Basic Terminology, Sample Spaces, and Events
Definition:
Random Experiment
Example
Examples of random experiments:
Definition:
Sample Space
Definition:
Event
Example
Roll a die once $\qquad \Rightarrow \quad S=\lbrace 1, 2, 3, 4, 5, 6 \rbrace$
Solution
Here are some possible events.
A Review of Set Operations
De Morgan's Rule
- 1. $(E_1\cup E_2\cup \cdots \cup E_n)^{\prime} \quad \Rightarrow\quad $ None of the events, $E_1, E_2, \dots, E_n$ occur, $$ (E_1\cup E_2\cup \cdots \cup E_n)^{\prime}=E_1^{\prime}\cap E_2^{\prime}\cap \cdots \cap E_n^{\prime} $$
- 2. $(E_1\cap E_2\cap \cdots \cap E_n)^{\prime} \quad \Rightarrow\quad $ $\geq 1$ of the events, $E_1, E_2, \dots, E_n$ does not occur, $$(E_1\cap E_2\cap \cdots \cap E_n)^{\prime}=E_1^{\prime}\cup E_2^{\prime}\cup \cdots \cup E_n^{\prime}$$
Mutually Exclusive Events
Definition:
Mutually Exclusive Events
Example 1
Four law students have to decide if they should return to the scene of a crime in order to retrieve the body of man they had just murdered. They agree to toss a coin, and if it lands on heads, then they would go back.
Calculating Probability
Law of Large Numbers
Kolmogorov's Axioms of Probability
- The probability of an event is always between 0 and 1, inclusive. $$0 \leq P(A) \leq 1$$
- The probability of the sample space is 1. $$P(S) = 1$$
- The probability of the union of two mutually exclusive events is the sum of their individual probabilities. $$P(A \cup B) = P(A) + P(B)$$
Counting Techniques
Definition:
Basic Principle of Counting
Example 1
Former Phillipines First Lady, Imelda Marcos, is known for her extravagance and love of shoes. She once spent $\$2000$ on chewing gum and forced a plane to return to Rome because she forgot to buy some cheese. When her husband fell from power, amid charges of corruption, the public were finally given a glimpse into the lavish lifestyle led by the Marcos'. In her closet were $15$ mink coats, $1000$ handbags, $580$ gowns, and $1060$ pairs of shoes.
How many different outfits does Imelda have? Assume that an outfit consists of a gown, one handbag, one pair of shoes, and a mink coat.
The total number of outfits that Imelda Marcos can put together is $15\times 1000\times 580\times 1060 = 9,222,000,000$
Solution
Example 2
FIJI Water once ran an ad campaign stating “The label says Fiji because its not bottled in Cleveland”. The city of Cleveland took offence and responded by testing both FIJI water and their own tap water. They found $6.3$ micrograms of arsenic per litre of FIJI water. The Cleveland tap water had none.
New designs for a water treatment tank have proposed three possible shapes, four possible sizes, three locations for input valves, and four locations for output valves. How many different product designs are possible?
Total number of configurations $= 3 \times 4 \times 3 \times 4 = 144$
Solution
Example 3
Steve Wozniak, the co-founder Apple, was the first the person to own the phone number 888-888-8888. But it proved impossible to use, because he was receiving well over 100 wrong numbers a day; mostly from babies playing with the phone.
Example
- A department needs to select three people out of 20 to serve on a committee. In how many ways can this be done?
- A student needs to choose two questions out of four in a final exam. How many selections can the student make?
Definition:
Combinations
Formula:
Combinations
Example 4
A department needs to select three people out of 20 to serve on a committee. In how many ways can this be done?
The number of ways of selecting $3$ people out of $20$ is given by: $$\binom{20}{3} = \frac{20!}{3!(20-3)!} = 1140$$
Solution
Example 5
A student needs to choose two questions out of four in a final exam. How many selections can the student make?
The number of ways of selecting $2$ questions out of $4$ is given by: $$\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6$$
Solution
Example 6
A student council consists of 12 members. A subcommittee of 4 members is to be chosen to organize a school event.
Example 7
A lottery requires players to select 6 numbers from a set of 49 numbers.
