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AP Stats · Mathematics
Chance and Probability · Probability

Properties of Probability

Before you start, make sure you know: Events (Algebra 2)

Emma surveys 200 seniors at her Denver high school. She finds that 70% use Instagram, 55% use TikTok, and 40% use both. Her friend says, "70% plus 55% is 125%. That's impossible!" Is something wrong with the data? And what percent of seniors use neither app? By the end of this page you will be able to answer both questions, and much harder ones.

The rules every probability must follow

You can't give an event just any number as its probability. A few basic rules must always hold, and every other formula on this page comes from them.

These three rules are known as Kolmogorov's axioms. Some problems call mutually exclusive events incompatible and events that can overlap compatible. Everything else on this page follows from them. In particular, every probability lies between $0$ and $1$.

Three boxes showing the probability rules: probabilities are never negative, the whole sample space has probability 1, and probabilities add for mutually exclusive events. P(A) ≥ 0 P(S) = 1 A, B mutually exclusive:P(A∪B) = P(A) + P(B)
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A video game developer is analyzing the probability of obtaining rare items in a surprise chest. The possible outcomes are: "Common Item" ($C$), "Rare Item" ($R$) and…

A video game developer is analyzing the probability of obtaining rare items in a surprise chest. The possible outcomes are: "Common Item" ($C$), "Rare Item" ($R$) and "Epic Item" ($E$). Indicate whether the following probability assignment is valid according to Kolmogorov's axioms and justify your answer:

$P(C) = 0,65$

$P(R) = 0,20$

$P(E) = 0,25$

Solve

The complement rule

The complement of an event $A$ is the event "$A$ does not happen." It is written $A^C$ (you will also see $\overline{A}$ or $A'$, and some problems use the bar). Every outcome is either in $A$ or in $A^C$, never both.

Venn diagram: an event A inside the sample space and its complement, the rest of the space. A P(A) = 0.35 Aᶜ P(Aᶜ) = 1 − P(A)
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In a study on the use of mobile apps among teenagers, it has been determined that the probability that a student has a video editing app installed is $P(V) = 0,35$.…

In a study on the use of mobile apps among teenagers, it has been determined that the probability that a student has a video editing app installed is $P(V) = 0,35$. Calculate the probability that a student chosen at random does not have this app installed.

E (Total = 1) V P(V) = 0.35 V' P(V') = ? 1 - P(V)
Solve
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In a box of materials for a technology project, there are 50 resistors. It is known that some are $100\ \Omega$ ($A$) and others are $220\ \Omega$ ($B$), and some are…

In a box of materials for a technology project, there are 50 resistors. It is known that some are $100\ \Omega$ ($A$) and others are $220\ \Omega$ ($B$), and some are new and others used. One is drawn at random. If $P(A) = 0,6$ and the events $A$ and $B$ are complementary, determine:

  1. a) The probability of drawing a $220\ \Omega$ resistor.
  2. b) What is the value of $P(A \cap B)$? Justify your answer based on the relationship between complementary events.
Solve

The addition rule

Rule 3 only works for mutually exclusive events. What if the events can overlap, like "uses Instagram" and "uses TikTok"? If you add $P(A)$ and $P(B)$, the outcomes in both events get counted twice. So you subtract the overlap once.

Venn diagram of two overlapping events A and B, with the overlap A and B marked as the part counted twice. A B A∩B subtract the overlap once: it was counted twice

Splitting an event into pieces

An event $A$ breaks into two mutually exclusive pieces: the part inside $B$ and the part outside $B$. That gives a very useful formula:

$$P(A \cap B^C) = P(A) - P(A \cap B)$$

This is the probability of "$A$ only" (also written $P(A - B)$, "$A$ happens but $B$ doesn't"). In Emma's survey, $P(\text{Instagram only}) = 0.70 - 0.40 = 0.30$.

Back to Emma's hook: 70% + 55% = 125% is not a mistake. The 40% who use both apps were counted twice, so the real total is $125\% - 40\% = 85\%$.

