Math 105: Topics in Mathematics 

Lesson 3 : Probability
Due Date:
See the Lecture Notes Site.
IntroductionThe concept of probability is prevalent at a very basic human intuitive and intellectual level. Other synonyms of probability include likelihood and chances. Probabilistic statements are made on daily basis without any awareness that there may be some intuitive mathematical calculation involved behind such statements. Most people are aware of different kinds of game of chances and gambling. Examples of such games include tossing coins, any dice rolling game and games of cards. It is universally accepted that when you toss a coin, likelihood of head showing up is fifty percent. When you roll a normal die, it is universally accepted that the likelihood of that a particular face will show up is one in six. There is also awareness of loaded coins and loaded dice, in which likelihood of an outcome (say head or the face six) is higher than that of other outcomes. It is a common sense that in a poker game it is extremely unlikely that one would get three aces in a particular deal. A lot of people would not buy a lottery ticket because they believe that it is extremely unlikely that he or she would ever win a million dollar or would not make money in the long run. These are intuitive or semimathematical understanding of probability of occurrence of certain events. Statements regarding chances of departure of a flights on time would not be so uncommon. Same is true regarding similar statements regarding chances of a thunderstorm, rain, snow or other whether related events. Statements regarding the chances of accidents would be another types probabilistic statements. One would not be surprised to hear a five year old making statements like "I will probably invite Aaron for my birthday". Such would mean that the child is aware of uncertainties of parental permission or his/her own indecisiveness. When such statement that includes word like "likelihood", "chances" or "probability" are made, one is essentially talking about what they have experienced in the past and trying to project that the same pattern will continue in future. Some of the early development of Probability as a mathematical theory originated in gambling. In the last century, this concept received further boost in genetics (more generally biosciences) and other branches of science.

Animation 3.1.1 
Similarly, our experience with rolling normal dice tells us that when we roll a die a particular face (say face six) would show up essentially once is six times. So, we hypothesize that probability that face six will show up is 1/6. Contrary to this, there are loaded dice. You may have a loaded die that you experimented with and determined face six shows up 40 percent of the times. So, you would hypothesize that probability that face six will show up for this loaded die is .40.
Similar data may be collected for road accidents and probabilistic hypothesis could be made regarding number of daily accidents on a street.
These examples explain the basic notion of probability. The probability of an event is hypothesized as the "relative frequency", the ratio of occurrences of the EVENT to the total number of times the EXPERIMENT is repeated (or experienced in the past).
This section provides basic definitions that will be needed for the rest of this lesson.
Definitions.
Examples. The following are examples of some experiments and their sample spaces.
S=  (1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)  
(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)  
(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)  
(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)  
(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)  
(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6) 
In brace notation, we can write the same as
S = {(i,j) : i = 1,2,3,4,5,6 and j = 1,2,3,4,5,6}.
5. Suppose that the experiment is rolling a die three times. Then the sam pleican be written as
S = {(i,j, k) : i = 1,2,3,4,5,6 and j = 1,2,3,4,5,6 and k = 1,2,3,4,5,6}.
6. Suppose the experiment is to determine the number of road accidents
in Lawrence on a particular day. So, the sample space is S = {0,1,2,3
... }.
7. Suppose the experiment is to determine the sex of an unborn chlid.
Then the sample space is S = {Female, Male}.
8. Suppose the experiment is to determine the blood group of a patient
in a lab. Then the sample space is S = {O,A,B,AB}.
9. Suppose the experiment is to observe the annual wheat production
in Kansas. Then the sample space is
S={x :
x is a nonnegative Number} = {x ∈ R
: x ≥ 0} =[0, ∞).
Definition. The sample space S is called a finite sample space if S has only a finite number of outcomes. If S has infinite elements, it is called an infinite sample space. Note that examples 1, 2, 3, 4, 6, 7 and 9 above have finite sample spaces, and 5 and 8 have infinite sample space.
Events
Definitions. Given an experiment and its sample space S, the following are some important definitions.
We say that E would have occurred, if the outcome of the experiment would be in E, when performed.
Remark. Often, we will describe events in "English," and we may have to identify them as a subset of the sample space and also conversely.
Examples. The following are some examples
of events.
The Theory of Probability
Given a sample space S, in the MATHEMATICS of probability we have hypotheses and rules for how to compute the probability of an event E. Although the MATHEMATICS of probability was modeled based on our past experiences, we do not derive anything from our intuitive ideas. We would be guided by the precise hypotheses, rules and laws that we set up.
For now we will be dealing with finite sample spaces.
Definition. Let
S = { e_{1}, e_{2}, ... ,e_{n} }.
be a finite sample space. The probability of a simple event {e} is a number (possibly given) denoted by P({e}) which has the following properties:
P(E)=  ∑  P({e}) 
_{e ∈ E} 
Remark. If we know the probabilities P({e})
of all the simple events {e}, we will be able to compute the probability
of any event E using 3. The probabilities of the simple events will
Probability with Equally Likely Outcomes
One of the most frequently used models to compute probabilities of
simple events is called EQUALLY LIKELY OUTCOMES.
Definition. Let S
= {e_{1}, ... , e_{N}} be a finite sample space. We
say that all the outcomes are equally likely
if all the outcomes have the same probability. So, in this case, we
have
P({e_{1}}) = P({e_{2}}) = … = P({e_{N}}) = 1/N.
Also, in this case, for an event E
P(E) =  ∑  P({e}) =  ∑  1/N 
_{e ∈ E}  _{e ∈ E} 
=(Number of Outcomes in E)/(Number of Outcomes in S)
If n(E) denotes the number of outcomes in E then
P(E) =  n(E)
n(S) 
. 
Problems on 3.2
Probability of Simple Events Given in a Table
Exercise 3.2.1. The following table gives the blood group distribution of a certain population.
Blood Group Distribution  
Blood Group  O  A  B  AB 

