Events and Probability Rules
In a school survey, 60% of students say they watch comedies and 50% say they watch thrillers. Does that mean 110% of students watch one of the two? Obviously not, but why? Probability needs a precise language for "this or that", "this and that" and "not this". This page builds that language and the basic rules that go with it. You will use it in every later topic of the course.
Sample space and events
When all outcomes in $S$ are equally likely, the theoretical probability of an event is the number of favorable outcomes divided by the number of possible outcomes:
$$P(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}$$Some problems call this Laplace's Rule. The outcomes must really be equally likely, so list them carefully.
A high school senior is organizing their digital note folders. In their "Economics" folder, they have 12 PDF files, 8 text documents, and 5 diagrams in image format. If…
A high school senior is organizing their digital note folders. In their "Economics" folder, they have 12 PDF files, 8 text documents, and 5 diagrams in image format. If they choose a file at random to review before an exam, determine:
- a) The sample space of the experiment.
- b) The probability that the chosen file is a diagram in image format.
- c) The probability that the chosen file is not a text document.
Combining events: or, and, not
Events are sets, so you combine them with set operations. Probability adds nothing new to the vocabulary here.
| Words | Notation | Meaning |
|---|---|---|
| $A$ or $B$ | $A \cup B$ (union) | $A$ happens, $B$ happens, or both |
| $A$ and $B$ | $A \cap B$ (intersection) | both happen at the same time |
| not $A$ | $A^C$ (complement, also written $\overline{A}$) | $A$ does not happen |
| $A$ but not $B$ | $A \cap B^C$ | $A$ happens and $B$ does not |
In math, "or" is inclusive: "$A$ or $B$" includes the case where both happen.
In a high school senior class, a survey on social media use has been carried out. It is known that event $A$ is "using Instagram" and event $B$ is "using TikTok". Define…
In a high school senior class, a survey on social media use has been carried out. It is known that event $A$ is "using Instagram" and event $B$ is "using TikTok". Define the following situations using event operations (union, intersection, or complement):
- a) The student uses both social networks.
- b) The student uses at least one of the two networks.
- c) The student does not use Instagram.
- d) The student uses Instagram but not TikTok.
Mutually exclusive events
For mutually exclusive events, nothing is counted twice, so you can simply add:
$$P(A \cup B) = P(A) + P(B)$$An esports team has players of three nationalities on its roster: 5 Spaniards, 3 Frenchmen and 2 Italians. If a captain is selected at random for the next tournament,…
An esports team has players of three nationalities on its roster: 5 Spaniards, 3 Frenchmen and 2 Italians. If a captain is selected at random for the next tournament, calculate:
- a) The probability that they are French.
- b) The probability that they are not Spanish.
- c) Are the events "being Spanish" and "being Italian" compatible or incompatible? Justify your answer.
Given two events $M$ and $N$ such that they are incompatible with each other, with $P(M) = 0,35$ and $P(N) = 0,45$, find the probability that: a) Event $M$ or event $N$…
Given two events $M$ and $N$ such that they are incompatible with each other, with $P(M) = 0,35$ and $P(N) = 0,45$, find the probability that:
- a) Event $M$ or event $N$ occurs.
- b) Neither of the two events occurs.
- c) $M$ and $N$ occur simultaneously.
The addition rule
When events can overlap, adding $P(A) + P(B)$ counts the overlap twice. Subtract it once:
Now you can answer the hook. Suppose 60% of students watch comedies, 50% watch thrillers, and 30% watch both.
You can also solve the rule for a missing piece: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$.
When outcomes are equally likely, you can also list the sets. Take the numbers 1 through 12 with $A$ = multiples of 3 $= \{3, 6, 9, 12\}$ and $B$ = even $= \{2, 4, 6, 8, 10, 12\}$. Then $A \cap B = \{6, 12\}$ and $A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$ has 8 elements, so $P(A \cup B) = \tfrac{8}{12}$. The formula agrees: $\tfrac{4}{12} + \tfrac{6}{12} - \tfrac{2}{12} = \tfrac{8}{12}$.
We draw a ball from an urn containing 10 balls numbered from 1 to 10. We consider the events: $A = \{ \text{drawing an even number} \}$…
We draw a ball from an urn containing 10 balls numbered from 1 to 10. We consider the events:
$A = \{ \text{drawing an even number} \}$
$B = \{ \text{drawing a multiple of 3} \}$
- a) Write the elements that make up the events $A \cup B$ and $A \cap B$.
- b) Calculate $P(A \cup B)$ using Laplace's Rule on the count of their elements.
