region of density must be 1, so that
" "
" œ ! ! OÐB#  C# Ñ .C .B œ O † #$ Ê O œ $#
Ê 0 ÐBß CÑ œ $# ÐB#  C# Ñ for ! Ÿ B Ÿ " and ! Ÿ C Ÿ ".
In order to find the probability T Ò\  ] "Ó, we identify the two dimensional region representing
\  ] " . This is generally found by drawing the boundary line for the inequality, which is
B  C œ " (or C œ "  B) in this case, and then determining which side of the line is represented in the
inequality. We can see that B  C " is equivalent to C "  B .
This is the shaded region in the graph below.
y
1
y  1 x
1
x
The probability T Ò\  ] "Ó is found by integrating the joint density over the two-dimensional region.
It is possible to represent two-variable integrals in either order of integration. In some cases one order of
integration is more convenient than the other. In this case there is not much advantage of one direction of
integration over the other.
Cœ"
" "
"
T Ò\  ] "Ó œ ! "B $# ÐB#  C# Ñ .C .B œ ! "# Ð$B# C  C$ 
"
œ ! "# Ð$B#  "  $B# Ð"  BÑ  Ð"  BÑ$ Ñ .B
Cœ"B
œ $% Þ
Ñ .B
Reversing the order of integration, we have B "  C , so that
" "
T Ò\  ] "Ó œ ! "C $# ÐB#  C# Ñ .B .C œ $% Þ
© ACTEX 2015

SOA Exam P - Probability
236
SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS
Example 8-3:
Continuous random variables \ and ] have a joint distribution with density function
BC
0 ÐBß CÑ œ B#  $ for !  B  " and
Find the conditional probability T Ò\ 
!C#.
"
# l]
 "# Ó.
Solution:
T ÒE∩FÓ
T ÒFÓ
We use the usual definition T ÒElFÓ œ
T Ò\  "# l]  "# Ó œ
T ÒÐ\ "# Ñ∩Ð]  "# ÑÓ
T Ò]  "# Ó
.
.
These regions are described in the following diagram
y
y
2
2
1
2
1
2
1
2
1
x
1
x
Y  12
 X  12   Y  12 
BC
"
#
%$
T ÒÐ\  "# Ñ ∩ Ð]  "# ÑÓ œ "Î# "Î# ÒB#  $ Ó .C .B œ '%
.
BC
#
"
#
"
T Ò]  "# Ó œ "Î# Ò! 0 ÐBß CÑ .BÓ .C œ "Î# ! ÒB#  $ Ó .B .C œ "$
"'
%$Î'%
"
"
%$
p T Ò\  # l]  # Ó œ "$Î"' œ &# .

Cumulative distribution function of a joint distribution: If random variables \ and ] have a joint
distribution, then the cumulative distribution function is J ÐBß CÑ œ T ÒÐ\ Ÿ BÑ ∩ Ð] Ÿ CÑÓ .
B
C
In the continuous case, J ÐBß CÑ œ ∞ ∞ 0 Ð=ß >Ñ .> .= ,
and in the discrete case, J ÐBß CÑ œ 
B
 0 Ð=ß >Ñ.
C
=œ∞ >œ∞
#
In the continuous case, `B` `C J ÐBß CÑ œ 0 ÐBß CÑ
Example 8-4:
The cumulative distribution function for the joint distribution of the continuous random variables \ and
] is J ÐBß CÑ œ ÐÞ#ÑÐ$B$ C  #B# C# Ñ , for ! Ÿ B Ÿ " and ! Ÿ C Ÿ " . Find 0 Ð "# ß "# Ñ .
Solution: 0 ÐBß CÑ œ
© ACTEX 2015
`#
`B `C
"(
J ÐBß CÑ œ ÐÞ#ÑÐ*B#  )BCÑ p 0 Ð #" ß #" Ñ œ #!
Þ

SOA Exam P - Probability
PRACTICE EXAM 7
469
PRACTICE EXAM 7
1.
A study of the relationship between blood pressure and cholesterol level showed the following
results for people who took part in the study:
(a) of those who had high blood pressure, 50% had a high cholesterol level, and
(b) of those who had high cholesterol level, 80% had high blood pressure.
Of those in the study who had at least one of the conditions of high blood pressure or high
cholesterol level, what is the proportion who had both conditions?
A) "$
2.
