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Chance and Probability · Probability

Independent and Dependent Events

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

Flip a coin, then roll a die: what the coin shows has no effect on the die. Now deal two cards from a standard deck. If the first card is an ace, is the second card just as likely to be an ace as it was before? Telling these two situations apart is the key to almost every compound probability calculation in AP Statistics. We will answer the card question at the end of the page.

What it means for events to be independent

Why does the product rule work? In general, $P(A \cap B) = P(A)\cdot P(B \mid A)$. If $A$ tells us nothing about $B$, then $P(B \mid A) = P(B)$, and the formula becomes the product.

Tree diagram showing that the die keeps probability one sixth for each face whether the coin lands heads or tails. coin: Hcoin: T die: 1/6 for each face die: 1/6 for each face (same die probabilities either way)
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In a random experiment, a fair coin (with faces $C$ and $X$) and a six-sided die numbered from 1 to 6 are flipped and rolled. The following events are considered: $A$:…

In a random experiment, a fair coin (with faces $C$ and $X$) and a six-sided die numbered from 1 to 6 are flipped and rolled. The following events are considered:

  • $A$: "Obtaining heads on the coin".

  • $B$: "Obtaining an even number on the die".

Reason whether events $A$ and $B$ are dependent or independent and calculate the probability that both occur at the same time, that is, $P(A \cap B)$.

∩ C Event A A = {C} P(A) = 1/2 Event B B = {2, 4, 6} P(B) = 3/6
Solve
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Let $A$ and $B$ be two events in a sample space such that we know their probabilities: $P(A) = 0,6$ and $P(B) = 0,4$. If we know that the events are independent,…

Let $A$ and $B$ be two events in a sample space such that we know their probabilities: $P(A) = 0,6$ and $P(B) = 0,4$. If we know that the events are independent, calculate:

  1. a) $P(A \cap B)$
  2. b) $P(A \cup B)$
  3. c) $P(A|B)$
Solve

The multiplication rule and "at least one"

Some problems write the complement of $A$ as $\bar{A}$. That is the same as $A^c$ ("not $A$").

The multiplication rule works for any events, but you must use $P(B \mid A)$ when the events are not independent. A sentence like "of those who..., 75% also..." is a conditional probability.

The picture below shows two independent free throws, each made with probability $0.7$. Each region's area is the product of the two probabilities.

Area model of two independent free throws, each made with probability 0.7, with regions of area 0.49, 0.21, 0.21 and 0.09. 0.49 0.21 0.21 0.09 1st: made 1st: miss 2nd: made 2nd: miss
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In a high school senior class, $40\%$ of the students regularly use the social network Instagram ($I$) and $25\%$ use TikTok ($T$). It is also known that the use of both…

In a high school senior class, $40\%$ of the students regularly use the social network Instagram ($I$) and $25\%$ use TikTok ($T$). It is also known that the use of both networks is independent of each other. If we choose a student at random, find the probability that they:

  1. a) Use both social networks.
  2. b) Use neither of the two.
Class (100%) Instagram P(I)=0.40 TikTok P(T)=0.25 a) Both (I ∩ T) b) Neither Independent events: P(I ∩ T) = P(I) · P(T)
Solve
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A basketball player has a probability of making a free throw of $0,8$. If they shoot two consecutive free throws and we assume that success in the second shot is…

A basketball player has a probability of making a free throw of $0,8$. If they shoot two consecutive free throws and we assume that success in the second shot is independent of the result of the first, calculate:

  1. a) The probability that they make both shots.
  2. b) The probability that they miss both shots.
  3. c) The probability that they make at least one.
Solve
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On a streaming* platform, the algorithm recommends a series based on two criteria: genre (event $G$) and popularity (event $P$). It is known that the probability that a…

On a streaming platform, the algorithm recommends a series based on two criteria: genre (event $G$) and popularity (event $P$). It is known that the probability that a user likes a series is $0,6$. The probability that it is recommended for its genre is $0,45$. Given that a user likes a series, the probability that the algorithm recommended it for its genre is $0,8$. Furthermore, the events "liking the series" and "being recommended for its popularity" are independent.