- c) Verify the previous result by applying the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
On a streaming platform, $70\%$ of users watch action series ($S$), $40\%$ watch documentaries ($D$) and $25\%$ watch both types of content. Calculate the probability…
On a streaming platform, $70\%$ of users watch action series ($S$), $40\%$ watch documentaries ($D$) and $25\%$ watch both types of content. Calculate the probability that a user chosen at random:
- a) Watches action series or documentaries.
- b) Watches neither of these two contents.
Let $A$ and $B$ be two events such that $P(A) = 0,5$, $P(\bar{B}) = 0,7$ and $P(A \cup B) = 0,65$. Calculate the probability of the intersection $P(A \cap B)$ and…
Let $A$ and $B$ be two events such that $P(A) = 0,5$, $P(\bar{B}) = 0,7$ and $P(A \cup B) = 0,65$. Calculate the probability of the intersection $P(A \cap B)$ and determine whether the events are compatible.
De Morgan's laws
Problems often ask for "neither", "not both" or "only one". It helps to see the four regions of a two-event Venn diagram.
In probability, the most useful consequences are:
-
$P(\text{neither}) = P(A^C \cap B^C) = 1 - P(A \cup B)$
-
$P(\text{not both}) = 1 - P(A \cap B)$
-
$P(\text{exactly one}) = P(A \cup B) - P(A \cap B)$
Consider two events $A$ and $B$ associated with a random experiment. If we know that $P(A) = 0,6$, $P(B) = 0,4$ and $P(A \cap B) = 0,2$, calculate the following…
Consider two events $A$ and $B$ associated with a random experiment. If we know that $P(A) = 0,6$, $P(B) = 0,4$ and $P(A \cap B) = 0,2$, calculate the following probabilities:
- a) $P(A \cup B)$
- b) $P(\bar{A})$ (probability of the complementary event of $A$).
- c) $P(\bar{A} \cap \bar{B})$ using De Morgan's Laws.
Let $A$ and $B$ be two events of the same random experiment for which the following probabilities are known: $P(A) = 0,6$, $P(B) = 0,5$ and…
Let $A$ and $B$ be two events of the same random experiment for which the following probabilities are known: $P(A) = 0,6$, $P(B) = 0,5$ and $P(\bar{A} \cap \bar{B}) = 0,2$.
Based on the properties of probability and De Morgan's laws, calculate:
- a) The probability of the union of the events, $P(A \cup B)$.
- b) The probability of the intersection, $P(A \cap B)$.
- c) Indicate with reasoning whether events $A$ and $B$ are compatible or incompatible.
In a study on the consumption of digital content among 16-year-olds, it has been detected that $70 \%$ use the platform StreamVibe* and $45 \%$ use CloudMusic*.…
In a study on the consumption of digital content among 16-year-olds, it has been detected that $70 \%$ use the platform StreamVibe and $45 \%$ use CloudMusic. Furthermore, it is known that $25 \%$ of the young people do not use either of the two platforms. If we choose a young person at random, calculate:
- a) The probability that they use both platforms.
- b) The probability that they use only StreamVibe.
- c) The probability that they use CloudMusic but not StreamVibe.
Conditional probability
Sometimes you know something already happened, and that changes the odds. $P(B \mid A)$ is read "the probability of $B$ given $A$". It is the probability of $B$ when you look only at the outcomes where $A$ happened.
A two-way table makes this visible. In a survey of 200 students:
| Job | No job | Total | |
|---|---|---|---|
| Sport | 30 | 70 | 100 |
| No sport | 40 | 60 | 100 |
| Total | 70 | 130 | 200 |
To fill missing cells of a partial table, use subtraction: each row and column adds to its total.
Conditional probability is also what happens when you pick without replacement. Suppose 15 of 25 students in a class have a pet, and you pick two students. If the first has a pet, 14 of the 24 remaining students have one. So $P(\text{2nd has a pet} \mid \text{1st has a pet}) = \dfrac{14}{24} = \dfrac{7}{12}$.
In a study on the use of social networks, two high school seniors are selected consecutively and without replacement. It is known that $60\%$ of the students at the…
In a study on the use of social networks, two high school seniors are selected consecutively and without replacement. It is known that $60\%$ of the students at the school use social network X.
If the probability that the second selected student uses social network X is $35/59$, determine the total number of high school seniors there are in that school.
The multiplication rule and tree diagrams
Rearranging the conditional probability formula gives a rule for "and":
A tree diagram organizes this. Multiply along a path to get the probability of that whole sequence. Add the paths that belong to the event.
Without replacement. After each pick the numbers change. A bag has 6 blue and 4 green marbles, and you take two without putting the first back. Then $P(\text{both blue}) = \dfrac{6}{10} \cdot \dfrac{5}{9} = \dfrac{1}{3}$, and $P(\text{at least one green}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$.