B) %*
C) &*
D) #$
E) (*
A study of international athletes shows that of the two performance-enhancing steroids Dianabol
and Winstrol, 5% of athletes use Dianabol and not Winstrol, 2% use Winstrol and not Dianabol,
and 1% use both. A breath test has been developed to test for the presence of the these drugs in an
athlete. Past use of the test has resulted in the following information regarding the accuracy of the
test. Of the athletes that are using both drugs, the test indicates that 75% are using both drugs, 15%
are using Dianabol only and 10% are using Winstrol only. In 80% of the athletes that are using
Dianabol but not Winstrol, the test indicates they are using Dianabol but not Winstrol, and for the
other 20% the test indicates they are using both drugs. In 60% of the athletes that are using
Winstrol but not Dianabol, the test indicates that they are using Winstrol only, and for the other
40% the test indicates they are using both drugs. For all athletes that are using neither Dianabol nor
Winstrol, the test always indicates that they are using neither drug.
Of those athletes who test positive for Dianabol but not Winstrol, find the percentage that are using
both drugs.
A) 1.2%
3.
B) 2.4%
C) 3.6%
D) 4.8%
E) 6.0%
The random variable R has the following characteristics:
(i) With probability :, R has a binomial distribution with ; œ !Þ& and 7 œ #.
(ii) With probability "  :, R has a binomial distribution with ; œ !Þ& and 7 œ %.
Which of the following is a correct expression for ProbÐR œ #Ñ?
A)
B)
C)
D)
E)
!Þ"#&:#
!Þ$(&  !Þ"#&:
!Þ$(&  !Þ"#&:#
!Þ$(&  !Þ"#&:#
!Þ$(&  !Þ"#&:
© ACTEX 2015
SOA Exam P - Probability
470
4.
PRACTICE EXAM 7
An insurance company does a study of claims that arrive at a regional office. The study focuses on
the days during which there were at most 2 claims. The study finds that for the days on which there
were at most 2 claims, the average number of claims per day is 1.2 . The company models the
number of claims per day arriving at that office as a Poisson random variable. Based on this model,
find the probability that at most 2 claims arrive at that office on a particular day.
A) .62
5.
B) .64
C) .66
D) .68
E) .70
An actuarial trainee working on loss distributions encounters a special distribution. The student
α
reads a discussion of the distribution and sees that the density of \ is 0 ÐBÑ œ Bα)
α" on the region
α)
\  ) , where α and ) must both be  !, and the mean is α"
if α  ".
The student is analyzing loss data that is assumed to follow such a distribution, but the values of α
and ) are not specified, although it is known that )  #!!. The data shows that the average loss for
all losses is 180, and the average loss for all losses that are above 200 is 300.
Find the median of the loss distribution.
A)
B)
C)
D)
E)
6.
Less than 100
At least 100, but less than 120
At least 120, but less than 140
At least 140, but less than 160
At least 160
An insurance claims administrator verifies claims for various loss amounts.
For a loss claim of amount B, the amount of time spent by the administrator to verify the claim is
uniformly distributed between 0 and "  B hours. The amount of each claim received by the
administrator is uniformly distributed between 1 and 2. Find the average amount of time that an
administrator spends on a randomly arriving claim.
A) 1.125
7.
B) 1.250
C) 1.375
D) 1.500
E) 1.625
A husband and wife have a health insurance policy. The insurer models annual losses for the
husband separately from the wife. \ is the annual loss for the husband and ] is the annual loss for
the wife. \ has a uniform distribution on the interval Ð!ß &Ñ and ] has a uniform distribution on the
interval Ð!ß (Ñ, and \ and ] are independent. The insurer applies a deductible of 2 to the combined
annual losses, and the insurer pays a maximum of 8 per year. Find the expected annual payment
made by the insurer for this policy.
A) 2
© ACTEX 2015
B) 3
C) 4
D) 5
E) 6
SOA Exam P - Probability
PRACTICE EXAM 7
477
PRACTICE EXAM 7 - SOLUTIONS
1.
We will use F to denote the event that a randomly chosen person in the study has high blood
pressure, and G will denote the event high cholesterol level.
The information given tells us that T ÐGlFÑ œ Þ&! and T ÐFlGÑ œ Þ)! .
We wish to find T ÐF ∩ GlF ∪ GÑ . This is
T ÒÐF∩GÑ∩ÐF∪GÑÓ
T ÐF∪GÑ
œ
œ
T ÒF∩GÑÓ
T ÐFÑT ÐGÑT ÐF∩GÑ
"
T ÐFÑT ÐGÑT ÐF∩GÑ
T ÐF∩GÑ
œ
"
"
"
Ò T ÐGlFÑ
 T ÐGlFÑ
"Ó
œ
2.