Percentage of Population 
47  42  8  3 
Find the probability that a random sample of blood will be of Blood
Group A or B or AB. (Here S={O, A, B, AB}
and we want to compute the probability P(E) of the event E={A, B, AB}.
Solution
Exercise 3.2.2. A student wants to pick a school based on its grade distribution. Following is the most recent grade distribution in a school:
Grade Distribution Unreal Data 

Grades  A  B  C  D  F 

Percentage of Students 
19  33  31  14  3 
Find the probability that a randomly picked student will have at least
a B average.
Solution
Exercise 3.2.3. The following table gives the probability distribution of a loaded die.
Probability Distribution for a Die  
Face  1  2  3  4  5  6 

Probability  0.20  0.15  0.15  0.10  0.05  0.35 
Find the probability that the face 2 or 3 or 6 will show up when you
roll the die.
Solution
Exercise 3.2.4. In a certain county, following is the distribution of population:
Population Distribution by enthnicity  
ethnicity  W  H  AA  A  O 

Percent  62  14  13  5  6 
Here W=White, H=Hispanic, AA= African American, A= Asian, O=Others. A jury is selected at random.
Solution:
Here, the sample space is
S = {W,H,AA,A,O}.
From the table, probability P(W)=.62, P(H)= .14, P(AA) = .13, P(A)=.05 and P(O)=.06.
Exercise 3.2.5. An arbitrary spot is selected in a swamp. The depth (in feet) of water in the swamp has the following probability distribution:
Depth Distribution  
depth  0+  1+  2+  3+  4+  5+  6+  7+  8+ 

Probability  .1  .2  .09  .17  .13  .11  .08  .07  .05 
Solution:
Here, the sample space is
S = {0+, 1+, 2+, 3+, 4+, 5+, 6+, 7+, 8+}.
From the table, probability P(0+)=.1, P(1+)= .2, P(2+) = .09 and so on.
Exercise 3.2.6. A Van pool can carry 7 people. Following is the distribution of number of riders in the van on a given day.:
Distribution of number of passengers  
number of passengers  1  2  3  4  5  6  7 

Probability  0  .12  .22  .23  .28  .08  .07 
Solution:
Here, the sample space is
S = {1, 2, 3, 4, 5, 6, 7}.
From the table, probability P(1)=0, p(2)=.12, P(3) = .22 and so on.
Exercise 3.2.7. Following is the distribution of hourly wages (in whole dollars) earned by workers in an industry:
Wage Distribution  
wages in dollars  7  8  9  10  11  12  13  14  15  16  17  18  19  20 

