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Probability · Probability

Definitions and Properties of Probability

Before you start, make sure you know: Concepts (Math 7)

Jayden surveys his high school. He finds that 60% of the students play a sport and 45% belong to a club. He adds them up: 60% + 45% = 105%. But no probability can be more than 100%. What went wrong?

By the end of this page you'll know the rules that probabilities must follow, and you'll be able to fix Jayden's mistake.

Outcomes, sample spaces, and events

A random experiment is something whose result you can't predict for sure: rolling a die, drawing a raffle ticket, picking a student at random. Each possible result is an outcome. The set of all outcomes is the sample space, written $S$.

An event is any set of outcomes, that is, a subset of $S$. We combine events with the same set language you know:

  • $A \cup B$ ("$A$ or $B$"): outcomes in $A$, in $B$, or in both.

  • $A \cap B$ ("$A$ and $B$"): outcomes in both.

  • $\bar{A}$ ("not $A$", the complement): outcomes in $S$ that are not in $A$.

  • $A - B$ means "$A$ but not $B$". It is the same as $A \cap \bar{B}$.

Two events are mutually exclusive if they can't happen at the same time, that is, $A \cap B = \emptyset$. (Some problems say incompatible for these, and compatible for events that can overlap.) The impossible event is $\emptyset$ and the certain event is $S$.

Three Venn diagrams showing the union, the intersection, and the complement of events inside a sample space. A ∪ B: A or B A ∩ B: A and B Ā: not A
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A content creator on TikTok decides to raffle three different prizes among their followers. To do this, they use a drum with balls numbered from $1$ to $20$. We consider…

A content creator on TikTok decides to raffle three different prizes among their followers. To do this, they use a drum with balls numbered from $1$ to $20$. We consider the random experiment of extracting a ball and noting its number.

Define the following elements:

  1. a) The sample space $E$.
  2. b) The event $A$: "obtaining a number that is a multiple of $5$".
  3. c) The event $B$: "obtaining a prime number greater than $10$".
  4. d) The event $C$: "obtaining a number less than $4$".

Write down all the elementary events that make up each of the previous sets.

E 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 C: < 4 A: Multiple of 5 B: Prime > 10
Solve
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We roll a six-sided die numbered from $1$ to $6$. We consider the events: $A = \{2, 4, 6\}$ (rolling an even number) $B = \{1, 2, 3\}$ (rolling less than or equal to…

We roll a six-sided die numbered from $1$ to $6$. We consider the events:

  • $A = \{2, 4, 6\}$ (rolling an even number)

  • $B = \{1, 2, 3\}$ (rolling less than or equal to $3$)

  • $C = \{5\}$ (rolling a five)

Determine whether the following pairs of events are compatible or incompatible:

  1. a) $A$ and $B$
  2. b) $A$ and $C$
  3. c) $B$ and $C$

Justify your answer by checking whether the intersection of each pair is the empty set $\emptyset$.

Solve

The rules of probability

Probability is a number we assign to each event. No matter how you assign it, it must obey three rules. (Some problems call them Kolmogorov's axioms.)

Two facts follow right away:

  • $P(\emptyset) = 0$. The impossible event has probability $0$.

  • Complement rule: $P(\bar{A}) = 1 - P(A)$. Why? $A$ and $\bar{A}$ are mutually exclusive and together make up all of $S$. So $P(A) + P(\bar{A}) = P(S) = 1$.

Rule 3 extends to any number of mutually exclusive events. In particular, the probabilities of all the individual outcomes must be non-negative and add up to $1$. That is the quick test for whether an assignment is valid.

