Formula
Addition Rule
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.
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:
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:
Definition
Conditional Probability
Remark
The order of events is important. $$\begin{align}P(A|B) &\neq P(B|A)\\ \because \frac{P(A\cap B)}{P(B)}&\neq \frac{P(A\cap B)}{P(A)}\end{align}$$
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
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
$$ \begin{array}{lccc} \textbf{Age} (years) & \textbf{Male} & \textbf{Female} \\ \hline $ 11 - 17 $ & $96 $ & $88$ \\ $ 18 - 24 $ & $204$ & $198$ \\ $ 25 - 31 $ & $206$ & $117$ \\ $ 32 - 38 $ & $54 $ & $12$ \\ \hline \end{array} $$
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:
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.
Theorem:
The Multiplication Rule
Suppose that $A$ and $B$ are events in $S$, then $$\begin{align} P(A\cap B) & = P(A|B)\cdot P(B) \\ & = P(B|A)\cdot P(A) \end{align}$$
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 $$P\left(E_n\cap E_{n-1}\cap \cdots \cap E_1\right) = \prod _{k=1}^{n} P\left(E_k \bigg| \bigcap_{j=1}^{k-1} E_j \right) $$
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 $$P\left(E_n\cap E_{n-1}\cap \cdots \cap E_1\right) = \prod _{k=1}^{n} P\left(E_k \bigg| \bigcap_{j=1}^{k-1} E_j \right) $$
Theorem:
The Rule of Total Probability
Suppose that $A$ and $B$ are disjoint events that form a partition in $S$. Then $$\begin{align} P(B) & = P(A\cap B)+P(A^{\prime} \cap B)\\ & =P(B|A)\cdot P(A) + P(B|A^{\prime})\cdot P(A^{\prime})\end{align}$$
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 $B$\ $$\begin{align} P(B)& = P(B\cap E_1)+ P(B\cap E_1)+\cdots +P(B\cap E_k)\\ &=P(B|E_1)\cdot P(E_1)+P(B|E_2)\cdot P(E_2)+\cdots+P(B|E_k)\cdot P(E_k) \end{align}$$
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 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.
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.
Theorem:
Bayes' Theorem
If $S=E_1\cup E_2\cup \cdots \cup E_n$ with $E_i\cap E_j=\emptyset \quad \forall i\neq j$ (i.e. we have a disjoint partition of the sample space) then $$\begin{align} P(E_i\,|\,B) &=\frac{P(E_i\cap B)}{P(B)}=\frac{P(E_i\cap B)}{\displaystyle\sum^n_{k=1} P(E_k\cap B)}\end{align}$$
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
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.
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.
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.
Definition
Independent Events
Remark
$$P(A)=P(A|B)\quad; \quad P(B)=P(B|A)$$ neither $A$ is informative of $B$, nor is $B$ informative of $A$
Remark
Disjoint events $A\cap B=\emptyset$ are not independent at all
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.
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.
Theorem:
Independence of Complementary Events
If $A$ and $B$ are independent events, then $A^{\prime}$ and $B^{\prime}$ are also independent.
Definition
Independence of $n-$Events
Remark
In the case of two or more events, pairwise independence does not imply mutual independence. For example, if $A$ and $B$ are independent, and $B$ and $C$ are independent, it does not imply that $A$ and $C$ are 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.
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.
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.