If there is one prayer that you should pray/sing every day and every hour, it is the
LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer.
A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka
For in GOD we live, and move, and have our being. - Acts 17:28
The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka
Solved Examples: Odds of Events
For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you
solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.
For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
Solve all questions.
Show all work.
Unless specified or implied otherwise, leave all applicable answers as simplified fractions or integers.
(1.) A modified roulette wheel has $44$ slots.
One slot is $0$, another is $00$, and the others are numbered $1$ through $42$ respectively.
In a roulette, $0$ and $00$ are neither odd nor even.
Timothy went to the casino and placed a bet that the outcome is an odd number.
(a.) What is Timothy's probability of winning?
(b.) What are Timothy's odds against winning?
(c.) The payoff odds for an odd number are $1:1$.
How much profit would Timothy make if he bets $\$16$ and wins?
(d.) If Timothy convinces the casino to change its payoff odds to be the same as the actual odds
against winning, how much profit would he make on the $\$16$ bet?
(2.) ACT The probability of Jamie being chosen to bat first in the lineup for his baseball team is
$\dfrac{1}{9}$.
What are the odds in favor of Jamie being chosen to bat first?
(Note: The odds in favor of an event are defined as the ratio of the probability that the event will
happen to the probability that the event will NOT happen.)
$
F.\:\: \dfrac{1}{8} \\[5ex]
G.\:\: \dfrac{1}{9} \\[5ex]
H.\:\: \dfrac{1}{10} \\[5ex]
J.\:\: \dfrac{8}{1} \\[5ex]
K.\:\: \dfrac{9}{1} \\[5ex]
$
Let $E$ be the event that Jamie is chosen to bat first in the lineup for his baseball team
(3.) Criminal cases are assigned to judges randomly.
The list below is a list of nine judges.
Joshua Judges
Ruth Kings
Samuel Daniel
Nehemiah Job
Esther Proverbs
Isaiah Psalms
Jeremiah Amos
Ezekiel Joel
Micah Nahum
(a.) Event $A$ is the event that the judge is a woman.
List the outcomes of event $A$
(b.) What is the probability that a case will be assigned to a female judge?
(c.) List the outcomes of the complement of event $A$
(4.) A $12-sided$ die is rolled one time.
Determine the probability of rolling a number greater than $12$
$
S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex]
n(S) = 12 \\[3ex]
E = \{ \} \\[3ex]
n(E) = 0 \\[3ex]
P(E) = \dfrac{n(E)}{n(S)} \\[5ex]
P(E) = \dfrac{0}{12} \\[5ex]
P(E) = 0 \\[3ex]
$
This is an impossible event.
It is impossible to obtain a number greater than $12$ when a $12-sided$ doe is rolled one time.
(5.) ACT Yulan will use a bag of $30$ solid-colored marbles for a game in which each player
randomly draws marbles from the bag.
The number of marbles of each color is shown in the table below.
Color
Number
Blue
Red
Black
White
Green
$10$
$8$
$6$
$4$
$2$
Yulan will randomly draw $2$ marbles from the bag, one after the other, without replacing the first marble.
What is the probability that Yulan will draw a black marble first and a green marble second?
(6.) ACT A basket contains $10$ solid-colored balls - $2$ blue, $3$ red, and $5$ green.
Each ball has a single number printed on it.
The blue balls are numbered $1$ and $2$ (each number is used once), the red balls are numbered $1 - 3$
(each number is used once), and the green balls are numbered $1 - 5$ (each number is used once).
A ball will be drawn at random from the basket.
What is the probability that the ball that is drawn will be red OR have a $3$ printed on it?
(8.) ACT A bag contains $16$ red marbles, $7$ yellow marbles, and $19$ green marbles.
How many additional red marbles must be added to the $42$ marbles already in the bag so that the
probability of randomly drawing a red marble is $\dfrac{3}{5}$?