"
"
"
Þ&  Þ) "
œ
"
#Þ#&
œ
%
*
.
Answer: B
We define the following events:
H - the athlete uses Dianabol
[ - the athlete uses Winstrol
X H - the test indicates that the athlete uses Dianabol
X [ - the test indicates that the athlete uses Winstrol
We are given the following probabilities
T ÐH ∩ [ w Ñ œ Þ!& ß
T ÐHw ∩ [ Ñ œ Þ!# ß
T ÐH ∩ [ Ñ œ Þ!" ,
T ÐX H ∩ X [ lH ∩ [ Ñ œ Þ(& ß
T ÐX H ∩ X [ w lH ∩ [ Ñ œ Þ"& ß
T ÐX H ∩ X [ lH ∩ [ w Ñ œ Þ# ß
T ÐX H ∩ X [ w lH ∩ [ w Ñ œ Þ) ß
T ÐX H ∩ X [ lHw ∩ [ Ñ œ Þ% ß
T ÐX Hw ∩ X [ lHw ∩ [ Ñ œ Þ' .
We wish to find T ÐH ∩ [ lX H ∩ X [ w Ñ œ
T ÐH∩[ ∩X H∩X [ w Ñ
T ÐX H∩X [ w Ñ
T ÐX Hw ∩ X [ lH ∩ [ Ñ œ Þ" ß
.
The numerator is
T ÐH ∩ [ ∩ X H ∩ X [ w Ñ œ T ÐX H ∩ X [ w lH ∩ [ Ñ † T ÐH ∩ [ Ñ œ ÐÞ"&ÑÐÞ!"Ñ œ Þ!!"& .
The denominator is
T ÐX H ∩ X [ w Ñ œ T ÐX H ∩ X [ w ∩ H ∩ [ Ñ  T ÐX H ∩ X [ w ∩ Hw ∩ [ Ñ
 T ÐX H ∩ X [ w ∩ H ∩ [ w Ñ  T ÐX H ∩ X [ w ∩ Hw ∩ [ w Ñ
We have used the rule T ÐEÑ œ T ÐE ∩ F" Ñ  T ÐE ∩ F# Ñ  â , where F" ß F# ß ÞÞÞ
forms a partition. The partition in this case is F" œ H ∩ [ ß F# œ Hw ∩ [ ß
F$ œ H ∩ [ w ß F% œ Hw ∩ [ w , since an athlete must be using both, one or neither of the drugs.
© ACTEX 2015
SOA Exam P - Probability
478
PRACTICE EXAM 7
2. continued
We have just seen that T ÐX H ∩ X [ w ∩ H ∩ [ Ñ œ Þ!!"& .
In a similar way, we have
T ÐX H ∩ X [ w ∩ Hw ∩ [ Ñ œ T ÐX H ∩ X [ w lHw ∩ [ Ñ † T ÐHw ∩ [ Ñ œ Ð!ÑÐÞ!#Ñ œ ! , and
T ÐX H ∩ X [ w ∩ H ∩ [ 'Ñ œ T ÐX H ∩ X [ w lH ∩ [ w Ñ † T ÐH ∩ [ w Ñ œ ÐÞ)ÑÐÞ!&Ñ œ Þ!% , and
T ÐX H ∩ X [ w ∩ Hw ∩ [ w Ñ œ T ÐX H ∩ X [ w lHw ∩ [ w Ñ † T ÐHw ∩ [ w Ñ œ Ð!ÑÐÞ*#Ñ œ !
(note that T ÐHw ∩ [ w Ñ œ "  T ÐH ∪ [ Ñ œ "  T ÐH ∩ [ w Ñ  T ÐHw ∩ [ Ñ  T ÐH ∩ [ Ñ œ Þ*#
Þ!!"&
Then, T ÐH ∩ [ lX H ∩ X [ w Ñ œ Þ!!"&!Þ!%!
œ Þ!$' , 3.6% .
Answer: C
3.
T ÐR œ #Ñ œ :T ÐR" œ #Ñ  Ð"  :ÑT ÐR# œ #Ñ œ :ÐÞ&Ñ#  Ð"  :Ñ'ÐÞ&Ñ% œ Þ$(&  Þ"#&: .
75
 5
We have used the binomial probabilities  7
.
Answer: E
5 ; Ð"  ;Ñ
4.
Suppose that the mean number of claims per day arriving at the office is -.
Let \ denote the number of claims arriving in one day.