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For the graduation trip, the high school seniors must choose between three possible destinations: Orlando, Cancun, or Hawaii. It is known that the probability that they…

For the graduation trip, the high school seniors must choose between three possible destinations: Orlando, Cancun, or Hawaii. It is known that the probability that they choose Orlando is $P(M) = 0,42$ and the probability that they choose Cancun is $P(L) = 0,38$. Knowing that the events are mutually exclusive (they can only choose one destination), calculate:

  1. a) The probability that the students travel to Orlando or Cancun.
  2. b) The probability that they choose neither of these two destinations and go to Hawaii.
Solve
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Let $X$ and $Y$ be two events such that $P(X) = 0,25$ and $P(Y) = 0,55$. If we know that these events are mutually exclusive, calculate the probability that: a) Event…

Let $X$ and $Y$ be two events such that $P(X) = 0,25$ and $P(Y) = 0,55$. If we know that these events are mutually exclusive, calculate the probability that:

  1. a) Event $X$ or event $Y$ occurs.
  2. b) Event $X$ does not occur.
  3. c) $X$ and $Y$ occur at the same time.
Solve
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According to the statistics of a high school, $75 \%$ of the students regularly use Instagram ($I$), $60 \%$ use TikTok ($T$) and $45 \%$ use both social networks…

According to the statistics of a high school, $75 \%$ of the students regularly use Instagram ($I$), $60 \%$ use TikTok ($T$) and $45 \%$ use both social networks simultaneously. Calculate the probability that a student, selected at random, uses Instagram or TikTok.

Sample Space (100%) Instagram P(I)=75% Both P(I∩T)=45% TikTok P(T)=60% 30% 45% 15% None 10% P(I ∪ T) = 30% + 45% + 15% = 90%
Solve

De Morgan's Laws and "neither"

Many problems give you the probability of "neither event happens" and ask for the union (or the other way around). De Morgan's Laws translate between "not" and "and/or."

Venn diagram of two events A and B; the region outside both circles represents neither A nor B. A B outside both circles = neither: 1 − P(A∪B)
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We consider the events $A$ and $B$ associated with a random experiment. If we know that $P(A) = 0,5$, $P(B) = 0,4$ and that the probability that neither $A$ nor $B$…

We consider the events $A$ and $B$ associated with a random experiment. If we know that $P(A) = 0,5$, $P(B) = 0,4$ and that the probability that neither $A$ nor $B$ occurs is $P(\bar{A} \cap \bar{B}) = 0,3$, apply De Morgan's Laws to find the probability of the union, $P(A \cup B)$.

E A B P(A̅ ∩ B̅) = 0.3 P(A) = 0.5 P(B) = 0.4 P(A ∪ B) = ?
Solve

Venn diagrams with probabilities

Two events split the sample space into four mutually exclusive regions: $A$ only, both, $B$ only, and neither. Their probabilities always add to $1$. If you can fill in all four, every question becomes a matter of adding the right regions.

Venn diagram for tacos T and burritos B with the four region probabilities 0.25, 0.15, 0.30 and 0.30. T B 0.25 0.15 0.30 0.30 neither
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From the data represented in the following Venn diagram, where the values indicate the probabilities of each region, calculate the requested probabilities: a) $P(A)$ b)…

From the data represented in the following Venn diagram, where the values indicate the probabilities of each region, calculate the requested probabilities:

  1. a) $P(A)$
  2. b) $P(B)$
  3. c) $P(A \cup B)$
E A B 0,2 0,1 0,4 0,3
Solve
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Observe the following Venn diagram representing the probabilities of different regions for two events $M$ and $N$ in a sample space $E$. From the data in the diagram,…

Observe the following Venn diagram representing the probabilities of different regions for two events $M$ and $N$ in a sample space $E$.