In a role-playing game, a player must pass two consecutive tests. The probability of passing the first test is $0,7$. If they pass the first, the probability of passing…
In a role-playing game, a player must pass two consecutive tests. The probability of passing the first test is $0,7$. If they pass the first, the probability of passing the second is $0,8$, but if they fail the first, the probability of passing the second drops to $0,4$.
- a) Draw the tree diagram that represents this situation.
- b) Calculate the probability that the player passes both tests.
- c) Calculate the probability that the player passes only one of the two tests.
A high school senior has prepared 18 of the 25 topics that make up the Geography syllabus. In the final exam, the teacher chooses two topics at random (without…
A high school senior has prepared 18 of the 25 topics that make up the Geography syllabus. In the final exam, the teacher chooses two topics at random (without replacement). Calculate:
- a) The probability that the student knows both topics.
- b) The probability that they do not know either.
- c) The probability that they know at least one of the two selected topics.
From a 40-card Spanish deck, two cards are drawn successively without replacement. Determine the following probabilities: a) That both cards are face cards (Jack,…
From a 40-card Spanish deck, two cards are drawn successively without replacement. Determine the following probabilities:
- a) That both cards are face cards (Jack, Knight, or King).
- b) That the first card is an Ace and the second is a King.
- c) That at least one of the two cards is a Coin.
Independent events
Why is the second form equivalent? Substitute $P(B \mid A) = P(B)$ in the multiplication rule. To test independence, check whether $P(A \cap B)$ equals $P(A) \cdot P(B)$.
If $A$ and $B$ are independent, then so are $A$ and $B^C$, $A^C$ and $B$, and $A^C$ and $B^C$. That gives a clean "at least one" shortcut.
We have the following partial contingency table showing the distribution of a group of 120 high school students according to their participation in extracurricular…
We have the following partial contingency table showing the distribution of a group of 120 high school students according to their participation in extracurricular activities (Sports or Music):
| Sports (D) | No Sports ($\bar{D}$) | Total | |
|---|---|---|---|
| Music (M) | 15 | 45 | |
| No Music ($\bar{M}$) | |||
| Total | 70 | 120 |
- a) Complete the contingency table in your notebook.
- b) Calculate the probability that a student chosen at random does not do either of the two activities.
- c) Are the events "Participating in Music" and "Participating in Sports" independent? Justify your answer mathematically.
Consider the events $A$ and $B$ such that $P(A) = 0,5$, $P(B|A) = 0,4$ and $P(A \cup B) = 0,65$. From this data: a) Find the value of the probability of the intersection…
Consider the events $A$ and $B$ such that $P(A) = 0,5$, $P(B|A) = 0,4$ and $P(A \cup B) = 0,65$. From this data:
- a) Find the value of the probability of the intersection $P(A \cap B)$.
- b) Calculate the value of $P(B)$.
- c) Determine with reasoning whether the events $A$ and $B$ are independent.
A cybersecurity system has two independent protection barriers, $B_1$ and $B_2$. The probability that an external attack overcomes the first barrier is $P(B_1) = 0,15$…
A cybersecurity system has two independent protection barriers, $B_1$ and $B_2$. The probability that an external attack overcomes the first barrier is $P(B_1) = 0,15$ and the probability that it overcomes the second is $P(B_2) = 0,10$.
Calculate the probability that an attack:
- a) Overcomes at least one of the two barriers.
- b) Overcomes exactly one of the barriers.
- c) Is not detected by either barrier, given that the attack has managed to overcome the first protection.
Independent or mutually exclusive?
These two ideas are often confused. They are opposites in spirit.
| Mutually exclusive | Independent | |
|---|---|---|
| Meaning | cannot happen together | one does not change the other |
| Test | $P(A \cap B) = 0$ | $P(A \cap B) = P(A) \cdot P(B)$ |
| $P(A \cup B)$ | $P(A) + P(B)$ | $P(A) + P(B) - P(A)P(B)$ |
If $A$ and $B$ are mutually exclusive and both have positive probability, they are not independent. Knowing that $A$ happened means $B$ cannot, so $P(B \mid A) = 0 \neq P(B)$.
Let $A$ and $B$ be two events such that $P(A) = 0,4$ and $P(B) = 0,3$. Calculate the probability of the union $P(A \cup B)$ in each of the following scenarios: a) $A$…
Let $A$ and $B$ be two events such that $P(A) = 0,4$ and $P(B) = 0,3$. Calculate the probability of the union $P(A \cup B)$ in each of the following scenarios:
- a) $A$ and $B$ are incompatible events.
- b) $A$ and $B$ are independent events.
- c) Event $B$ is contained in event $A$ ($B \subset A$).
Prove, using the properties of probability and De Morgan's laws, the truth or falsity of the following statement: "If $A$ and $B$ are two incompatible events with…
Prove, using the properties of probability and De Morgan's laws, the truth or falsity of the following statement:
"If $A$ and $B$ are two incompatible events with non-zero probabilities, then it is always true that $P(\bar{A} | B) = 1$".