Let the number of additional red marbles = $r$
$
\underline{Old} \\[3ex]
n(S) = 42 \\[3ex]
n(Red) = 16 \\[3ex]
\underline{New} \\[3ex]
n(Red) = 16 + r \\[3ex]
n(S) = 42 + r \\[3ex]
P(Red) = \dfrac{16 + r}{42 + r} \\[5ex]
P(Red) = \dfrac{3}{5} \\[5ex]
\therefore \dfrac{16 + r}{42 + r} = \dfrac{3}{5} \\[5ex]
Cross\:\: Multiply\:\: method \\[3ex]
5(16 + r) = 3(42 + r) \\[3ex]
80 + 5r = 126 + 3r \\[3ex]
5r - 3r = 126 - 80 \\[3ex]
2r = 46 \\[3ex]
r = \dfrac{46}{2} \\[5ex]
r = 23 \\[3ex]
$
$23$ red marbles need to be added so that the probability of drawing a red marble is $\dfrac{3}{5}$
(9.) ACT A professional baseball team will play $1$ game Saturday and $1$ game Sunday.
A sportswriter estimates the team has a $60\%$ chance of winning on Saturday but only a $35\%$ chance
of winning on Sunday.
Using the sportswriter's estimates, what is the probability that the team will lose both games?
(Note: Neither game can result in a tie.)
(10.) WASSCE A number is selected at random from each of the sets $\{2, 3, 4\}$ and $\{1, 3, 5\}.$
Find the probability that the sum of the two numbers is greater than $3$ and less than $7$.
It is much better to represent this information in a table.
Because the question asked for the probability of the "sum", we shall be adding the elements.
$(+)$
$2$
$3$
$4$
$1$
$3$
$\color{blue}{4}$
$\color{blue}{5}$
$3$
$\color{blue}{5}$
$\color{blue}{6}$
$7$
$5$
$7$
$8$
$9$
Numbers in blue are greater than $3$ and less than $7$
Let $x$ be those numbers
$
n(S) = 9 \\[3ex]
n(3 \lt x \lt 7) = 4 \\[3ex]
P(3 \lt x \lt 7) = \dfrac{n(3 \lt x \lt 7)}{n(S)} \\[5ex]
P(3 \lt x \lt 7) = \dfrac{4}{9}
$
(11.) ACT The Harrisburg Recreation Center recently changed its hours to open $1$ hour later and
close $3$ hours later than it had previously.
Residents of Harrisburg age $16$ or older were given a survey, and $560$ residents replied.
The survey asked each resident his or her student status (high school, college, or nonstudent) and what he
or she thought about the change in hours (approve, disapprove, or no opinion).
The results are summarized in the table below.
Student status
Approve
Disapprove
No opinion
High school
College
Nonstudent
$30$
$14$
$85$
$4$
$10$
$353$
$11$
$6$
$47$
Total
$129$
$367$
$64$
Suppose a person will be chosen at random from these $560$ residents.
Which of the following values is closest to the probability that the person chosen will NOT be a high
school student and will NOT have replied with no opinion?
The person chosen will NOT be a high school student and will NOT have replied with no opinion
means that the person is a college student who approved or disapproved OR the person is a nonstudent
who approved or disapproved. We shall add all of them.
Let $E$ be the event of selecting such a person.
This includes:
All college students who approved = $14$
All college students who disapproved = $10$
All nonstudents who approved = $85$
All nonstudents who disapproved = $353$
(12.) Samuel, Dominic, and Chukwuemeka work for SNHU.
The university wants to send two employees to a Statistics conference.
To be fair, the university decided that the two individuals who would attend will have their names
drawn from a hat. This is similar to obtaining a sample random sample of size, $2$.
(a.) Determine the sample space of the experiment. In other words, list all the possible simple
random samples of size, $n = 2$.
(b.) What is the probability that Dominic and Chukwuemeka attend the conference?
(c.) What is the probability that Chukwuemeka stays home?
$
Let\:\:Samuel = A \\[3ex]
Dominic = D \\[3ex]
Chukwuemeka = C \\[3ex]
(a.)\:\:S = \{AD, AC, DC\} \\[3ex]
n(S) = 3 \\[3ex]
(b.)\:\:n(DC) = 1 \\[3ex]
P(DC) = \dfrac{n(DC)}{n(S)} = \dfrac{1}{3} \\[5ex]
$
If Chukwuemeka stays home, this means that Samuel and Dominic will attend
(13.) ACT The $16-member$ math club needs to choose a student government representative.
They decide that the representative, who will be chosen at random, CANNOT be any of the $3$ officers
of the club.
What is the probability that Adrian, who is a member of the club but NOT an officer, will be chosen?