E M N 0,35 0,15 0,25

From the data in the diagram, calculate the following properties:

  1. a) $P(M \cup N)$
  2. b) $P(M - N)$, that is, the probability that $M$ occurs but not $N$.
  3. c) $P(\bar{M} \cup \bar{N})$ using De Morgan's Laws.
  4. d) Are the events $M$ and $N$ mutually exclusive? Justify your answer.
Solve

Conditional probability

Sometimes you learn something before you compute. "What's the probability a food truck customer orders a burrito, given that they ordered tacos?" Now the sample space shrinks: you only look inside the $T$ circle.

Venn diagram where the circle T is outlined: given T, only the T circle counts, and the part also in B is 0.15 out of 0.40. T 0.25 0.15 0.30 given T, only the T circle counts: 0.15 out of 0.40
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On a streaming content platform, $60 \%$ of young subscribers watch science fiction series (event $S$), $45 \%$ watch nature documentaries (event $D$) and $20 \%$ of…

On a streaming content platform, $60 \%$ of young subscribers watch science fiction series (event $S$), $45 \%$ watch nature documentaries (event $D$) and $20 \%$ of users watch both types of content.

Calculate the probability that a user chosen at random:

  1. a) Watches science fiction series or nature documentaries.
  2. b) Watches neither of the two genres.
  3. c) Watches only one of the two genres.
  4. d) Given that they do not watch science fiction series, what is the probability that they watch nature documentaries?
Solve
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In a study on sustainable mobility among high school students, it was found that the probability that a student uses a bicycle to go to school is $P(B) = 0,25$, the…

In a study on sustainable mobility among high school students, it was found that the probability that a student uses a bicycle to go to school is $P(B) = 0,25$, the probability that they use public transportation is $P(T) = 0,60$ and the probability that they use neither of these two means is $P(\bar{B} \cap \bar{T}) = 0,30$.

  1. a) Are the events "using a bicycle" and "using public transportation" compatible?
  2. b) Calculate the probability that a student uses only a bicycle.
  3. c) What is the probability that they use public transportation if we know that they do not use a bicycle?
Solve
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In a university entrance exam, $75\%$ of students pass Math ($M$) and $60\%$ pass History ($H$). It is known that $10\%$ of students pass neither of the two subjects. a)…

In a university entrance exam, $75\%$ of students pass Math ($M$) and $60\%$ pass History ($H$). It is known that $10\%$ of students pass neither of the two subjects.

  1. a) Calculate the probability that a student passes both subjects.
  1. b) Given that a student has passed History, what is the probability that they have also passed Math?
  1. c) What is the probability that a student passes Math but not History? Explain the relationship between this result and the difference of events $P(M - H)$.
Solve

Independent events

Two events are independent if knowing that one happened does not change the probability of the other. Flip a coin and roll a die: the die doesn't care what the coin did.

Area model for independent events with P(A) = 0.5 and P(B) = 0.4: each cell's area equals its row probability times its column probability. A (0.5) Aᶜ (0.5) B Bᶜ 0.20 0.20 0.30 0.30 independent: each cell = row × column
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Let $A$ and $B$ be two events associated with a random experiment. It is known that the probability that neither of the two occurs is $P(\bar{A} \cap \bar{B}) = 0,2$.…

Let $A$ and $B$ be two events associated with a random experiment. It is known that the probability that neither of the two occurs is $P(\bar{A} \cap \bar{B}) = 0,2$. Furthermore, it is known that $P(A) = 0,5$ and that $P(B) = 0,4$.

Calculate with reasoning the following probabilities:

  1. a) $P(A \cup B)$
  2. b) $P(A \cap B)$
  3. c) $P(\bar{A} | B)$
  4. d) Determine whether the events $A$ and $B$ are independent.
Solve
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Consider two events $X$ and $Y$ such that they are independent. It is known that $P(X) = 0,6$ and $P(X \cup Y) = 0,8$. Using the properties of probability and the…

Consider two events $X$ and $Y$ such that they are independent. It is known that $P(X) = 0,6$ and $P(X \cup Y) = 0,8$. Using the properties of probability and the definition of independence, calculate:

  1. a) The probability of event $Y$.
  2. b) The probability that $X$ occurs but $Y$ does not occur.
  3. c) The probability that neither of the two occurs.
Solve
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A task management app classifies notifications according to their urgency ($U$) and their type: personal ($P$) or work ($T$). It is known that $P(U) = 0,35$,…

A task management app classifies notifications according to their urgency ($U$) and their type: personal ($P$) or work ($T$). It is known that $P(U) = 0,35$, $P(P) = 0,55$ and that $P(U \cup P) = 0,75$.