A number line from 0 to 1 showing impossible, even chance, and certain events. impossible even chance certain 0 0.5 1 every probability lives between 0 and 1
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State with reasons whether the following probability assignments for a sample space $E = \{x_1, x_2, x_3\}$ are valid according to Kolmogorov's axioms: a)…

State with reasons whether the following probability assignments for a sample space $E = \{x_1, x_2, x_3\}$ are valid according to Kolmogorov's axioms:

  1. a) $P(x_1) = 0,4; \quad P(x_2) = 0,3; \quad P(x_3) = 0,3$
  2. b) $P(x_1) = 0,7; \quad P(x_2) = -0,1; \quad P(x_3) = 0,4$
  3. c) $P(x_1) = 0,5; \quad P(x_2) = 0,5; \quad P(x_3) = 0,2$
  4. d) $P(x_1) = 1; \quad P(x_2) = 0; \quad P(x_3) = 0$
Solve
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In a high school, $25\%$ of the students are enrolled in guitar lessons, and $15\%$ are enrolled in piano lessons. If we know that there are no students studying both…

In a high school, $25\%$ of the students are enrolled in guitar lessons, and $15\%$ are enrolled in piano lessons. If we know that there are no students studying both instruments at the same time:

  1. a) What are these types of events called in probability?
  2. b) Calculate the probability that a student chosen at random studies guitar or piano.
  3. c) Calculate the probability that a student does not study either of these two instruments.
Total = 100% Guitar 25% Piano 15% Incompatible 0% Neither = ?
Solve

Equally likely outcomes

If all $n$ outcomes of $S$ are equally likely, each one has probability $\tfrac{1}{n}$. (They are mutually exclusive and add to $1$.) Adding up the outcomes of an event gives theoretical probability:

Try it with an urn. Draw several times and see how the proportion of red balls compares with $\tfrac{5}{8}$.

Card problems. A standard deck has 52 cards: 4 suits (hearts, diamonds, spades, clubs) with 13 ranks each. Some problems use a 40-card deck instead: four suits (cups, coins, swords, clubs) with 10 cards each, namely an ace, the numbers 2 through 7, and three face cards (jack, knight, king). That deck has 12 face cards. The method is the same: count the favorable cards and divide by the total number of cards, here $40$.

With the 40-card deck, only the total changes. For example, $P(\text{king}) = \dfrac{4}{40} = \dfrac{1}{10}$ and $P(\text{a card of one particular suit}) = \dfrac{10}{40} = \dfrac{1}{4}$.

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In an urn there are $12$ colored balls: $5$ are red, $4$ are blue and $3$ are yellow. We extract a ball at random and observe its color. Using Laplace's rule, calculate…

In an urn there are $12$ colored balls: $5$ are red, $4$ are blue and $3$ are yellow. We extract a ball at random and observe its color. Using Laplace's rule, calculate the probability of the following events:

  1. a) $P(R)$: "the ball is red".
  2. b) $P(A \cup M)$: "the ball is blue or yellow".
  3. c) $P(\bar{R})$: "the ball is not red".
  4. d) $P(V)$: "the ball is green".
Solve
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A card is drawn from a Spanish deck of $40$ cards. Calculate the probability of obtaining: a) A king. b) A card of cups. c) The ace of coins. d) A face card (jack,…

A card is drawn from a Spanish deck of $40$ cards. Calculate the probability of obtaining:

  1. a) A king.
  2. b) A card of cups.
  3. c) The ace of coins.
  4. d) A face card (jack, knight or king).
  5. e) A card that is not an ace.
Solve

Counting outcomes with combinations

When a group is chosen at random, every possible group is equally likely. So the probability is (number of favorable groups) $\div$ (number of possible groups). The order of the chosen people doesn't matter, so we count with combinations:

A group of fifteen students, nine juniors and six seniors, from which three are chosen at random. 9 juniors 6 seniors Choose 3 at random: C(15, 3) = 455 possible groups
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In a prestigious esports tournament, a digital doping control committee made up of 4 members must be chosen. The group of candidates consists of 8 data analysts and 6…

In a prestigious esports tournament, a digital doping control committee made up of 4 members must be chosen. The group of candidates consists of 8 data analysts and 6 cybersecurity experts. If the selection is made at random, calculate the probability of the following events:

  1. a) The committee is made up exclusively of cybersecurity experts.
  2. b) There is at least one representative of each profile on the committee.
  3. c) The number of data analysts is strictly greater than that of cybersecurity experts.
Solve

The addition rule

If you add $P(A)$ and $P(B)$, the outcomes in $A \cap B$ are counted twice. Subtract them once:

Now back to Jayden's survey. Let $A$ = "plays a sport" and $B$ = "in a club", with $P(A \cap B) = 0.25$.