Calculate, justifying your steps:

  1. a) The probability that a series is recommended for its genre and the user also likes it.
  2. b) The probability that a user likes a series or it has been recommended for its popularity.
  3. c) Are the events "being recommended for its genre" and "liking the series" independent? Justify the answer with calculations.
E (Sample space) G L G∩L G: Recomendada por Género L: Gustar la serie P: Recomendada por Popularidad Data P(G)=0.45 P(L)=0.60 P(G|L)=0.8 L and P are independent
Solve

Drawing with and without replacement

Comparison of two draws with and without replacement from a jar of 7 green and 3 yellow jelly beans. with replacement 7/10 → 7/10 independent without replacement 7/10 → 6/9 dependent

Cards work the same way. Draw two cards from a standard $52$-card deck without replacement. After one king is removed, $3$ kings remain among $51$ cards, so $P(\text{king, then king}) = \dfrac{4}{52}\cdot\dfrac{3}{51} = \dfrac{1}{221}$. If a problem uses a deck of a different size, the method does not change: take one king and one card away from the totals for the second draw.

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An urn contains 5 orange balls and 3 purple balls. Two balls are drawn consecutively with replacement (the first ball is returned to the urn before drawing the second).…

An urn contains 5 orange balls and 3 purple balls. Two balls are drawn consecutively with replacement (the first ball is returned to the urn before drawing the second).

Calculate the probability that the two drawn balls are orange. Does the result of the first draw influence the probability of the second?

WITH REPLACEMENT N₁ V₁ N₂ V₂ N₂ V₂ 5/8 3/8 5/8 3/8 5/8 3/8 P(N₁ ∩ N₂) = 5/8 × 5/8 = 25/64 The 1st draw does not influence the 2nd (Independent)
Solve
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Two cards are drawn successively and without replacement from a 40-card Spanish deck. Let the events be: $A$: "The first card is a King". $B$: "The second card is a…

Two cards are drawn successively and without replacement from a 40-card Spanish deck. Let the events be:

  • $A$: "The first card is a King".

  • $B$: "The second card is a King".

Calculate the probability of $B$ conditional on $A$, that is, $P(B|A)$. Is this probability equal to the probability of obtaining a King in the first draw? Justify whether the events are dependent or independent.

4/40 36/40 P(B|A) 3/39 36/39 Start King (A) Not King King (B) Not King 40 cards 4 Kings 39 cards 3 Kings
Solve
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In a pencil case there are 10 markers: 6 are blue and 4 are red. Two markers are drawn successively and without replacement. a) What is the probability that both are…

In a pencil case there are 10 markers: 6 are blue and 4 are red. Two markers are drawn successively and without replacement.

  1. a) What is the probability that both are blue?
  2. b) What is the probability that the second is red if the first was blue?
  3. c) Are the events "the first marker is blue" and "the second marker is blue" independent? Justify your answer by comparing the conditional probabilities.
1st Draw: Blue WITHOUT REPLACEMENT 5 Azules 4 Rojos Estado para la 2ª extracción: 9 rotuladores en total
Solve

Testing whether two events are independent

To decide, you must compute and compare. Whether the events "sound" unrelated does not prove anything. Use whichever check fits the information you are given:

Check Independent if…
Product test $P(A \cap B) = P(A)\cdot P(B)$
Conditional on $B$ $P(A \mid B) = P(A)$
Conditional on $A$ $P(B \mid A) = P(B)$

If any one of these holds, all of them hold. If any one fails, the events are dependent.

Decision diagram comparing P(A) times P(B) with P(A and B) to decide whether events are independent or dependent. P(A)·P(B) =? P(A∩B) equal →independent not equal →dependent

A two-way table gives the same information. Here are $100$ students surveyed about having a job and playing a sport:

Plays a sport No sport Total
Has a job 18 42 60
No job 12 28 40
Total 30 70 100
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The following information is available about two events $X$ and $Y$: $P(X) = 0,5$, $P(Y) = 0,3$ and $P(X \cap Y) = 0,15$. Prove mathematically whether events $X$ and $Y$…

The following information is available about two events $X$ and $Y$:

$P(X) = 0,5$, $P(Y) = 0,3$ and $P(X \cap Y) = 0,15$.

Prove mathematically whether events $X$ and $Y$ are independent. To do this, check if the necessary condition relating the probability of the intersection to the product of the individual probabilities is met.