(14.) ACT A bag contains $10$ pieces of favored candy: $4$ lemon, $3$ strawberry, $2$ grape,
and $1$ cherry.
One piece of candy will be randomly picked from the bag.
What is the probability the candy picked is NOT grape flavored?
(16.) ACT The $13-member$ math club needs to choose a student government representative.
They decide that the representative, who will be chosen at random, CANNOT be any of the $3$ officers
of the club.
What is the probability that Samara, who is a member of the club but NOT an officer, will be chosen?
The table shows the outcome when a die is thrown a number of times.
If the probability of obtaining a $3$ is $0.225$:
(a.) how many times was the die thrown?
(b.) calculate the probability that a trial chosen at random gives a score of an even number or a
prime number.
The number of times the die was thrown is the total frequency
The total frequency is the cardinality of the sample space
$
n(S) = 25 + 30 + x + 28 + 40 + 32 \\[3ex]
n(S) = 155 + x \\[3ex]
P(3) = 0.225 \\[3ex]
P(3) = \dfrac{x}{155 + x} \\[5ex]
\rightarrow \dfrac{x}{155 + x} = 0.225 \\[5ex]
x = 0.225(155 + x) \\[3ex]
x = 34.875 + 0.225x \\[3ex]
x - 0.225x = 34.875 \\[3ex]
0.775x = 34.875 \\[3ex]
x = \dfrac{34.875}{0.775} \\[5ex]
x = 45 \\[3ex]
n(S) = 155 + x \\[3ex]
n(S) = 155 + 45 \\[3ex]
(a.)\:\: n(S) = 200 \\[3ex]
$
Let the event of getting an even number be $E$
Let the event of getting a prime number be $P$
(18.) ACT Mr. Chiang announced the grade distribution for this week's book reports.
Of the $24$ students in the class, $8$ received $A's$ for their book reports, $11$ received $B's$,
and $5$ received $C's$.
When a student is chosen at random to be the first one to read his or her book report to the class,
what is the probability that the student chosen had received an $A$ for the book report?
(19.) The table below shows the "Math Guys" and the "Art Guys" among high school students
($10^{th}, 11^{th}, \:\:and\:\: 12^{th} \:\:grade\:\:students$) in a certain school in the
City of Truth or Consequences, New Mexico
$10^{th}$ Grade
$11^{th}$ Grade
$12^{th}$ Grade
Total
Math Guys
$232$
$442$
$59$
$733$
Art Guys
$1036$
$68$
$53$
$1157$
Total
$1268$
$510$
$112$
$1890$
(a.) If a student is selected at random, what is the probability that the student is a Math Guy?
(b.) If a student is selected at random, what is the probability that the student is in $11^{th}$ grade?
(c.) If a student is selected at random, find the probability that the student is in $11^{th}$ grade or $12^{th}$ grade.
(d.) If a student is selected at random, determine the probability that the student is in $11^{th}$ grade and is a Math Guy.
(e.) If a student is selected at random, find the probability that the student is in $11^{th}$ grade or is a Math Guy.
(f.) If an $11^{th}$ grade student is selected at random, what is the probability that the student is a Math Guy?
(g.) If a $12^{th}$ grade student is selected at random, what is the probability that the student is a Math Guy?
(h.) If a $10^{th}$ grade student is selected at random, determine the probability that the student is a Math Guy
(i.) If two $11^{th}$ grade students are selected at random, determine the probability that both students are Math Guys
(20.) ACT Bella will pick $1$ jelly bean at random out of a bag containing $28$ jelly beans
that are in the colors and quantities shown in the table below.
Each of the jelly beans is $1$ color only.
Color
Quantity
Green
Black
Red
Orange
Yellow
Blue
$6$
$3$
$5$
$2$
$4$
$8$
What is the probability that Bella will pick a blue or yellow jelly bean?
Let the event of picking a blue jelly bean be $B$
and the event of picking a yellow jelly bean be $Y$
(21.) About $12\%$ of the population of the U.S state of Alabama smile at strangers.
If two Alabamians are randomly selected:
(a.) What is the probability that both smile at strangers?
(b.) What is the probability that both do not smile at strangers?
(c.) What is the probability that only one smiles at strangers?
(d.) What is the probability that at least one smiles at strangers?
(22.) $11$ of the $50$ laptops sold in a certain store in the town of Coward, South Carolina on
a certain Black Friday were defective.