Probability  .04  .06  .07  .09  .11  .12  .14  .11  .09  .08  .04  .03  .01  .01 
Solution:
Here, the sample space is
S = {7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}.
From the table, probability P(7) = .04, p(8)=.06, P(9) = .07 and so on.
Exercise 3.2.8. In a school district, the distribution of number of students in a class is as follows:
distribution of number of students  
number of students  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23 

Probability  .03  .04  .06  .07  .10  .12  .13  .11  .09  .07  .06  .04  .03  .02  .02  .01 
Solution:
Here, the sample space is
S = {8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}.
From the table, probability P(8) = .03, p(9)=.04, P(10) = .06 and so on.
Exercise 3.2.9. An urn contains 7 apples and 3 oranges and 5 pears. One piece of fruit is picked at random. Find the probability that
Solution:
Here, the sample space is
S = {A1, A2, A3, A4, A5, A6, A7, O1, O2, O3, P1, P2, P3, P4, P5}.
So, n(S) = 7 + 3 + 5 =15
Exercise 3.2.10. A die is rolled twice. Find the probability that
Solution:
Here, the sample space is
S=  (1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)  
(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)  
(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)  
(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)  
(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)  
(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6) 
G=  
(2,1)  
(3,1)  (3,2)  
(4,1)  (4,2)  (4,3)  
(5,1)  (5,2)  (5,3)  (5,4)  
(6,1)  (6,2)  (6,3)  (6,4)  (6,5) 
Exercise 3.2.11. A letter is chosen at random from the letters of the English alphabet. Find the probability that
Solution:
Here, the sample space is
S = {A, B, C, D, E, …, X, Y, Z}.
So, n(S) = 26.
In the above sections, we defined probability for finite sample spaces. Some of the laws of probability will be discussed in this section. In fact, probability laws are similar to that of laws of area, volume or weight computations. Before that newer events will be constructed by combining other events.
Notations. We always work in the context of a particular statistical experiment and its sample space. Let S denote the sample space and E, F be two events.
(E or F) = {x ∈ S : x ∈ E or x ∈ F}.
(E and F) = {x ∈ S : both x ∈ E and x ∈ F}.
(not E) = {x ∈ S : x ∉ E}.
Read the notation ∉ as "is NOT in".Laws of Probability
Following are some of the laws of probability.
First, probability behaves like area and the laws of probability are like that of area.
Some formulas:
Let S be
sample space and let E and F be two events.
P( E or F) = P(E)+P(F)  P(E and F)
We subtract P(E and F) because we counted it twice in the sum:
once in P(E) and once in P(F).
if E and F are mutually exclusive then
P(E or F) = P(E) + P(F)
P(not E) = 1  P(E).
P(E) = a/(a+b)
Remark: This concept of ODDS is used often in gambling and horse races. When the odds in favor of a horse are 2 to 3, essentially this means that the probability the horse will win is 2/5. We say "essentially" because in actual betting, the probability is actually slightly less than 2/3, so that in the long run the gambling establishment makes more money than it gives. (This instructor is not particularly experienced in such betting or horse races.)
Problems on 3.3: Laws of Probability
Exercise 3.3.1. Let E, F, G be three events. It is given
P(E)=0.3
P(F)=0.7 P(G)=0.6
P( E and F) = 0.2 P( E or G) = 0.7
Solution:
Exercise 3.3.2. The probability that a Christmas
tree is taller than 6 feet is .30; the probability that a Christmas
tree weighs more than sixty pounds is 0.25; and the probability that
a Christmas tree is either taller than 6 feet or more than sixty pounds
is .4.
Solution:
Let E be the event that the selected tree is taller than 6 feet.
Let F be the event that the selected tree is heavier than sixty pounds.
We are given P(E) = 0.30, P(F) = 0.25, and P(E or F) = 0.4
Exercise 3.3.3. The probability that a student majors in liberal arts is .44; the probability that a student majors in business is .33; and the probability that a student majors in either liberal arts or business is .65. Find the probabilities
Solution:
Let E be the event that the selected student majors in liberal arts .