Solve
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A digital content creator has two channels: a gaming one ($V$) and a cooking one ($C$). $60\%$ of their subscribers follow channel $V$, $40\%$ follow channel $C$, and…

A digital content creator has two channels: a gaming one ($V$) and a cooking one ($C$). $60\%$ of their subscribers follow channel $V$, $40\%$ follow channel $C$, and $20\%$ follow both channels.

  1. a) If we select a subscriber at random who we know follows the cooking channel, what is the probability that they also follow the gaming one?
  2. b) In light of the data, would you say that following one channel influences the probability of following the other? Justify whether the events are dependent or independent.
Solve
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Traffic accidents in a city have been classified according to the type of vehicle and whether the driver was under 25 years old. The data is collected in the following…

Traffic accidents in a city have been classified according to the type of vehicle and whether the driver was under 25 years old. The data is collected in the following table:

Car ($T$) Moped ($M$) TOTAL
Under 25 years old ($J$) $12$ $18$ $30$
Over 25 years old ($V$) $48$ $22$ $70$
TOTAL $60$ $40$ $100$
  1. a) If we choose an accident at random, what is the probability that the driver is under 25 years old?
  2. b) If we know that the accident involved a moped, what is the probability that the driver is under 25 years old?
  3. c) Using the previous results, explain with reasoning whether the events "being under 25 years old" and "driving a moped" are independent.
Solve

Mutually exclusive is not independent

Two separate circles A and B that do not overlap, showing mutually exclusive events. A B No overlap: if A happens, B cannot

These are nearly opposite ideas. If $P(A) > 0$ and $P(B) > 0$ and the events are mutually exclusive, then $P(A \cap B) = 0$ but $P(A)\cdot P(B) > 0$. The product test fails, so they are dependent. Intuitively, once you know $A$ happened, you know $B$ cannot: $P(B \mid A) = 0 \neq P(B)$.

This changes the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:

  • mutually exclusive: the last term is $0$, so $P(A \cup B) = P(A) + P(B)$;

  • independent: the last term is $P(A)\cdot P(B)$.

Working backward from partial information

Many harder problems give you a mix of unions, complements, and conditional probabilities. Write each fact in terms of $P(A)$, $P(B)$, and $P(A \cap B)$, then solve.

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Let $A$ and $B$ be two events such that $P(A) = 0,4$ and $P(B) = 0,3$. a) If $A$ and $B$ are independent events, calculate $P(A \cup B)$ and $P(A | B)$. b) If $A$ and…

Let $A$ and $B$ be two events such that $P(A) = 0,4$ and $P(B) = 0,3$.

  1. a) If $A$ and $B$ are independent events, calculate $P(A \cup B)$ and $P(A | B)$.
  2. b) If $A$ and $B$ are dependent events and we know that $P(A \cap B) = 0,2$, calculate $P(B | A)$.
  3. c) In case b), are $A$ and $B$ mutually exclusive events? Justify your answer.
Solve
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In a study on the consumption of streaming content among young people, it has been observed that the probability that a student is subscribed to platform A is…

In a study on the consumption of streaming content among young people, it has been observed that the probability that a student is subscribed to platform A is $P(A) = 0,6$, while the probability that they are subscribed to platform B is $P(B) = 0,5$. Furthermore, it is known that the probability that they are not subscribed to either of the two is $P(\bar{A} \cap \bar{B}) = 0,2$.

  1. a) Calculate the probability that a student is subscribed to both platforms.
  2. b) Determine whether the events "being subscribed to platform A" and "being subscribed to platform B" are independent events. Justify your answer by comparing $P(A \cap B)$ with $P(A) \cdot P(B)$.
E Platform A P(A) = 0.6 Platform B P(B) = 0.5 P(A∩B) ? Neither platform P(A̅ ∩ B̅) = 0.2
Solve
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Let $A$ and $B$ be two events of a random experiment such that $P(A^c \cap B^c) = 0,2$, $P(A | B) = 0,4$ and $P(B | A) = 0,5$. a) Calculate the probabilities of events…

Let $A$ and $B$ be two events of a random experiment such that $P(A^c \cap B^c) = 0,2$, $P(A | B) = 0,4$ and $P(B | A) = 0,5$.