Example 8
A standard deck of 52 playing cards is shuffled, and players are dealt hands of 5 cards.
Definition:
Permutation
Formula:
Permutation of $n$ Distinct Objects
Formula:
Permutation of $n$ Distinct Objects Taken $r$ at a Time
Example 9
At the race track, an exacta wager is a bet where the bettor picks the horse that will finish in first and second place. A trifecta is a bet where the bettor picks the horses that finish in first, second, and third place. If 12 horses enter a race,
Example 10
A credit card company requires that user set up a four-digit PIN number for security purposes.
Definition:
Permutation of Similar Objects
Formula:
Permutation of Similar Objects
Example 11
How many different ways can the letters in the word 'MISSISSIPPI' be arranged?
The number of ways that the letters in the word 'MISSISSIPPI' can be arranged is: $$\frac{11!}{4!4!2!1!}=34,650$$
Solution
Example 12
If all the letters of the word MINIMUM are to be used, how many different arrangements are there?
The number of ways that the letters in the word 'MINIMUM' can be arranged is: $$\frac{7!}{3!2!}=420$$
Solution
Example 13
If all arrangements are equally likely, what is the probability that a randomly selected arrangement of the word, MINIMUM begins with MMM?
$E=$ arrangement begins with MMM. $n(E)=\frac{3!}{3!}\cdot \frac{4!}{2!}=12$ $$P(E)=\frac{n(E)}{n(S)}=\frac{12}{420}=\frac{1}{35}$
Solution
Example 14
Five red and four blue marbles are arranged in a row. What is the probability that both end marbles are blue?
$B=$ both end marbles are blue. $n(S)=\frac{9!}{5!4!}=126$
$n(B)=\frac{2!}{2!}\cdot\frac{7!}{5!2!}=21$ $$P(B)=\frac{n(B)}{n(S)}=\frac{21}{126} $$
Solution
Example 15
The Price is Right is a game show which offers contestants an opportunity to win prizes such as cars, trips, home appliances, and cash (mundane by today standards). Back in 1956, a contestant on the show won a grand piano, and as a joke the producers thought it would be funny to offer him a live elephant as the bonus prize; the real prize was $4000$ in cash. The joke backfired and the show was forced to ship in an elephant from Kenya.
Consider the word ELEPHANT.
Addition Rule
Formula:
Addition Rule
Other Useful Results
Example 1
Suppose that $P(A) = 0.4$, $P(B) = 0.3$, and $P(A\cap B) = 0.1$. Find $P(A\cup B)$.
$P(A\cup B) = P(A) + P(B) - P(A\cap B) = 0.4 + 0.3 - 0.1 = 0.6$
Solution
Example 2
In a survey of 120 students, 70 students liked coffee, 50 students liked tea, and 90 students liked both coffee and tea. Find the probability that a student likes coffee and tea.
$P(C\cup T) = P(C) + P(T) - P(C\cap T) = \frac{70}{120} + \frac{50}{120} - \frac{90}{120} = \frac{30}{120} =\frac{1}{4} $
Solution
Example 3
In a survey of 120 students, 70 students liked coffee, 50 students liked tea, and 90 students liked both coffee and tea. What is the probability that a randomly selected student likes only coffee?
Let $CO$ = student likes coffee only. Then $P(CO) = P(C) - P(C\cap T) = \frac{70}{120}-\frac{30}{120} = \frac{40}{120} = \frac{1}{3}$
Solution
Example 4
In a group of 200 science students, 80 of them are taking an optional math course, and 50 of them are taking an optional physics course. Due to scheduling contraints, students cannot take both optional math and physics courses. What is the probability that a randomly selected student is taking either an optional math or physics course?
Let $M$ = student is taking optional math course, $P$ = student is taking optional physics course. Then $P(M\cup P) = P(M) + P(P) = \frac{80}{200} + \frac{50}{200} = \frac{130}{200} = \frac{13}{20}$
Solution
Example 5
In a group of 200 science students, 80 of them are taking an optional math course, and 50 of them are taking an optional physics course. Due to scheduling contraints, students cannot take both optional math and physics courses. What is the probability that a randomly selected student is taking neither of the an optional math or physics courses?