Venn diagram of sport and club with region probabilities 35 percent, 25 percent, 20 percent, and 20 percent outside both. Sport (60%) Club (45%) 35% 25% 20% neither: 20%
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A music app has analyzed the preferences of a group of young people. It is known that the probability of a user liking Trap is $P(T) = 0,65$, the probability of them…

A music app has analyzed the preferences of a group of young people. It is known that the probability of a user liking Trap is $P(T) = 0,65$, the probability of them liking Pop is $P(P) = 0,40$ and the probability of them liking both genres is $P(T \cap P) = 0,15$.

Calculate the probability that, when choosing a user at random:

  1. a) They like Trap or Pop.
  2. b) They do not like Trap.
  3. c) They like Trap but not Pop ($P(T - P)$).
Solve
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On a streaming* platform, $65\ \%$ of the subscribers have watched the latest season of a science fiction series ($S$), $40\ \%$ have watched a nature documentary ($D$)…

On a streaming platform, $65\ \%$ of the subscribers have watched the latest season of a science fiction series ($S$), $40\ \%$ have watched a nature documentary ($D$) and $25\ \%$ have watched both. If we select a user at random, calculate the probability of the following events:

a) That they have watched the series or the documentary.

b) That they have watched neither the series nor the documentary.

c) That they have watched the series but not the documentary.

E 100% Series P(S)=65% Only S Documentary P(D)=40% Only D S∩D 25% Ninguno: (S∪D)ᶜ
Solve
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In a charity raffle organized by students for the end-of-year trip, tickets numbered from $000$ to $999$ are sold. We consider the events: $A =$ "The number ends in an…

In a charity raffle organized by students for the end-of-year trip, tickets numbered from $000$ to $999$ are sold. We consider the events:

$A =$ "The number ends in an even digit (0, 2, 4, 6, 8)".

$B =$ "The number is a multiple of 25".

a) Calculate $P(A)$ and $P(B)$ by applying Laplace's rule.

b) Are $A$ and $B$ compatible or incompatible events? Justify your answer by finding the event $A \cap B$.

c) Calculate the probability that the winning number ends in an even digit or is a multiple of 25.

Solve

Venn diagrams and three events

A Venn diagram is a map of probabilities: all the regions together add up to $P(S) = 1$. Fill the diagram from the middle outward. Start with the region where everything overlaps, then work out the regions around it by subtraction.

With three events, the addition rule gets one more term:

Three-circle Venn diagram for soccer, basketball, and track with the percentage in each region. Soccer (S) Basketball (B) Track (T) none: 5% 25% 20% 15% 15% 10% 5% 5%
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Observe the following Venn diagram representing the events $A$ and $B$ within a sample space $E$. The numbers represent the quantity of elementary events in each region.…

Observe the following Venn diagram representing the events $A$ and $B$ within a sample space $E$. The numbers represent the quantity of elementary events in each region.

E A B 8 3 5 4

If the experiment consists of choosing at random one of the elementary events of $E$, all of them equally likely, calculate:

  1. a) $P(A)$
  2. b) $P(A \cap B)$
  3. c) $P(A \cup B)$
  4. d) $P(\bar{A} \cap \bar{B})$ (that it is neither in $A$ nor in $B$).
Solve
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A market study on the consumption of digital platforms among 16-year-olds reveals that $70\%$ use Spotify, $55\%$ use TikTok and $45\%$ use Netflix. Furthermore, it is…

A market study on the consumption of digital platforms among 16-year-olds reveals that $70\%$ use Spotify, $55\%$ use TikTok and $45\%$ use Netflix. Furthermore, it is known that $30\%$ use Spotify and TikTok, $25\%$ use Spotify and Netflix, and $20\%$ use TikTok and Netflix. Only $10\%$ of the young people use all three platforms simultaneously.