Determine:

  1. a) Whether the events $U$ and $P$ are independent.
  2. b) The probability that a notification is urgent given that it is personal.
  3. c) The probability that a notification is personal but not urgent.
  4. d) If the notifications are either personal ($P$) or work ($T$), calculate the probability that a notification is not urgent and is for work, that is, $P(\bar{U} \cap T)$.
Solve

Complements, and more than two events

If $A$ and $B$ are independent, so are their complements. It makes sense: if $B$ gives no information about $A$, then "not $B$" shouldn't either. Here is a proof for one of the cases.

Three events

The same rules extend to more events when you can split them into pieces that don't overlap. If $C$ is mutually exclusive with both $A$ and $B$, then $C$ doesn't overlap $A \cup B$ either, so:

$$P(A \cup B \cup C) = P(A \cup B) + P(C)$$
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An eSports team is analyzing its performance on two different maps of a video game: Inferno* (event $I$) and Mirage* (event $M$). The statistics show that they win on…

An eSports team is analyzing its performance on two different maps of a video game: Inferno (event $I$) and Mirage (event $M$). The statistics show that they win on Inferno with a probability of $0,7$ and that the probability of winning on Mirage is $0,5$. It has been verified that the probability of winning on at least one of the two maps is $0,85$.

  1. a) Calculate the probability that they win on both maps.
  2. b) Is winning on Inferno independent of winning on Mirage?
  3. c) Calculate the probability that they win on Mirage given that they have lost on Inferno.
Solve
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Prove formally, using Kolmogorov's axioms and the properties of probability, the following statement: "If $A$ and $B$ are two independent events, then their…

Prove formally, using Kolmogorov's axioms and the properties of probability, the following statement: "If $A$ and $B$ are two independent events, then their complementary events $\bar{A}$ and $\bar{B}$ are also independent".

Hint: You must start from the definition of independence $P(A \cap B) = P(A) \cdot P(B)$ and arrive at $P(\bar{A} \cap \bar{B}) = P(\bar{A}) \cdot P(\bar{B})$.

Solve
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Let $A$, $B$ and $C$ be three events such that $A$ and $B$ are independent of each other, and $C$ is an event mutually exclusive with $A$ and with $B$. The following…

Let $A$, $B$ and $C$ be three events such that $A$ and $B$ are independent of each other, and $C$ is an event mutually exclusive with $A$ and with $B$. The following probabilities are known: $P(A) = 0,4$, $P(B) = 0,3$ and $P(C) = 0,1$. Calculate with reasoning:

  1. a) $P(A \cup B \cup C)$
  1. b) $P((A \cap B) \cup C)$
  1. c) $P(\bar{A} | \bar{B})$
Ω A P=0.4 B P=0.3 C P=0.1 A∩B
Solve

Tree diagrams and total probability

When events happen in stages (first shot, then second shot), a tree diagram keeps things organized. On each branch you write a conditional probability. To find the probability of a path, multiply along it. To find the probability of an event that can happen along several paths, add those paths.

Tree diagram for two free throws: first shot make 0.6 or miss 0.4; after a make the second is a make 0.8 or miss 0.2; after a miss the second is a make 0.5 or miss 0.5. 0.6 0.4 0.8 0.2 0.5 0.5 1st: make 1st: miss 2nd make: 0.48 2nd miss: 0.12 2nd make: 0.20 2nd miss: 0.20
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In an esports tournament, a player has a probability $p$ of winning the first match. If they win the first, their confidence increases and the probability of winning the…

In an esports tournament, a player has a probability $p$ of winning the first match. If they win the first, their confidence increases and the probability of winning the second is $0,7$. If they lose the first, the probability of winning the second drops to $0,3$. Knowing that the probability of winning the second match is $0,54$, calculate:

  1. a) The value of the probability $p$.
  1. b) The probability that the player won the first match given that they lost the second.
p 1-p 0.7 0.3 0.3 0.7 G₁ P₁ G₂ P₂ G₂ P₂ P(G₂) = 0.54
Solve
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A music streaming app classifies its users into three categories according to their listening time: Free* ($F$), Premium* ($P$) and Family* ($S$). $50\%$ are Free*…

A music streaming app classifies its users into three categories according to their listening time: Free ($F$), Premium ($P$) and Family ($S$). $50\%$ are Free users, $30\%$ are Premium and the rest are Family. It is known that the percentage of users who regularly listen to podcasts is $20\%$ in the Free category, $60\%$ in the Premium category and $40\%$ in the Family category.

  1. a) If we choose a user at random who listens to podcasts, what is the probability that they belong to the Premium plan?
  1. b) If a new condition is added: "No Family user who listens to podcasts is under 18 years old". If we choose a user at random, what is the probability that they are a Family user, listen to podcasts and are under 18 years old? Justify your answer using the impossible event.
Solve

Turning word clues into equations

Harder problems hide numbers in sentences. A good routine:

  1. Name the unknown (for example $x = P(B)$, or the parameter $k$).

  2. Translate each clue into an equation using the rules above: complement, addition, De Morgan, independence.

  3. Solve, then check that every probability is between $0$ and $1$ and the four Venn regions add to $1$.

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In a study on social media use among high school seniors, two platforms are analyzed: InstaPlus* (event $A$) and TikToker* (event $B$). It is known that the probability…

In a study on social media use among high school seniors, two platforms are analyzed: InstaPlus (event $A$) and TikToker (event $B$). It is known that the probability that a student uses at least one of the two is $0,8$. Furthermore, the probability that they use the first but not the second is $0,3$. If it is known that the probability that they use both platforms is half the probability that they use neither of the two, calculate:

  1. a) The probability that a student, chosen at random, uses the TikToker platform.
  1. b) Are the events $A$ and $B$ independent? Justify your answer mathematically by comparing $P(A \cap B)$ with the product of their individual probabilities.
  1. c) Calculate the probability that they use only one of the two platforms.
E P(E) = 1 A InstaPlus B TikToker 0.3 x ? 2x None P(A ∪ B) = 0.8
Solve
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Let $A$ and $B$ be two events of a random experiment such that their probabilities are given in terms of a parameter $k$. It is known that $P(A) = k$, $P(B) = 0,5$ and…

Let $A$ and $B$ be two events of a random experiment such that their probabilities are given in terms of a parameter $k$. It is known that $P(A) = k$, $P(B) = 0,5$ and $P(A \cup B) = 0,7$.

  1. a) Determine the value of $k$ so that the events $A$ and $B$ are incompatible.
  1. b) Determine the value of $k$ so that the events $A$ and $B$ are independent.
  1. c) If $k = 0,4$, calculate the probability that $B$ occurs given that $A$ has not occurred, that is, $P(B | \bar{A})$.
Solve
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A cybersecurity system monitors two types of attacks: Phishing* ($A$) and Malware* ($B$). The metrics for the last month indicate that $P(\bar{A} \cap \bar{B}) = 0,15$…

A cybersecurity system monitors two types of attacks: Phishing ($A$) and Malware ($B$). The metrics for the last month indicate that $P(\bar{A} \cap \bar{B}) = 0,15$ and that $P(A \cap B) = 0,25$. If the probability of suffering a Phishing attack is twice that of suffering a Malware attack ($P(A) = 2P(B)$), find:

  1. a) The probability of suffering a Malware attack.
  1. b) The probability that, having suffered a Malware attack, it is not a Phishing attack.
  1. c) Using De Morgan's Laws, show whether the union of the complementary events $P(\bar{A} \cup \bar{B})$ or the complement of the intersection $\overline{P(A \cap B)}$ is more probable.
Solve

Summary

See all the exercises in this section