Let $M$ = student is taking optional math course, $P$ = student is taking optional physics course. Then $P(M^{\prime}\cap P^{\prime}) = 1 - P(M\cup P) = 1 - \frac{13}{20} = \frac{7}{20}$
Solution
Example 6
The color most often associated with the Dutch is orange. During a diplomatic dispute between Turkey and the Netherlands, protesters in Ankara symbolically stabbed oranges in the street. They also mistakenly burned French flags thinking that were Dutch ones (because both are red, white, and blue). In a survey of the protesters, $15\%$ said that they neither stabbed an orange, nor burned a flag, $73\%$ admitted to stabbing an orange, and $49\%$ said that they burned a flag. What is the probability that a randomly selected protester who answered the survey,\
Example 7
Kanye North and Kanye South (but not Kanye West) are electoral districts in Bostwana. The number of registered voters in 2002 and their gender are shown in the table below. $$ \begin{array}{|c|c|c|} \hline & \text{Female} & \text{Male} \\ \hline \text{Kanye North} & 8,705 & 9795 \\ \text{Kanye South} & 8,430 & 9,570 \\ \hline \end{array} $$ If a voter is selected at random, find the probability that the
Example 8
In 2011, scientists in the US used $\$660,000$ in federal research money to examine whether distant prayer could heal AIDS, $\$406,000$ to see if injecting brewed coffee into someone's intestines would help with pancreatic cancer, and $\$1.25$ million to examine whether massages made people with advanced cancer feel better. The distant prayers and coffee enemas did not work, but the massages did.
Oncologists at a local hospital are studying the effectiveness of two medical treatments against massages, in the fight against pancreatic cancer. The table below shows the number of patients in each group, and the number of individuals who showed a complete (positive) response after 24 weeks of treatment. $$\begin{array}{|l|c|c|} \hline & \textbf{Complete Response} & \textbf{Total}\\ \hline \textbf{Proton beam} & 16 & 21\\ \textbf{Nanoliposomal irinotecan} & 6 & 19\\ \textbf{Massages} & 0 & 20\\ \hline \end{array}$$ Let $PR$ denote the event that the patient was treated with the proton beam, and $CR$ denote the event that the patient showed a complete response.
Example 9
According to SplashData, 123456, was the most frequently used password for 2021, followed by 123456789 in second place. The company estimates that about $10\%$ of all accounts can be logged in with one of passwords in their top 25 - making it easy for hackers to access and steal private information. Other not-so-secret passwords which make the list regularly are: 111111, letmein, and trustno1.
A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercaseletters (a-z) or 26 uppercase letters (A-Z) or 10 integers (0-9). ,br/.
Assume all passwords are equally likely. Let $OL$ and $OI$ denote the events that consist of passwords with only letters or only integers, respectively. Determine the following probabilities:
Conditional Probability
Roll a fair die once.
Conditional Probability
A conditional probability measures the probability of an event given that (by assumption, evidence, or partial information) another event has occurred.
When you condition on an event, you are entering the universe where that event has taken place. Mathematically, if you condition on $B$, then $B$ becomes your new sample space.
Definition:
Conditional Probability
Remark
Example 1
In the US, there is a legal tactic known as the ``Chewbacca Defense`` whereby the goal is to deliberately confuse the jury rather than to factually refute the case of the other side. The term was coined in an episode of ``South Park`` that satirized the closing argument of the O.J Simpson trial and is now widely used.