  1. a) Calculate the probabilities of events $A$ and $B$. (Hint: use the definition of conditional probability to set up a system of equations).
  2. b) Determine whether events $A$ and $B$ are independent. Justify your answer.
  3. c) Calculate $P(A^c \cup B)$.
Ω A B P(A∩B) = x P(Aᶜ ∩ Bᶜ) = 0.2 LAWS AND DEFINITIONS P(Aᶜ ∩ Bᶜ) = P((A ∪ B)ᶜ) 0.2 = 1 - P(A ∪ B) P(A|B) = x / P(B) = 0.4 P(B|A) = x / P(A) = 0.5
Solve

Dependent events and tree diagrams

When the second stage depends on the first, a tree diagram organizes the calculation. Multiply along a branch; add across the branches that lead to the same event.

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A sportswear brand knows that $70\%$ of its customers buy running shoes. Of those who buy running shoes, $40\%$ also purchase an athletic t-shirt. On the other hand,…

A sportswear brand knows that $70\%$ of its customers buy running shoes. Of those who buy running shoes, $40\%$ also purchase an athletic t-shirt. On the other hand, among those who do not buy running shoes, only $15\%$ purchase the athletic t-shirt.

  1. a) If we choose a customer at random, what is the probability that they buy the athletic t-shirt?
  2. b) Are the events "buying running shoes" and "buying a t-shirt" independent? Reason your answer by analyzing whether the fact of buying running shoes modifies the probability of buying the t-shirt.
Solve
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In a high school, $45\%$ of the senior class students are in the Social Sciences track ($CS$), $35\%$ in Science and Technology ($CT$), and the rest in Arts ($A$). It is…

In a high school, $45\%$ of the senior class students are in the Social Sciences track ($CS$), $35\%$ in Science and Technology ($CT$), and the rest in Arts ($A$). It is known that the probability that a student likes statistics is $0,8$ if they are in $CS$, $0,6$ if they are in $CT$, and $0,3$ if they are in $A$.

  1. a) Draw the tree diagram representing this situation.
  2. b) Is liking statistics an event independent of the high school track studied? Prove it mathematically for the case of the Social Sciences track.
0,45 0,35 0,20 CS CT Arts 0,8 0,2 0,6 0,4 0,3 0,7 Stat. No Stat. Stat. No Stat. Stat. No Stat.
Solve
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Consider two events $A$ and $B$ of a random experiment for which the following probabilities are known: $P(A) = 0,5$, $P(B | A) = 0,4$ and $P(B | \bar{A}) = 0,4$. a)…

Consider two events $A$ and $B$ of a random experiment for which the following probabilities are known: $P(A) = 0,5$, $P(B | A) = 0,4$ and $P(B | \bar{A}) = 0,4$.

  1. a) Calculate $P(A \cap B)$ and $P(\bar{A} \cap B)$.
  2. b) From the previous results, calculate $P(B)$.
  3. c) Prove whether events $A$ and $B$ are independent. What conclusion can you draw about the probability of $B$ when it does not vary even if we know that $A$ has occurred or has not occurred?
Solve

Reversing the condition: Bayes' rule

Trees tell you $P(\text{result} \mid \text{cause})$. Often you want the reverse: seeing a result, how likely is each cause?

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A diagnostic test for a rare disease (which affects $0,5\%$ of the population) yields $2\%$ of false positives (healthy people who test positive) and $1\%$ of false…

A diagnostic test for a rare disease (which affects $0,5\%$ of the population) yields $2\%$ of false positives (healthy people who test positive) and $1\%$ of false negatives (sick people who test negative).

  1. a) If a person takes the test and tests positive, and immediately afterward takes a second identical and independent test which also tests positive, what is the probability that they are actually sick?
  2. b) Comment on the result obtained by comparing it with the probability after a single positive test.
Solve
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A courier company uses three routes for its deliveries: North, Central, and South. $20\%$ of the packages go via the North route, $50\%$ via the Central, and $30\%$ via…

A courier company uses three routes for its deliveries: North, Central, and South. $20\%$ of the packages go via the North route, $50\%$ via the Central, and $30\%$ via the South. Delays depend on the route: on the North there are $10\%$ delays, on the Central $5\%$, and on the South $x\%$. Given that the probability that a package chosen at random arrives on time is $0,923$:

  1. a) Determine the value of $x$ (percentage of delays on the South route).
  2. b) If a package has arrived late, which route is most likely to have been the cause?
North Central South 0.2 0.5 0.3 0.10 0.90 0.05 0.95 x 1 - x Delay (R) On time (T) Delay (R) On time (T) Delay (R) On time (T) Total probability: P(T) = 0.923
Solve
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In a sociological study on the use of renewable energies, individuals are classified as "proactive" ($A$) or "passive" ($B$). It is known that $30\%$ of the population…

In a sociological study on the use of renewable energies, individuals are classified as "proactive" ($A$) or "passive" ($B$). It is known that $30\%$ of the population is proactive. The probability that a proactive individual installs solar panels is $0,8$, while for a passive individual this probability is reduced to $0,2$. Two people are chosen at random independently.

  1. a) Calculate the probability that at least one of the two people installs solar panels.
  2. b) Given that only one of the two people has installed solar panels, what is the probability that this person belongs to the proactive group?
Solve

Chains of dependent stages

Some experiments have several stages where each one changes the next: moving an object between containers, a system of nodes where one failure affects another, a game with a second chance. The tools are the same:

  • Chain rule: $P(A \cap B \cap C) = P(A)\cdot P(B \mid A)\cdot P(C \mid A \cap B)$.

  • Probabilities leaving the same node must add to $1$.

  • Multiply along a path, then add the paths that give the same event.

  • To reverse the condition, divide one path by the sum of all paths that give the event.

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We have two urns with the following compositions: urn $U_1$ contains 4 red balls and 6 black balls; urn $U_2$ contains 3 red balls and 2 black balls. The experiment…

We have two urns with the following compositions: urn $U_1$ contains 4 red balls and 6 black balls; urn $U_2$ contains 3 red balls and 2 black balls. The experiment consists of drawing a ball at random from $U_1$ and placing it into $U_2$, without looking at its color. Next, two balls are drawn simultaneously from $U_2$.

  1. a) Calculate the probability that the two balls drawn from $U_2$ are of different colors.
  2. b) If the two balls drawn from $U_2$ turned out to be black, what is the probability that the ball transferred from $U_1$ was red?
U₁ 4 Red 6 Black U₂ 3 Red 2 Black 1 ball ? 2 balls ? ?
Solve
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An industrial cybersecurity system uses three verification nodes ($N_1, N_2, N_3$) arranged in series, so that the system only allows access if all three nodes validate…

An industrial cybersecurity system uses three verification nodes ($N_1, N_2, N_3$) arranged in series, so that the system only allows access if all three nodes validate the entry. The probability that node $N_1$ fails is $0,02$. Node $N_2$ is more sensitive: if $N_1$ has validated the entry, the probability that $N_2$ fails is $0,05$, but if $N_1$ has failed, node $N_2$ blocks and fails with probability $1$. Finally, node $N_3$ works independently of the rest with a success probability of $0,98$.

  1. a) What is the probability that a legitimate entry is rejected by the system?
  2. b) If the system has rejected an entry, calculate the probability that the failure occurred exclusively in node $N_2$.
Solve
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In a video game tournament, a player must pass two stages. The probability of passing the first stage is $0,7$. If they pass the first, the probability of passing the…

In a video game tournament, a player must pass two stages. The probability of passing the first stage is $0,7$. If they pass the first, the probability of passing the second is $0,5$. If they do not pass the first, they are given a "second chance" opportunity which they pass with probability $0,3$. If they pass the second chance, they enter the second stage, but with a success probability of only $0,2$ due to accumulated fatigue.

  1. a) Calculate the probability that the player finally passes the tournament (passing the second stage).
  2. b) If the player has passed the tournament, what is the probability that they achieved it without needing to go through the second chance?
Solve

Back to the start: with a standard $52$-card deck, $P(\text{ace on the 1st card}) = \dfrac{4}{52} \approx 0.077$. If the first card was an ace and is not replaced, $P(\text{ace on the 2nd} \mid \text{ace on the 1st}) = \dfrac{3}{51} \approx 0.059$. The probability changed, so the draws are dependent. With the coin and the die, nothing changes, so they are independent.

Summary

See all the exercises in this section