# Then the probability of at most 2 claims in one day is T Ð\ Ÿ #Ñ œ /-  -/-  - /# .
The conditional probability of 0 claims arriving on a day given that there are at most 2 for the day is
-
T Ð\œ!Ñ
"
T Ð\ œ !l\ Ÿ #Ñ œ T Ð\Ÿ#Ñ œ - /- -#/- œ
# .
"- -#
/ -/  #
The conditional probability of 1 claim arriving on a day given that there are at most 2 for the day is
-
T Ð\œ"Ñ
T Ð\ œ "l\ Ÿ #Ñ œ T Ð\Ÿ#Ñ œ - -/- -#/- œ
# .
"- -#
/ -/  #
The conditional probability of 2 claims arriving on a day given that there are at most 2 for the day is
-# /-
T Ð\œ#Ñ
-#
#
#
T Ð\ œ #l\ Ÿ #Ñ œ T Ð\Ÿ#Ñ œ # .
# - œ
"- -#
/ -/-  - /#
The expected number of claims per day, given that there were at most 2 claims per day is
Ð!ÑÐ
"
#
"- -#
Ñ  Ð"ÑÐ
#
"- -#
Ñ  Ð#ÑÐ
-#
#
#
"- -#
ќ
--#
#
"- -#
.
We are told that this is 1.2 .
#
Therefore -  -# œ Ð"Þ#ÑÐ"  -  -# Ñ , which becomes the quadratic equation
Þ%-#  Þ#-  "Þ# œ ! . Solving the equation results in - œ # or  "Þ& , but we ignore the
negative root. The probability of at most 2 claims arriving at the office on a particular day is
# #
T Ð\ Ÿ #Ñ œ /#  #/#  # /# œ Þ'('( .
© ACTEX 2015
Answer: D
SOA Exam P - Probability
PRACTICE EXAM 7
5.
479
α
C
C
)α
The distribution function will be J ÐCÑ œ ) 0 ÐBÑ .B œ ) Bα)
α" .B œ "  C α .
The median 7 occurs where J Ð7Ñ œ "# . If α and ) were known, we could find the median.
α)
The average loss for all losses is α"
œ ")! , but both ) and α are not known.
The conditional distribution of loss amount B given that \  #!! is
)
α#!!
0 ÐBl\  #!!Ñ œ T Ð\#!!Ñ œ Bα)
α"  #!!α œ Bα" .
0 ÐBÑ
α
α
α
α
This random variable has a mean of #!!
α" . We are given that this mean is 300,
α
so #!!
α" œ $!! , and therefore α œ $ .
α)
Then, from α"
œ ")! , we get $#) œ ")! , so that ) œ "#! .
)α
"#! $
The median 7 satisfies the relation "# œ J Ð7Ñ œ "  7
α œ "  Ð 7 Ñ , so that 7 œ "&"Þ# .
Answer: D
6.
\ œ amount of loss claim, uniformly distributed on Ð"ß #Ñ , so 0\ ÐBÑ œ " for "  B  # .
] œ amount of time spent verifying claim.
We are given that the conditional distribution of ] given \ œ B is uniform on Ð!ß "  BÑ ,
"
so 0 ÐClBÑ œ "B
for !  C  "  B .
We wish to find IÒ] Ó Þ The joint density of \ and ] is
"
0 ÐBß CÑ œ 0 ÐClBÑ † 0\ ÐBÑ œ "B
for !  C  "  B and "  B  # .
There are a couple of ways to find IÒ] Ó :
(i)
IÒ] Ó œ   C 0 ÐBß CÑ .C .B or IÒ] Ó œ   C 0 ÐBß CÑ .B .C , with careful setting of the
(ii)
integral limits, or
IÒ] Ó œ  C 0] ÐCÑ .C , where 0] ÐCÑ is the pdf of the marginal distribution of ] .
# "B
IÒ] Ó œ " ! C †
(iii) The double expectation rule, IÒ] Ó œ IÒ IÒ] l\Ó Ó .
If we apply the first approach for method (i), we get
IÒ] Ó œ " !
#
"B
# Ð"BÑ#
#
"
&
C † "B
.C .B œ " #Ð"BÑ .C œ " "B
# .B œ % .
If we apply the second approach for method (i), we must split the double integral into
# #
$ #
IÒ] Ó œ ! " C † " .B .C  # C" C † " .B .C
"B
"B
#
The first integral becomes ! C 68Ð $# Ñ .C œ # 68Ð $# Ñ .