As the O.J. trial was coming to a close, three hundred people were asked if they thought that Simpson was guilty of the crime that he was being tried for. Below are the responses broken down into gender: $$ \begin{array}{lccc} & \textbf{Guilty} & \textbf{Not Guilty} & \textbf{Undecided}\\ \hline \textbf{Female} & $90$ & $15$ & $10$\\ \textbf{Male} & $45$ & $110$ & $30$ \\ \end{array}$$
If a person is chosen at random from the group of people surveyed, what is the probability
Example 2
World of Warcraft is a massive multi-player online role-playing game. At its peak, it had 12 million subscribers - most of them young adults who spent countless hours playing the game instead of studying. One father in China was so vexed with his son's obsession with the game that he hired virtual assassins to kill off his son's avatar. The table below shows the age distribution of a random sample of War Craft players from China.
$$ \begin{array}{lccc} \textbf{Age} (years) & \textbf{Male} & \textbf{Female} \\ \hline $ 11 - 17 $ & $96 $ & $88$ \\ $ 18 - 24 $ & $204$ & $198$ \\ $ 25 - 31 $ & $206$ & $117$ \\ $ 32 - 38 $ & $54 $ & $12$ \\ \hline \end{array} $$
What is the probability that a randomly selected person from this group
Example 3
A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters (a-z) or 26 uppercase letters (A-Z) or 10 integers (0-9). Let $S$ denote the set of all possible passwords.
Suppose that all passwords in $S$ are equally likely. Determine the probability for each of the following:
Example 4
Two dice are tossed.
Example 5
$76\%$ of AC flights depart on time (within $5$ minutes of schedule). $68\%$ of AC flights arrive on time. $89\%$ of the flights that depart on time arrive on time.
The Multiplication Rule, Rule of Total Probability, and Bayes' Theorem
Multiplication Rule
Theorem:
The Multiplication Rule
In general if $E_1, E_2, E_3, \dots E_n$ are $n$ events in $S$ then $$\begin{align} &P(E_1\cap E_2\cap E_3\cap \cdots \cap E_n)\\ &=P(E_n|E_{n-1}\cap \cdots \cap E_1)\cdots P(E_3|E_2\cap E_1)\cdot P(E_2|E_1)\cdot P(E_1)\\ &=P(E_1)\cdot P(E_2|E_1)\cdot P(E_3|E_2\cap E_1)\cdot \cdots \cdot P(E_n|E_{n-1}\cap \cdots \cap E_1) \end{align}$$
The above can be expressed more compactly as $$P\left(E_n\cap E_{n-1}\cap \cdots \cap E_1\right) = \prod _{k=1}^{n} P\left(E_k \bigg| \bigcap_{j=1}^{k-1} E_j \right) $$
The Rule of Total Probability
Theorem:
The Rule of Total Probability
More generally, if $S=E_1\cup E_2\cup \cdots \cup E_k$ with $E_i\cap E_j=\emptyset \quad \forall i\neq j$ such that $P(E_i)>0\quad \forall i$. Then for any event $B$\ $$\begin{align} P(B)& = P(B\cap E_1)+ P(B\cap E_1)+\cdots +P(B\cap E_k)\\ &=P(B|E_1)\cdot P(E_1)+P(B|E_2)\cdot P(E_2)+\cdots+P(B|E_k)\cdot P(E_k) \end{align}$$
Example 1
A bag contains 5 red candies and 3 blue candies. If two candies are drawn without replacement, what is the probability that both candies are red?
Let $R_1$ be the event that the first candy is red and $R_2$ be the event that the second candy is red. Then $P(R_1)=\frac{5}{8}$ and $P(R_2|R_1)=\frac{4}{7}$. Thus, $$\begin{align*} P(R_1\cap R_2) & = P(R_2|R_1)\cdot P(R_1) \\ & = \frac{4}{7}\cdot \frac{5}{8} \\ & = \frac{5}{14} \end{align*}$$
Solution
Example 2
A student must pass both a written exam and an oral exam to qualify for a scholarship. The probability of passing the written exam is $0.8$, and the probability of passing the oral exam given that the written exam was passed is $0.7$. What is the probability of qualifying for the scholarship?
Let $W$ be the event that the written exam is passed and $O$ be the event that the oral exam is passed. Then $P(W)=0.8$ and $P(O|W)=0.7$. Thus, $$\begin{align*} P(W\cap O) & = P(O|W)\cdot P(W) \\ & = 0.7\cdot 0.8 \\ & = 0.56 \end{align*}$$
Solution
Example 3
A fair six-sided die is rolled twice. What is the probability of rolling a 4 on the first roll and a number greater than 3 on the second roll?