Accompany your reasoning with an original numerical example where $P(A) = 0,3$ and $P(B) = 0,4$.
Splitting an event into pieces
Any event $A$ can be split by another event $B$ into the part inside $B$ and the part outside $B$. These two parts never overlap, so you can add them:
The main strategy for tough problems: turn every conditional probability into an intersection with the multiplication rule, then use the addition rule and complements.
In an esports championship, the plays of a pro player are analyzed. The events $K$ (getting a kill) and $W$ (winning the round) are defined. The analyst determines that:…
In an esports championship, the plays of a pro player are analyzed. The events $K$ (getting a kill) and $W$ (winning the round) are defined. The analyst determines that:
-
The probability of winning the round having gotten a kill is $P(W|K) = 0,8$.
-
The probability of winning the round without having gotten any kills is $P(W|\bar{K}) = 0,3$.
-
The probability that in a round the player wins or gets a kill is $0,85$.
Using the algebra of events, calculate:
- a) The probability that the player gets a kill in a round, $P(K)$.
- b) The probability that the player wins the round and does not get any kills.
- c) Show whether the events "winning the round" and "getting a kill" are dependent or independent.
On a streaming platform, it has been observed that $65\%$ of users under 20 years old watch animated series (event $A$) and $40\%$ watch nature documentaries (event…
On a streaming platform, it has been observed that $65\%$ of users under 20 years old watch animated series (event $A$) and $40\%$ watch nature documentaries (event $D$). Furthermore, it is known that the probability that a user does not watch animated series given that they do not watch nature documentaries is $0,5$.
Calculate with reasoning:
- a) The probability that a user watches both types of content.
- b) The probability that a user watches animated series but not nature documentaries.
- c) Are events $A$ and $D$ independent? Justify your answer using the previous results.
Let $A$ and $B$ be two events of a random experiment such that their probabilities are $P(A) = x$ and $P(B) = y$. It is known that the union event has a probability of…
Let $A$ and $B$ be two events of a random experiment such that their probabilities are $P(A) = x$ and $P(B) = y$. It is known that the union event has a probability of $P(A \cup B) = 0,75$ and that the intersection event satisfies $P(A \cap B) = 0,15$.
If it is also known that the probability of $A$ occurring given that $B$ does not occur is $P(A | \bar{B}) = 0,5$, determine the values of $x$ and $y$. Are $A$ and $B$ compatible events? Are they independent?
Three events
For three events, the addition rule needs one more correction term. Add the singles, subtract the pairs, add back the triple overlap:
A Venn diagram with three circles helps. Fill it from the center out: first the triple overlap, then the pairs, then each event alone.
If two events are mutually exclusive, such as $A$ and $C$, their overlap is empty. That also removes $A \cap B \cap C$. If $A$ and $B$ are independent, use $P(A \cap B) = P(A) P(B)$.
Consider three events $A$, $B$ and $C$ such that $A$ and $B$ are independent, and $A$ and $C$ are incompatible. The following probabilities are known: $P(A) = 0,4$;…
Consider three events $A$, $B$ and $C$ such that $A$ and $B$ are independent, and $A$ and $C$ are incompatible. The following probabilities are known:
$P(A) = 0,4$; $P(B) = 0,5$; $P(C) = 0,2$ and $P(B \cap C) = 0,1$.
Find the probability of the following compound events:
- a) $P(A \cup B \cup C)$
- b) $P((A \cap B) \cup C)$
- c) $P(A | \bar{C})$
A team of video game developers is analyzing the errors (bugs*) of their latest version. They have classified the errors into three categories: Critical ($C$), Graphical…
A team of video game developers is analyzing the errors (bugs) of their latest version. They have classified the errors into three categories: Critical ($C$), Graphical ($G$) and Sound ($S$). After a massive test, the following data has been obtained:
-
$10\%$ of the errors are of all three types at the same time.
-
The probability that an error is Critical is $P(C) = 0,4$.
-
The probability that it is Graphical is $P(G) = 0,5$.
-
The probability that it is Sound is $P(S) = 0,35$.
-
The probability that an error is Critical and Graphical is $0,2$.
-
The probability that it is Graphical and Sound is $0,15$.
-
The probability that an error does not belong to any of these three categories is $0,2$.
Calculate the probability that an error selected at random is:
- a) Critical and Sound.
- b) Exclusively a Graphical error.
- c) Sound, given that it is neither Critical nor Graphical.
Back to the start: 60% watch comedies and 50% watch thrillers, which adds to 110%, but 30% watch both and were counted twice. By the addition rule, $60\% + 50\% - 30\% = 80\%$ watch at least one, and $20\%$ watch neither.