Determine the probability of randomly selecting a non-defective laptop.
(23.) WAEC The probabilities that two athletes $P$ and $Q$ will win a gold medal in a competition are
$0.75$ and $0.60$ respectively.
What is the probability that in the competition,
(i) both $P$ and $Q$ will win gold medals,
(ii) neither of them will win a gold medal,
(iii) at least one of them will win a gold medal?
(24.) Let the sample space, $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
Assume all the outcomes are equally likely.
Compute the probability of the event, $E = \{3, 10\}$
(25.) A drug testing company located in the City of Surprise, Arizona tested several job
applicants for marijuana.
$146$ applicants tested positive, but among them were $26$ false positive results.
$160$ applicants tested negative, but among them were $3$ false negative results.
(a.) Represent this information in a table.
(b.) How many job applicants were tested?
(c.) How many applicants did not use marijuana?
(d.) How many applicants used marijuana?
(e.) Determine the probability that a randomly selected applicant did not use marijuana.
(f.) Determine the probability that a randomly selected applicant used marijuana.
(g.) Determine the probability that a randomly selected applicant tested negative OR did not use
marijuana.
Recall:
In testing for marijuana and for most applicable tests:
A False Positive means that the applicant did not use it but tested positive for it.
A False Negative means that the applicant used it but tested negative for it.
A True Positive means that the applicant used it and tested positive for it.
A True Negative means that the applicant did not use it and tested negative for it.
(26.) The hours of sleep that citizens get on a typical night in the City of Boring, Oregon is
shown in the table below.
Hours of Sleep
Number of Citizens (in millions)
$4\:\:or\:\:less$
$11$
$5$
$31$
$6$
$74$
$7$
$89$
$8$
$81$
$9$
$10$
$10\:\:or\:\:more$
$4$
A researcher in sleep deprivation finds that the average human needs at least six hours of sleep a
night to function properly.
Determine the probability of a citizen getting at least six hours of sleep a night.
Let the event of finding a citizen who gets at least six hours of sleep a night = $E$
(27.) ACT If a marble is randomly chosen from a bag that contains exactly $8$ red marbles,
$6$ blue marbles, and $6$ white marbles, what is the probability that the marble will NOT be
white?
$
1 - YES \\[3ex]
1.37 - NO \\[3ex]
0.26 - YES \\[3ex]
-0.52 - NO \\[3ex]
0.05 - YES \\[3ex]
0 - YES \\[3ex]
\dfrac{25}{12} = 2.08333 - NO \\[3ex]
120\% = \dfrac{120}{100} = 1.2 - NO \\[3ex]
$
The probabilities of events could be: $0, 0.05, 0.26, 1$
The probability of an impossible event is $0$
The probability of an event that must occur (a surely certainty event) is $1$
The probability of any event is from $0$ through $1$
(29.) ACT The stem-and-leaf plot below shows the number of rebounds a basketball player with the
Connecticut Suns grabbed in each of $17$ games.
$$
\begin{array}{c|c c}
Stem & Leaf \\
\hline
0 & 8 & 9 \\
1 & 1 & 1 & 3 & 3 & 4 & 4 & 5 & 5 & 5 & 6 & 9 \\
2 & 1 & 1 & 2 & 4
\end{array}
$$
(Note: For example, $13$ rebounds would have a stem value of $1$ and a leaf value of $3$)
(30.) A bag of $100$ marbles contains $35$ red marbles, $30$ yellow marbles, and $35$ purple marbles.
What is the probability that a randomly selected marble is purple?
$
Let\:\: red = R \\[3ex]
yellow = Y \\[3ex]
purple = P \\[3ex]
n(S) = 100 \\[3ex]
n(P) = 35 \\[3ex]
P(P) = \dfrac{n(P)}{n(S)} \\[5ex]
P(P) = \dfrac{35}{100} \\[5ex]
P(P) = \dfrac{7}{20}
$
(31.) ACT A bag contains several marbles.
On $3$ successive draws with replacement, a red marble is drawn from the bag each time.
Which of the following statements must be true about the marbles in the bag?
F. At least $1$ marble is red.
G. Exactly $1$ marble is red.
H. Exactly $3$ marbles are red.
J. All the marbles are red.
K. The bag contains more red marbles than marbles of other colors.
Let us analyze each of these options.