$
$
The second integral becomes # C Ò68$  68 CÓ .C œ &# 68$  # C 68 C .C .
$
The integral # C 68 C .C is found by integration by parts.
© ACTEX 2015
SOA Exam P - Probability
480
PRACTICE EXAM 7
6. continued
Let  C 68 C .C œ E Þ
Let ? œ C and .@ œ 68 C .C , then @ œ C 68C  C (antiderivative of 68 C), and then
E œ  C 68 C .C œ CÐC 68C  CÑ   ÐC 68 C  CÑ .C œ C# 68 C  C#  E  # ,
C#
so that E œ  C 68 C .C œ "# C# 68 C  % .
C#
C# $
$
Then # C 68 C .C œ "# C# 68 C  %  œ *# 68$  %*  Ð #% 68 #  "Ñ œ #* 68$  # 68#  %& .
#
$
$
&
Finally, IÒ] Ó œ # 68Ð Ñ  68$  # C 68 C .C
œ # 68 $  # 68 # 
#
#
&
*
# 68$  Ð #
68$  # 68#  &% Ñ œ &% .
The first order of integration for method (i) was clearly the more efficient one.
(ii)
This method is equivalent to the second approach in method
(i)
because we find 0] ÐCÑ from the relationship 0] ÐCÑ œ  0 ÐBß CÑ .B . The two-dimensional
region of probability for the joint distribution is "  B  # and !  C  "  B . This is
illustrated in the graph below
y
y  1 z
3
2
1
2
1
x
"
For !  C  # , 0] ÐCÑ œ " 0 ÐBß CÑ .B œ " "B
.B œ 68Ð #$ Ñ,
#
#
"
and for # Ÿ B  $ , 0] ÐCÑ œ C" 0 ÐBß CÑ .B œ C" "B
.B œ 68 $  68 C .
#
#
#
$
Then IÒ] Ó œ ! C 68Ð $# Ñ .C  # C Ò68$  68 CÓ .C , which is the same as the second part
of method (i).
(iii) According to the double expectation rule, for any two random variables Y and [ , we have
IÒY Ó œ IÒ IÒY l[ Ó Ó . Therefore, IÒ] Ó œ IÒ IÒ] l\Ó Ó .
We are told that the conditional distribution of ] given \ œ B is uniform on the interval
Ð!ß "  BÑ , so IÒ] l\Ó œ "\
# .
"
"
"
" $
&
Then IÒ IÒ] l\Ó Ó œ IÒ "\
# Ó œ #  # IÒ\Ó œ #  # Ð # Ñ œ % , since \ is uniform on
Ð"ß #Ñ and \ has mean $# .
Answer: B
© ACTEX 2015
SOA Exam P - Probability
PRACTICE EXAM 7
7.
481
"
The joint distribution of \ and ] has pdf 0 ÐBß CÑ œ "& † "( œ $&
on the rectangle
!  B  & and !  C  ( . The insurer pays \  ]  # if the combined loss \  ] is  #.
The maximum payment of 8 is reached if \  ]  # ) , or equivalently, if \  ] "! .
Therefore, the insurer pays \  ]  # if #  \  ] Ÿ "! (the lighter shaded region in the
diagram below) , and the insurer pays ) if \  ]  "! (the darker shaded region in the diagram
below). The expected amount paid by the insurer is a combination of two integrals:
  ÐB  C  #Ñ † " .C .B , where the integral is taken over the region #  B  C Ÿ "!
$&
(the lightly shaded region), plus
  ) † " .C .B , where the integral is taken over the region \  ]  "! (the darker region).
$&
)
The second integral is $&
† Ð#Ñ œ "'
$& , since the area of the darkly shaded triangle is 2 (it is a 2 ‚ #
right triangle).
y
7
6
X  Y  10
5
4
3
2
X Y  2
1
1
2
3
4
x
5
The first integral can be broken into three integrals:
(
$ (
& "!B
"
"
"
!# #B
ÐB  C  #Ñ † $&
.C .B  # ! ÐB  C  #Ñ † $&
.C .B  $ !
ÐB  C  #Ñ † $&
.C .B
#
# ÐB&Ñ#
$ (Ð#B$Ñ
&
"
œ $&
† Ò ! # .B  #
.B  $ '!%BB
.B Ó
#
#
"
"(*
"#%
œ $&
† Ò "!*
$  #)  $ Ó œ $& .
"#%
"%!
The total expected insurance payment is "'
$&  $& œ $& œ % .
© ACTEX 2015
Answer: C
SOA Exam P - Probability