Let F be the event that the selected student majors in business.
We are given P(E) = 0.44, P(F) = 0.33, and P(E or F) = 0.65
Exercise 3.3.4.
In a restaurant menu, entrees are served with rice product or potato product or others. Probability that an entree is served with rice product is .35 , probability that an entree is served with potato product is .40, probability that an entree is served with both is .15. What is the probability that an entree is served with either rice product or potato product?
Solution:
Let E be the event that an entree is served with rice product and F
be the event that an entree is served with potato product. We are given
P(E) = .35, P(F) = .40 and P(E and F) = .15.
We need to compute P(E or F).
P(E or F) = P(E) + P(F)  P(E and F) = .35 + .40  .15 = .60
Exercise 3.3.5.
Usual infections can be bacterial or viral. Probability that a person will get a bacterial infection (next winter) is .35, probability that a person will get a viral infection is .65, probability that a person will get either a bacterial or a viral infection is .85.
Solution:
Let E be the event that a person will get a bacterial infection and F
be the event a person will get a viral infection. We are given
P(E) = .35, P(F) = .65 and P(E or F) = .85.
Exercise 3.3.6.
You go for an examination of upper stomach (EGD) and lower stomach (colonoscopy). Probability that some problem in upper stomach would be found is .15, probability that some problem in lower stomach would be found is .20 and probability that some problem both in lower and upper stomach would be found is .07.
Solution:
Let E be the event that some problem would be found in the upper stomach and F
be the event that some problem would be found in the lower stomach. We are given
P(E) = .15, P(F) = .20 and P(E and F) = .07.
Exercise 3.3.7.
Probability that a person owns a domestic car is .55, probability that a person owns an import is .55 and the probability that a person owns both is .20.
Solution:
Let E be the event that a person owns a domestic and F
be the event that a person owns an import. We are given
P(E) = .55, P(F) = .55 and P(E and F) = .20.
Exercise 3.3.8.
In a county, 38 percent of the community is a minority. What
is the probability that a randomly selected jury will not be a minority?
Solution:
Let E be the event that the jury will be a minority. Then P(E) =
.38
Therefore, the answer is P(not E)=1  P(E)=1  .38=.62
Exercise 3.3.9.
In a school district, probability that a student will be in a class of less than 10 students is .27.
The probability that a student will be in a class of less than 20 students is .38.
What is the probability that a randomly selected student will be a class 10 or more?
Solution:
Let E be the event that the student will be in a class of less that 10 students. Then, P(E) = .27.
So, the answer is
P(not E)=1  P(E)=1 .27=.73.
Exercise 3.3.10.
In a swamp, probability that the depth at a random spot is higher than 4 feet is .17. What is the probability that at a random spot, the depth is four feet or less?
Solution:
Let E be the event that at a random spot the depth is higher than 4 feet. So, P(E) = .17. So, the answer is
P(not E)=1  P(E)=1  .17=.83.
Exercise 3.3.11.
It is known that 43 percent of the work force in a town earns more than $37,000 annually. What is the probability that a randomly selected working person would make at most $37,000 annually?
Solution:
Let E be the event that a randomly selected working person would make more than $37,000 annually. Therefore, P(E) = .43.
So, the answer is
P(not E)=1  P(E)= 1  .43=.57.
Exercise 3.3.12.
It is known that you can get an empty seat in the bus 64 percent of the rides. What is the probability that on a particular ride would not get a seat?
Solution:
Let E be the event that you get an empty seat in the bus. So, P(E) = .64. Therefore, the answer is
P(not E)=1  P(E)=1  .64=.36
Counting techniques are important and useful. The following are some interesting examples:
Notations. Let n be a positive integer. Then the n! (read as nfactorial) is defined as
n!= 1 . 2 . … (n2) . (n1). n
0!=1.
nfactorial is the product of all integers from 1 up to n.