F. At least $1$ marble is red.
At least $1$ means "$1$ or more", $\ge 1$
This is a correct option because it is possible that there are one or more red marbles in the bag.
But, let us look at the other options before making a conclusion.
G. Exactly $1$ marble is red.
We are not really sure of this option.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement
in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than only $1$ is red.
At least $1$ means $1$ or more
That also includes $1$
H. Exactly $3$ marbles are red.
We are not sure of this option either.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement
in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than only $3$ are red.
At least $1$ means $1$ or more
That also includes $3$
J. All the marbles are red.
We are not sure of this option as well.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement
in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than $all$ are red.
At least $1$ means $1$ or more
That also includes $all\:\: the\:\: marbles$
K. The bag contains more red marbles that marbles of other colors.
We are not sure of this option.
What if the bag contained only red marbles?
What if the bag contained more green marbles than red marbles and one just picked only a red marble each time
because of the "with replacement" condition?
So, it is not okay to go with this option given the nature of the question.
The correct option is F. At least $1$ marble is red.
(32.) ACT A bag contains $64$ marbles, all solid colored.
Each marble is either red, yellow, or green.
A marble is randomly removed from the bag and then returned to the bag.
The probability that this marble is red is $\dfrac{5}{8}$
The probability that this marble is yellow is $\dfrac{1}{4}$
How many green marbles are in the bag?
A marble is randomly removed from the bag and then returned to the bag.
This is a case of "with replacement" - independent events
(33.) Does the table represent a probability model?
How would you describe the event of selecting a brown color?
Color
Probability
red
$0.25$
green
$0.2$
blue
$0.15$
brown
$0$
yellow
$0.35$
orange
$0.1$
$
\Sigma Probability = 1.05 \\[3ex]
\Sigma Probability \ne 1 \\[3ex]
$
Therefore, it does not represent a probability model.
The event of selecting a brown color is an impossible event.
(34.) A multiple-choice test has five possible answers.
However, only one answer is correct for each question.
Determine the probability of:
(a.) Answering a question correctly
(b.) Answering a question incorrectly
(35.) A loaded die is a die in which a certain outcome is more likely.
A fair die is a die in which all outcomes are equally likely.
A die is rolled $400$ times.
The outcome of the experiment is listed in the table below.
Do you think the die is loaded or not? Give reasons for your answer.
Value of Die
Frequency, $F$
$1$
$71$
$2$
$62$
$3$
$66$
$4$
$63$
$5$
$70$
$6$
$68$
$\Sigma F = 400$
The die is not loaded (it is a fair die) because each value has an approximately equal chance of occurrence.
The frequency of the values are very close to each other.
(36.) ACT If a bag contains $5$ blue marbles, $4$ red marbles, and $3$ green marbles, what
is the probability that a marble randomly picked from the bag will be red?
$
Let \\[3ex]
blue\:\:marble = B \\[3ex]
red\:\:marble = R \\[3ex]
green\:\:marble = G \\[3ex]
S = \{5B, 4R, 3G\} \\[3ex]
n(S) = 5 + 4 + 3 = 12 \\[3ex]
n(R) = 4 \\[3ex]
P(R) = \dfrac{n(R)}{n(S)} \\[5ex]
P(R) = \dfrac{4}{12} \\[5ex]
P(R) = \dfrac{1}{3}
$
(37.) A loaded die is a die in which a certain outcome is more likely.
A fair die is a die in which all outcomes are equally likely.
A die is rolled $400$ times.
The outcome of the experiment is listed in the table below.
Do you think the die is loaded or not? Give reasons for your answer.
Value of Die
Frequency, $F$
$1$
$43$
$2$
$100$
$3$
$44$
$4$
$114$
$5$
$50$
$6$
$49$
$\Sigma F = 400$
The die is loaded (it is a loaded die) because each value does not have an approximately equal chance of occurrence.
The frequency of the values of $2$ and $4$ are very high.
(38.) JAMB The letters of the word MATRICULATION are cut and put into a box.
One of the letters is drawn at random from the box.
Find the probability of drawing a vowel.
(40.) ACT A bag contains $4$ red jelly beans, $5$ green jelly beans, and $3$ white jelly
beans.
if a jelly bean is selected at random from the bag, what is the probability that the jelly bean
selected is green?