Let $R4$ be the event that a 4 is rolled on the first roll and $>3$ be the event that a number greater than 3 is rolled on the second roll. Then $P(R4)=\frac{1}{6}$ and $P(>3)=\frac{3}{6}$. Thus, $$\begin{align*} P(R4\cap >3) & = P(R4 \cap >3)=P(R4)\cdot P(>3) \\ & = \frac{1}{6}\cdot \frac{1}{2} \\ & = \frac{1}{12} \end{align*}$$
Solution
Example 4
A company has three factories producing a product: Factory A produces $40\%$ of the total, Factory B produces $35\%$, and Factory C produces $25\%$. The probability of a defective product is $5\%$ for Factory A, $4\%$ for Factory B, and $6\%$ for Factory C. What is the probability that a randomly selected product is defective?
Let $A$, $B$, and $C$ be the events that the product is produced by Factories A, B, and C, respectively. Let $D$ be the event that the product is defective. Then $P(A)=0.4$, $P(B)=0.35$, $P(C)=0.25$, $P(D|A)=0.05$, $P(D|B)=0.04$, and $P(D|C)=0.06$. Thus, $$\begin{align*} P(D) & = P(D\cap A)+P(D\cap B)+P(D\cap C) \\ & = P(D|A)\cdot P(A)+P(D|B)\cdot P(B)+P(D|C)\cdot P(C) \\ & = 0.05\cdot 0.4+0.04\cdot 0.35+0.06\cdot 0.25 \\ & = 0.049 \end{align*}$$
Solution
Example 5
A student is preparing for a test. On days when they study, the probability of passing is 0.9; on days when they don't study, the probability of passing is 0.5. If the student studies 70% of the time, what is the probability that they pass the test?
Let $S$ be the event that the student studies and $P$ be the event that the student passes the test. Then $P(S)=0.7$, $P(P|S)=0.9$, and $P(P|S^{\prime})=0.5$. Thus, $$\begin{align*} P(P) & = P(P\cap S)+P(P\cap S^{\prime}) \\ & = P(P|S)\cdot P(S)+P(P|S^{\prime})\cdot P(S^{\prime}) \\ & = 0.9\cdot 0.7+0.5\cdot 0.3 \\ & = 0.78 \end{align*}$$
Solution
Example 6
IBM's Watson, is capable of answering questions posed in natural language. Researchers accidentally taught it how to swear after uploading the Urban Dictionary to its training corpus; not knowing that the definitions were laced with profanity and risque innuendos. Before appearing on Jeopardy! Watson's vocabulary was sanitized and a smart filter was added to prevent any slip ups on television.
In the process of training Watson, researchers generated random questions that the computer had to answer within a fixed amount of time. The probability that the first question is answer correctly is $0.8$ Whenever a question is answered correctly, the next question is more difficult, and the probability of getting it correct drops by $0.1$. Whenever a question is answered incorrectly, the difficulty level of the next question remains the same.
Bayes' Theorem
Theorem:
Bayes' Theorem
Example 7
At the beginning of World War I, an unofficial ceasefire was declared along the Western Front. In the weeks leading up to Christmas, German and British soldiers took the opportunity to exchange gifts and play soccer with each other. On the German side, troops put up a sign which read ``Gott mit uns`` (``God with us``). In response, the British put up a cheeky sign which said ``We've got mittens``
At a local store, $85\%$ of the mittens are made in Germany while the rest are made in England. Suppose that $4\%$ of the British mittens have an imperfection in them, while the same could be said about $6\%$ of the German ones. What is the probability that a randomly selected pair of mittens
Example 8
The Shaggy defense is a legal strategy in which the defendant flatly denies guilt even though there is overwhelming evidence against them - particularly in the form of a video recording. The hallmarks of the defence involves a strenuous denial that it was them who committed the act. The term is derived from Shaggy's 2000 single ``It Wasn't Me`` - a song which describes a man asking his friend what to do after his girlfriend caught him with another woman. The friend's advice: simply deny that it was him.