If we select a young person at random, calculate and explain the probability that they:

  1. a) Use any of the three platforms.
  2. b) Use none of these platforms.
  3. c) Use only one of the three platforms.
Ω Spotify TikTok Netflix 10% Probabilities: P(S) = 70% P(T) = 55% P(N) = 45% P(S∩T) = 30% P(S∩N) = 25% P(T∩N) = 20% P(S∩T∩N) = 10%
Solve

De Morgan's Laws

Problems often give you "neither" and ask about "both", or the other way around. De Morgan's Laws turn "not" over an "or" or "and":

Two Venn diagrams with the same shaded region: everything except the overlap of A and B. not (A ∩ B) (not A) ∪ (not B) Both panels shade everything except A ∩ B
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A survey has been carried out among students at a high school about the use of social media. It is known that the probability that a student uses Instagram ($I$) is…

A survey has been carried out among students at a high school about the use of social media. It is known that the probability that a student uses Instagram ($I$) is $P(I) = 0,82$ and the probability that they use TikTok ($T$) is $P(T) = 0,55$. If the probability that they use neither of the two is $P(\bar{I} \cap \bar{T}) = 0,10$, use De Morgan's Laws to calculate:

a) The probability that they use at least one of the two social media platforms.

b) The probability that they use both simultaneously.

c) Are the events "using Instagram" and "using TikTok" incompatible? Justify your answer.

Solve
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Given two events $A$ and $B$ associated with the same random experiment, we know that the probability that neither of them occurs is $P(\bar{A} \cap \bar{B}) = 0,2$.…

Given two events $A$ and $B$ associated with the same random experiment, we know that the probability that neither of them occurs is $P(\bar{A} \cap \bar{B}) = 0,2$. Furthermore, it is known that $P(A) = 0,5$ and that the probability that $A$ occurs but $B$ does not occur is $P(A - B) = 0,3$.

Calculate with reasoning, using the properties of probability and De Morgan's laws:

  1. a) $P(A \cup B)$
  2. b) $P(A \cap B)$
  3. c) $P(B)$
  4. d) $P(\bar{A} \cup \bar{B})$
E P(A̅ ∩ B̅) = 0.2 A B P(A-B) = 0.3 ? P(A) = 0.5
Solve

Conditional probability and tree diagrams

Sometimes you get extra information. "A student is chosen given that they're a junior" shrinks the sample space to the juniors. The probability of $A$ given $B$ is written $P(A \mid B)$.

With a two-way table you can just look at the right row or column. The "given" part sets which total goes in the denominator.

Plays a sport No sport Total
Junior 24 16 40
Senior 20 20 40
Total 44 36 80

A tree diagram organizes the multiplication rule. You multiply along the branches and add the outcomes you want. For dependent draws, such as drawing without replacement, the second-stage probabilities change after the first draw.

Tree diagram for drawing two marbles without replacement from a bag of 4 red and 6 blue marbles. R B 4/10 6/10 3/9 6/9 4/9 5/9 RR: 4/10 × 3/9 = 12/90 RB: 4/10 × 6/9 = 24/90 BR: 6/10 × 4/9 = 24/90 BB: 6/10 × 5/9 = 30/90
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In a language academy, students are divided according to the level they are studying and the language chosen. The data is collected in the following contingency table: |…

In a language academy, students are divided according to the level they are studying and the language chosen. The data is collected in the following contingency table:

English ($A$) French ($B$) TOTAL
Level B1 ($M$) $45$ $75$
Level B2 ($N$) $20$
TOTAL $100$

a) Complete the missing data in the table.

b) If we choose a student at random, calculate the probability that they study French or are in level B2.

c) Given that the chosen student is in level B1, what is the probability that they study English?