One of the main tools for counting is the following principle:
The Basic Counting Principle. Suppose we have an experiment that is a combination of r subexperiments, performed one after the other, such that
Then our original experiment will have n_{1}n_{2} ... n_{r} outcomes.
Remark. Here we have used the word "experiment"
in a slightly different sense than the statistical experiments. The
basic counting principle will be used to count the number of outcomes
in sample spaces and events.
Animation 3.1.2 Tree Diagrams as an Example 
Tree Diagrams and Counting 
Examples.
3.4.1. Count the number of words of length five that can be constructed from the English alphabet. Answer: 26x 26x26x26x26
We use the counting principle by dividing this job of constructing a word of length five into five subjobs:
Stage  Job to do  Number of Ways 
1.  Pick the first letter  26 
2.  Pick the 2nd letter  26 
3.  Pick the 3rd letter  26 
4.  Pick the 4th letter  26 
5.  Pick the 5th letter  26 
Answer = Product =  11,881,376 words  
One could make a tree diagram as above, with 26 branches coming out at each stage. 
3.4.2. Count the number of ways you can assign the 11 seats in the first
row in a concert hall to 231 guests.
The job of assigning 11 seats can be divided into
11 jobs of assigning each 11 seats.
Stage  Job to do  Number of Ways 
1.  Assign seat 1  231 
2.  Assign seat 2  230 
3.  Assign seat 3  229 
4.  Assign seat 4  228 
5.  Assign seat 5  227 
6.  Assign seat 6  226 
7.  Assign seat 7  225 
8.  Assign seat 8  224 
9.  Assign seat 9  223 
10.  Assign seat 10  222 
11.  Assign seat 11  221 
Answer = Product =  221*222*...*230*231  
One could make a tree diagram as above. But number of branches that come will keep dropping by one at each stage. Also list of who come out will not be the same; it will exclude the predecessors. 
3.4.3. Contrast: How many ways can you form
a committee of 11 members from a group of 231 people? Unlike assigning
seats, here the order of selection of the members will be ignored. The
11 members, when permuted around, will have different seat assignments
but in the same committee. Forming the committee is a "combination"
problem that comes below.
Remark. The difference between assigning 11
seats in a row and forming a committee of 11 is that in the first case
the order of assignment is important. Assigning
the first row to the same 11 guests in two different ways will count
as two different outcomes. When we form a committee, the order in which
we pick 11 members does not make any difference.
Definition. Suppose we have n objects. We
pick r of them one by one (without ever puttting them back) and arrange
them in a row. Such an ordered arrangement will be called a permutation
of n objects taken r at a time. The number of permutations
of n objects taken r at a time is denoted by _{n}P_{r}.
It follows from the basic counting principle that
_{n}P_{r}
= n (n1) (n2) ... (nr+1) = n!/(n r)!
Number of permutations _{n}P_{r}
= product of r integers starting from n
downward.
Again, permutations can be generated by a tree diagram.
Definition.
In contrast, we can pick r objects from a collection of n objects one
by one but place the object back in the collection before the next pick,
and arrange all of them in a row. Such selection and arrangement is
called selection with replacment. Constructing
a formal word of length 5 is an experiment of picking with replacement.
Definition. Suppose we have n objects in a
container. We pick r of them all at a time. In this case the order of
selection does not come into consideration. Such a selection is called
a combination of n
objects taken r at a time. The number of
combinations of n objects taken r at a time is denoted by _{n}C_{r}
and is given by
_{n}C_{r} =  n! 
(r! (nr)!) 
Tree diagrams will not work to generate combinations. In this case, select all of them in batch.
Examples. 1. Count the number of ways you
can form a committee of 11 from a group of 231 people. Answer:
_{231}C_{11}
2. Count the number of ways you can deal a hand of 13 cards from a deck
of 52 cards. Answer: _{52}C_{13}.
Remarks:
Animation 3.1.3 
Remark: In this course you are expected to work with
only permutations and combinations.