Lawyer, Saul Goodman, uses the Shaggy defence in $30\%$ of the cases that he is assigned to. The rest of the time, he uses the Chewbacca defence. Whenever he uses the Shaggy defence, it results in a not guilty verdict $40\%$ of the time, and a guilty verdict the rest of the time. If he were to use the Chewbacca defence, then the probability of getting a not guilty verdict for his client is $80\%$ and a guilty verdict the rest of the time.
Example 9
For years, North and South Korea have been bombarding each other with balloons filled with propaganda flyers and household items. South Korean activists have sent balloons filled with chocolate snack cakes and anti-Kim leaflets, while the North have sent back propaganda pamphlets. This is a step up for North Korea. Up until 2017, they were sending over balloons filled with cigarette butts and used toilet paper.
South Korea orders all of their balloons from one company which makes them in three sizes: small, medium, and large. In one box, $40\%$ of the balloons are small, $25\%$ are medium, and the rest are large. It was also discovered that $1\%$ of the small, $2\%$ of the medium, and $5\%$ of the large balloons in the box were manufactured too thin, and prone to popping prematurely.
Independence
Definition:
Independent Events
Equivalently, $$P(A|B)=P(A) \qquad \text{or} \qquad P(B|A)=P(B) $$
Remark
Remark
Mutually exclusive (or disjoint) events are not independent because the occurrence of one event directly affects the probability of the other event occurring. Let's break this down with definitions and reasoning.
Mutually Exclusive Events: Two events $A$ and $B$ are mutually exclusive if they cannot occur at the same time: $$P(A \cap B)=0$$
Independent Events: Two events $A$ and $B$ are independent if the occurrence of one event does not affect the probability of the other event occurring: $$P(A \cap B)=P(A)P(B)$$
The only way $P(A \cap B)=P(A) \cdot P(B)=0$ is if at least one of the events has a probability of 0 , meaning it cannot occur at all. Therefore, if two events are mutually exclusive, they cannot be independent.
Example 1
Draw a card, observe, and put it back. Reshuffle, draw another card, and observe. What is the probability that we observe two aces? What is the probability that we observe two aces if the first card is not replaced before drawing the second card?
Let $A_1=$ the first card is an ace, $A_2=$ the second card is an ace. Then with replacement: $$P(A_1\cap A_2)=P(A_1)P(A_2)=\frac{4}{52}\cdot \frac{4}{52}$$ Without replacement: $$P(A_1\cap A_2)=P(A_1)P(A_2|A_1)=\frac{4}{52}\cdot \frac{3}{51}=\frac{1}{221}$$
Solution
Theorem:
Independence of Complementary Events
Definition:
Independence of $n-$Events
Remark
$A=$ the first coin is heads
$B=$ the third coin is heads
$C=$ there are an equal number of heads and tails
Then, $n(S)=2^4=16, P(A)=\frac{1}{2}, P(B)=\frac{1}{2},$ and $P(C)=\frac{6}{16}$ $$\begin{align} P(A\cap B)&=P(A)P(B)=\frac{1}{4}\\ P(A\cap C)&=P(A)P(C)=\frac{1}{8}\\ P(B\cap C)&=P(B)P(C)=\frac{1}{8}\\ P(A\cap B\cap C)&=P(A)P(B)P(C)=\frac{1}{16} \end{align}$$ Since $P(A\cap B\cap C)=P(A)P(B)P(C)$, the events are pairwise independent. However, $P(A\cap B\cap C)\neq P(A)P(B)P(C)$, so the events are not mutually independent.
Example 5
In June 1982, Mount Galunggung in Indonesia erupted, throwing a massive cloud of volcanic ash into the atmosphere. British Airways Flight 9, then en route from London to Auckland, accidentally flew into it. After losing all four engines, the pilot calmly addressed passengers in what could only be described as a masterpiece of understatement with ``Ladies and gentlemen, this is your captain speaking. We have a small problem. All four engines have stopped. We are doing our damnedest to get them going again. I trust you are not in too much distress.''
The probability that an engine, clogged with volcanic ash will restart working is $0.97$. Assuming that all four engines on Flight 9 operate independently, determine the following.