Solve
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A box contains 10 wireless earphones, of which 3 are defective. If we extract two earphones consecutively and without replacement, calculate the probability that: a)…

A box contains 10 wireless earphones, of which 3 are defective. If we extract two earphones consecutively and without replacement, calculate the probability that:

a) Both earphones work correctly.

b) At least one of the earphones is defective.

c) The second earphone is defective given that the first one was not.

Correct (C) Defective (D) 1st Extraction 2nd Extraction 7/10 3/10 6/9 3/9 7/9 2/9 C₁ D₁ C₂ D₂ C₂ D₂ * Without replacement: the total is reduced to 9 in the 2nd extraction
Solve
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An esports team participates in a tournament with two independent knockout stages. The probability of passing the first stage is $0,7$ and the probability of passing the…

An esports team participates in a tournament with two independent knockout stages. The probability of passing the first stage is $0,7$ and the probability of passing the second is $0,6$. Represent the situation using a tree diagram and calculate:

a) The probability that the team wins both stages.

b) The probability that they are eliminated in either of the two stages.

c) The probability that they pass exactly one of the two stages.

Stage 1 Stage 2 Outcomes 0,7 0,3 0,6 0,4 0,6 0,4 S₁ E₁ S₂ E₂ S₂ E₂ S₁ ∩ S₂ = 0,42 S₁ ∩ E₂ = 0,28 E₁ ∩ S₂ = 0,18 E₁ ∩ E₂ = 0,12 Passes stage (S) Eliminated (E)
Solve

Independent events

Events are independent if knowing that one happened doesn't change the probability of the other. Flipping a coin twice is the standard case: the second flip doesn't care about the first.

A unit square where the overlap of a band of width 0.3 and a band of height 0.4 has area 0.12. P(A) = 0.3 P(B) = 0.4 0.12 Independent: the overlap is the product of the two bands: 0.3 × 0.4 = 0.12

Mutually exclusive is not the same as independent. If two events with positive probabilities are mutually exclusive, then $P(A \cap B) = 0 \ne P(A)\,P(B)$, so they are dependent: if one happens, the other can't. The only exception is when one of the events has probability $0$.

Also, $P(A \cup B) = P(A) + P(B)$ only tells you $P(A \cap B) = 0$. It does not make the events independent. And if $P(A) + P(B) > 1$, the events can't be mutually exclusive, because $P(A \cup B)$ can't exceed $1$.

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Let $A$ and $B$ be two events associated with a random experiment whose probabilities are known: $P(A) = 0,4$, $P(B) = 0,5$ and $P(A \cup B) = 0,7$. a) Calculate…

Let $A$ and $B$ be two events associated with a random experiment whose probabilities are known: $P(A) = 0,4$, $P(B) = 0,5$ and $P(A \cup B) = 0,7$.

a) Calculate $P(A \cap B)$.

b) Determine whether $A$ and $B$ are independent events.

c) Calculate the probability of the union of the complementary events, that is, $P(\bar{A} \cup \bar{B})$.

Solve
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Determine, with reasoning, whether the following statements are true or false, providing a numerical counterexample if they are false: a) If two events $A$ and $B$ are…

Determine, with reasoning, whether the following statements are true or false, providing a numerical counterexample if they are false:

  1. a) If two events $A$ and $B$ are mutually exclusive ($A \cap B = \emptyset$), then they must necessarily be dependent.
  2. b) If $P(A \cup B) = P(A) + P(B)$, then the events $A$ and $B$ are independent.
  3. c) If $P(A) = 0,6$ and $P(B) = 0,5$, then the events $A$ and $B$ cannot be mutually exclusive.
Solve
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In a high school, $65\%$ of the students play soccer ($F$) and $50\%$ play basketball ($B$). Knowing that the probability that a student plays neither of the two sports…

In a high school, $65\%$ of the students play soccer ($F$) and $50\%$ play basketball ($B$). Knowing that the probability that a student plays neither of the two sports is $P(\bar{F} \cap \bar{B}) = 0,15$:

  1. a) Determine whether the events $F$ and $B$ are independent.
  2. b) If the chess club ($A$) is an event such that $A \subset F$ and $P(A) = 0,2$, calculate the probability that a student chosen at random plays soccer but not chess.
  3. c) Calculate $P(\bar{F} \cup \bar{B})$ and interpret its meaning in the context of the problem.
Solve

Total probability and Bayes' Theorem

Sometimes an event can happen through several different routes. If $B_1, B_2, B_3$ are mutually exclusive and cover all of $S$ (a partition), you add the probabilities of the routes. This is the Law of Total Probability:

A rectangle split into bus, car, and walk groups, each with a shaded bottom part showing the late rate. Bus 50% Car 30% Walk 20% late 6% late 10% late 5% Shaded = late. Height shows the late rate in each group.
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In a youth health center, $30\ \%$ of the patients attend for nutritional reasons ($N$), $50\ \%$ for mental health ($S$) and $20\ \%$ for routine check-ups ($R$). It is…

In a youth health center, $30\ \%$ of the patients attend for nutritional reasons ($N$), $50\ \%$ for mental health ($S$) and $20\ \%$ for routine check-ups ($R$). It is known that $10\ \%$ of those attending for nutrition, $20\ \%$ for mental health and $5\ \%$ for check-ups are over 18 years old ($M$).

a) Calculate the probability that a patient chosen at random is over 18 years old.

b) If we choose a patient and they turn out to be over 18 years old, what is the probability that they attended for mental health reasons?

Solve

"At least one" and unequal outcomes

"At least one" covers many cases. It's much easier to use the complement:

Drag the slider to see how fast $1 - (1-p)^n$ climbs as the number of trials $n$ grows.

If you know $P(\text{at least one})$ and want $p$, isolate $(1-p)^n$ and take the $n$th root of both sides.

Outcomes that aren't equally likely. For a biased coin or die, you can't count outcomes. Instead, write each outcome's probability using one unknown, then use "all probabilities add to $1$". To find the probability of an event, add up the probabilities of its outcomes. For repeated rolls, list the pairs systematically and multiply the probabilities within each pair.

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A computer security system has three independent protection layers ($C_1, C_2, C_3$). The probability that an attacker manages to breach the first layer is $0,15$; the…

A computer security system has three independent protection layers ($C_1, C_2, C_3$). The probability that an attacker manages to breach the first layer is $0,15$; the probability of breaching the second is $0,10$, and that of the third is $0,05$. The system is only considered compromised if the attacker manages to breach all three layers.

  1. a) What is the probability that the system is compromised?
  2. b) What is the probability that the attacker breaches at least one protection layer?
  3. c) What is the probability that the attacker breaches exactly two of the three layers?
C₁ C₂ C₃ Breaches 0,15 0,10 0,05 0,85 Blocked 0,90 0,95 Attacker SYSTEM
Solve
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A six-sided die is rolled which has been biased in such a way that the even faces all have the same probability of appearing, and the odd faces also have the same…

A six-sided die is rolled which has been biased in such a way that the even faces all have the same probability of appearing, and the odd faces also have the same probability among themselves. However, it is known that the probability of obtaining an even number is three times that of obtaining an odd number.

  1. a) Determine the probability of each of the six faces of the die.
  2. b) If the die is rolled twice, what is the probability that the sum of the scores obtained is a prime number?
  3. c) Calculate the probability of obtaining at least one number greater than 4 in three consecutive rolls.
Solve
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In a biological laboratory, DNA samples are being analyzed. It has been observed that in an experiment with 5 independent samples, the probability that at least one…

In a biological laboratory, DNA samples are being analyzed. It has been observed that in an experiment with 5 independent samples, the probability that at least one sample presents a specific mutation is $0,67232$.

  1. a) Calculate the probability that an individual sample chosen at random presents this mutation (assuming it is the same for all samples).
  2. b) If 3 samples are analyzed, what is the probability that exactly two present the mutation?
  3. c) How many samples would have to be analyzed, as a minimum, for the probability of finding at least one mutation to be greater than $0,95$?
Solve

Summary

See all the exercises in this section