Use of Calculators (TI84): 

Computing nfactorial, _{n}P_{r}, _{n}C_{r}

Problems on 3.4: Counting Techniques and Probability
Exercise 3.4.1. Find 5!
Solution
Exercise 3.4.2. Suppose in the World Cup soccer tournament, group A has 8 teams. Each team of group A has to play all the other teams in the group. How many games will be played among the group A teams. Answer: 8^{C}_{2}
Exercise 3.4.3. How many ways can you deal
a hand of 13 cards from a deck of 52 cards?
Answer:_{52}C_{13}
Exercise 3.4.4. We have 13 students in a
class. How many ways can we assign the 4 seats in the first row?
Answer: _{13}P_{4}
Solution
Exercise 3.4.5. A committee of 9 is selected at random from a group of 11 students, 17 mothers and 13 fathers.
Exercise 3.4.6. Three scholarships of unequal
value will be awarded from a group of 15 male students and 10 female
students.
What is the probability that all three scholarships will be given to female students.
Answer= (_{10}P_{3})/(_{25}P_{3})
We used permuations because scholarships are of unequal values. So, the order od selection counts.
Alternately, if the scholarships were of equal value,
then the order of selection would not count.
In that case,
Answer= (_{10}C_{3})/(_{25}C_{3})
Sometimes it is understandable that two events E and F do not influence the occurrence of each other. For example, if you roll a die twice and E is the event that the first roll will show an oddnumber face and F is the event that the second roll will show 1 or 2, then it is reasonable to assume that the occurrence of E will not influence the occurrence of F. (Describe E and F in brace notation.)
P(E and F) = P(E)P(F). 
If two events are not independent, then they are said to be dependent.
Remark. Let us also describe what we mean by independence of 3 or more events. For events E_{1},E_{2}, … , E_{n}, we say they are independent if the "multiplication rule" applies for any number of them. For example E,F,G,H are independent if all of the following holds:
2 events
P(E and F) = P(E)P(F),
P(E and G) = P(E)P(G),
P(E and H) = P(E)P(H), P(F and G) = P(F)P(G),
P(F and H) = P(F)P(H), P(G and H) = P(G)P(H)
3 events
P(E and F and G)
= P(E)P(F)P(G), P(E and F and H) = P(E)P(F)P(H),
P(E and G and H) = P(E)P(G)P(H), P(F and G and H) = P(F)P(G)P(H)
4 events
P(E and F and G and H) = P(E)P(F)P(G)P(H)
Problems on 3.5: On Independent Events
Exercise 3.5.1. Consider the following two
circuit diagram:
As you can see, current flows through two switches A and B to the radio
and back to the battery. It is given that the probability that the switch
A is closed is 0.91 and the probability that the switch B is closed
is 0.83. Assume that the two switches function independently. Find the
probability that the radio is playing.
Solution:
Let E = event that switch A is on (closed) and
F = event that switch B is on (closed).
Then, the even that the radio plays is (E and F).
We have P(E)= .91 and P(F) = .83
Assuming independence,
P(E and F) = P(E) P(F) = .91*.83.
Exercise 3.5.2. An airplane has two engines. The probability that engine 1 fails is 0.023 and the probability that engine 2 fails is 0.06. Assume that the engines function independently. What is the probability that both engines fail?
Solution:
Let E = event that the engine 1 fails and
F = event that the engine 2 fails.
Then, the even that bothe fails is (E and F).
We have P(E)= .023 and P(F) = .06
Assuming independence,
P(E and F) = P(E) P(F) = .023*.06 = .00138
Exercise 3.5.3. The probability that you
will receive a wrong number call this week is 0.3; the probability that
you will receive a sales call this week is 0.8; and that the probability
that you will receive a survey call this week is 0.5. What is the probability
that you will receive one of each this week? (Assume that all these
calls are independent.)
Solution:
Let E = event that you will receive a wrong number call this week,
F = event that you will receive a sells number call this week,
G = event that you will receive a survey number call this week.
Then, (E and F and G) = the event that you will one of each this week.
We have P(E) = .3, P(F) = 0.8 and P(G) = 0.5.
So, P(E and F and G) = P(E) P(F) P(G) = .3*.8*.5 = .12.
Exercise 3.5.4. Suppose you went for a job interview in Lawrence and another one in Kansas City. Probability of that you will get the job in Lawrence is .25 and the probability of that you will get the job in Kansas City is .33. It is reasonable to assume independence.
Solution:
Let E = event that you will get the job in Lawrence and
F = event that you will get the job in Kansas City.
Exercise 3.5.5. You are taking the Topic course in KU and your brother is taking the same course in MU. The probability that you will get an A is .18 and the probability that your brother will get an A is .21.
Solution:
Let E be the event that you will get an A and
F be the event that your brother will get an A.
Exercise 3.5.6. Probability that you will receive a call from a sibling this week is .35 and the that you will receive a call from a parent this week is .43.(Assume independence.)
Solution:
Let E be the event that you will receive a call a sibling this week and
F be the event that you will receive a call a parent.
Exercise 3.5.7. Probability that it will rain in Lawrence today is .22 and probability that it will rain today at your home town is .40. (Assume independence.)
Solution:
Let E be the event that it will rain in Lawrence today and F be the event that it will rain at your home town today.
Exercise 3.5.8. According to the poll, probability that a person would vote for Candidate A is .43.
Solution:
Let E be the event that you will vote for Candidate A and let F be the event that I will vote for Candidate A.