(41.) A group of airline passengers responded to a question about their favorite seat on a plane.
$501$ passengers chose the window seat, $10$ chose the middle seat, and $297$ chose the aisle seat.
(a.) Determine the probability that a passenger prefers the middle seat.
Round your answer to three decimal places as needed.
(b.) Is it unlikely for a passenger to prefer the middle seat? Why?
$
(a.) \\[3ex]
Let \\[3ex]
Window\:\:seat = W \\[3ex]
Middle\:\:seat = M \\[3ex]
Aisle\:\:seat = A \\[3ex]
n(W) = 501 \\[3ex]
n(M) = 10 \\[3ex]
n(A) = 297 \\[3ex]
n(S) = 501 + 10 + 297 = 808 \\[3ex]
P(M) = \dfrac{n(M)}{n(S)} = \dfrac{10}{808} = 0.0123762376 \\[5ex]
P(M) \approx 0.012 \\[3ex]
(b.) \\[3ex]
$
Rare Event Rule
Yes, it is unlikely for a passenger to prefer the middle seat because the probability that a
passenger prefers the middle seat, $0.012$ is less than $0.05$
This may be due to the fact the middle seat lacks easy access to the aisle and the outside view
among others.
(42.) JAMB Bola chooses at random a number between $1$ and $300$.
What is the probability that the number is divisible by $4?$
Divisibility by Numbers
A number is divisible by $4$ if it is an even number AND if the last two digits are divisible by $4$
Between $1$ and $300$, there are $75$ numbers that are divisible by $4$
Student: Samdom For Peace, how did you know?
Why not list them and count them?
Teacher: Good question.
This is a JAMB question.
You should solve this question accurately on time without a calculator
Between $1$ and $12$, list the numbers that are divisible by $4$
Student: The numbers are: $4, 8, 12$
Teacher: How many are there?
Student: Three numbers, Sir
Teacher: Correct!
Between $1$ and $20$, list the numbers that are divisible by $4$
Student: The numbers are: $4, 8, 12, 16, 20$
Teacher: How many are there?
Student: Five numbers, Sir
Teacher: Correct!
Student: Okay, I understand
$
12 \div 4 = 3 \\[3ex]
20 \div 4 = 5 \\[3ex]
300 \div 4 = 75 \\[3ex]
$
Let the set of numbers that are divisible by $4 = E$
(43.) JAMB What is the probability that a number chosen at random from the integers between
$1$ and $10$ inclusive is either a prime or a multiple of $3$?
$
OR \implies \cup \\[3ex]
AND \implies \cap \\[3ex]
S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex]
n(S) = 10 \\[3ex]
Let\:\:A\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:prime \\[3ex]
A = \{2, 3, 5, 7\} \\[3ex]
n(A) = 4 \\[3ex]
P(A) = \dfrac{n(A)}{n(S)} = \dfrac{4}{10} \\[5ex]
Let\:\:B\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:multiple\:\:of\:\:3 \\[3ex]
B = \{3, 6, 9\} \\[3ex]
n(B) = 3 \\[3ex]
P(B) = \dfrac{n(B)}{n(S)} = \dfrac{3}{10} \\[5ex]
A \cap B = \{3\} \\[3ex]
n(A \cap B) = 1 \\[3ex]
P(A \cap B) = \dfrac{n(A \cap B)}{n(S)} = \dfrac{1}{10} \\[5ex]
P(A \cup B) = P(A) + P(B) - P(A \cap B)...Addition\:\:Rule \\[3ex]
P(A \cup B) = \dfrac{4}{10} + \dfrac{3}{10} - \dfrac{1}{10} \\[5ex]
P(A \cup B) = \dfrac{4 + 3 - 1}{10} \\[5ex]
P(A \cup B) = \dfrac{6}{10} \\[5ex]
P(A \cup B) = \dfrac{3}{5} \\[5ex]
$
Between the numbers $1$ and $10$ inclusive, there is a $\dfrac{3}{5}$ probability of picking a
prime number or a multiple of $3$
(44.) ACT Martin has an empty bag and puts in $3$ red marbles.
He now wants to put in enough green marbles so the probability of drawing a red marble at random from
the bag is $\dfrac{1}{4}$
How many green marbles should he put in?
For ACT Students