Example 5
Pilgrims en route to the Catholic shrine in Lourdes, France were left disappointed after their GPS system erroneously directed them to the less celebrated village of Lourde in the foothills of the Pyrenees. Unfortunately for them, the hamlet was: 57 miles off course from their intended target, does not have a shrine to the Virgin Mary, nor have a single hotel to stay in for the night.
In Lourde, cars approaching the main crossroad, must go in one of three directions: left,right, or straight on. As a train engineer, you observe that of the vehicles approaching from the north, $45\%$ turn left, $20\%$ turn right, and $35\%$ go straight on. Assuming that each driver acts independently from each other, what is the probability that of the next three drivers.
Exercises
Question 1
Given a sample space $S$ with $n(S)=100$, and the following events:
$A:n(A)=40$
$B: n(B)=30$
$A \cap B: n(A\cap B)=10$
Question 2
A box contains 5 red, 3 yellow and 2 green chips. A chip is drawn at random from the box. What is the probability that the chip is:
Question 3
A manufacturing plant produces three types of devices: Type A, Type B, and Type C. The table below shows the total number of devices produced and the number of failures for each type in the last month: $$ \begin{array}{|l|l|l|}\hline \text { Device Type } & \text { Total Produced } & \text { Failures } \\ \hline \text { Type A } & 500 & 25 \\ \hline \text { Type B } & 300 & 15 \\ \hline ext { Type C } & 200 & 10 \\ \hline \end{array} $$
Question 4
A fair six-sided die is rolled twice. Let:
Event $A$ : The sum of the two rolls is greater than 8
Event $B$ : The first roll is an even number.
Question 5
A box contains three bags. Each bag contains colored marbles:
Bag 1: 3 red, 2 blue
Bag 2: 4 red, 1 blue
Bag 3: 2 red, 3 blue.
A bag is chosen at random, and then a ball is drawn at random from the selected bag.
Question 6
A recommendation system for an e-commerce platform categorizes users into three groups based on their preferences:
Tech Enthusiasts (40\% of users), who click on a recommended item $60 \%$ of the time. Fashion Lovers ( $30 \%$ of users), who click on a recommended item $70 \%$ of the time. Casual Shoppers ( $30 \%$ of users), who click on a recommended item $40 \%$ of the time.
Question 7
An autonomous vehicle uses three sensors to detect obstacles:
Radar detects obstacles $80 \%$ of the time (true positive rate) and falsely detects obstacles $10 \%$ of the time (false positive rate)
Lidar detects obstacles $90 \%$ of the time and falsely detects obstacles $5 \%$ of the time.
Camera detects obstacles $85 \%$ of the time and falsely detects obstacles $8 \%$ of the time. The sensors operate independently.
Question 8
A spam filter is trained to classify emails as spam or not spam. The following statistics are observed:
$40 \%$ of emails are spam.
The filter identifies $90 \%$ of spam emails correctly (true positive rate).
The filter misclassifies $5 \%$ of non-spam emails as spam (false positive rate).
Question 9
In a wildlife reserve, cameras are placed to detect three animal species: Deer (D), Fox (F), and Rabbit(R). Observations in one day show the following probabilities:
A deer is detected with a probability of $0.6$
A fox is detected with a probability of $0.4$
A rabbit is detected with a probability of $0.5$
The detection of each species is independent.
Question 10
A particle detector identifies particles passing through it as either Alpha (A), Beta (B), or Gamma (G) particles. The probabilities of detecting these particles in a given trial are as follows:
Detecting Alpha: $0.4$
Detecting Beta: $0.3$
Detecting Gamma: $0.2$
Detecting no particle: $0.1$
Question 11
In a chemical reaction, there are three possible pathways for a molecule to react:
Pathway A occurs with a probability of $0.5$
Pathway B occurs with a probability of $0.3$
Pathway C occurs with a probability of $0.2$
The pathways are mutually exclusive.
Question 12
A standard deck of 52 cards is shuffled. You draw 3 cards without replacement.
Question 13
You roll 4 dice.
Question 14
A group of 10 students is to be split into two teams of 5 each. The group includes 6 boys and 4 girls.
Question 15
A box contains 6 red toys, 4 blue toys, and 5 green toys. You draw 3 toys without replacement.