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Prep 3 SS · Mathematics
Chance and probability · Probability

Dependent and Independent Events

Before you start, make sure you know: Conditional probability (High School Year 2 (semesters 3 and 4))
Before you start, make sure you know: Events (High School Year 2 (semesters 3 and 4))

You flip a coin and then roll a die: what shows up on the coin changes nothing about what the die can show. But if you draw a card from a deck and don't put it back, the second card comes from a different deck. How do we tell one case from the other with numbers? That is the question of this page, and the answer is the foundation of almost every multi-stage probability calculation.

Independent events

Why the formula? Because $P(A \cap B) = P(A)\,P(B|A)$ always holds; if $B$ does not depend on $A$, then $P(B|A) = P(B)$ and the product is what remains.

Tree diagram: the die keeps a probability of one sixth per face no matter whether the coin landed heads or tails. coin: headscoin: tails die: 1/6 per face, always the same die: 1/6 per face, always the same
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In a random experiment, a fair coin (with sides $C$ and $X$) and a six-sided die numbered 1 to 6 are rolled. Consider the following events: $A$: "Getting heads on the…

In a random experiment, a fair coin (with sides $C$ and $X$) and a six-sided die numbered 1 to 6 are rolled. Consider the following events:

  • $A$: "Getting heads on the coin".

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

Reason whether the 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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A basketball player has a probability of $0.8$ of making a free throw. If they shoot two free throws in a row and we assume that success in the second shot is…

A basketball player has a probability of $0.8$ of making a free throw. If they shoot two free throws in a row and we assume that success in the second shot is independent of the outcome 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 shot.
Solve
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In a 2nd Baccalaureate class, $40\%$ of students regularly use the social network Instagram ($I$) and $25\%$ use TikTok ($T$). It is also known that the use of both…

In a 2nd Baccalaureate class, $40\%$ of 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, calculate the probability that:

  1. a) They use both social networks.
  2. b) They use neither of them.
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

Draws: with and without replacement

Comparison of two draws from an urn with 5 orange balls out of 8: with replacement the second probability is 5/8; without replacement, 4/7. with replacement 5/8 → 5/8 independent without replacement 5/8 → 4/7 dependent
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An urn contains 5 orange balls and 3 purple balls. Two balls are drawn one after the other with replacement (the first ball is returned to the urn before the second one…

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

Urn with 5 orange balls and 3 purple balls and an arrow indicating that the drawn ball is returned to the urnWith replacement5 orange3 purple
  1. a) Calculate the probability that both balls drawn are orange.
  1. b) Does the result of the first draw affect the probability of the second? Justify your answer.
Solve
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Using the same urn as in the previous exercise (5 orange and 3 purple balls), two balls are now drawn without replacement. a) Calculate the probability that the second…

Using the same urn as in the previous exercise (5 orange and 3 purple balls), two balls are now drawn without replacement.

  1. a) Calculate the probability that the second ball is orange given that the first was purple.
  2. b) Are these events independent? Justify your answer by comparing this result with the probability of drawing an orange ball on the first draw.
Solve
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In a pencil case there are 10 felt-tip pens: 6 are blue and 4 are red. Two felt-tip pens are drawn successively without replacement. a) What is the probability that both…

In a pencil case there are 10 felt-tip pens: 6 are blue and 4 are red. Two felt-tip pens are drawn successively without replacement.

  1. a) What is the probability that both are blue?
  2. b) What is the probability that the second is red given that the first was blue?
  3. c) Are the events "the first felt-tip pen is blue" and "the second felt-tip pen 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

Checking whether two events are independent

Decision diagram: P(A)·P(B) is compared with P(A∩B); if they match the events are independent, and if not, dependent. P(A)·P(B) =? P(A∩B) they match →independent they don't match →dependent

Another way to check is with conditional probabilities: if $P(B|A) = P(B|A^c)$, the outcome of $A$ does not change $B$. If they are different, there is dependence.

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Let $A$ and $B$ be two events in a sample space such that their probabilities are known: $P(A) = 0.6$ 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 their probabilities are known: $P(A) = 0.6$ 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
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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 the events $X$ and…

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 the events $X$ and $Y$ are independent. To do so, check whether the necessary condition relating the probability of the intersection to the product of the individual probabilities is satisfied.

Solve
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In a study on streaming content consumption among young people, it was observed that the probability that a student is subscribed to platform A is $P(A) = 0.6$, whereas…

In a study on streaming content consumption among young people, it was observed that the probability that a student is subscribed to platform A is $P(A) = 0.6$, whereas the probability that they are subscribed to platform B is $P(B) = 0.5$. Furthermore, it is known that the probability of not being subscribed to either platform 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 $P(A) \times P(B)$.
S 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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Consider two events $A$ and $B$ of a random experiment with the following known probabilities: $P(A) = 0.5$, $P(B | A) = 0.4$ and $P(B | \bar{A}) = 0.4$. a) Calculate…

Consider two events $A$ and $B$ of a random experiment with the following known probabilities: $P(A) = 0.5$, $P(B | A) = 0.4$ 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) Show whether the events $A$ and $B$ are independent. What conclusion can you draw about the probability of $B$ when it does not change whether or not we know that $A$ has occurred?
Solve

Dependent events

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Two cards are drawn successively and without replacement from a Spanish deck of 40 cards (there are 4 Kings). Let the events be: $A$: "The first card is a King". $B$:…

Two cards are drawn successively and without replacement from a Spanish deck of 40 cards (there are 4 Kings). Let the events be:

  • $A$: "The first card is a King".

  • $B$: "The second card is a King".

The tree diagram shows the situation. The probabilities of the first draw are already written; the ones for the second draw, marked with «?», you have to determine yourself.

Tree diagram of two draws without replacement: King or Not King on the first card, with probabilities 4/40 and 36/40, and King or Not King on the second card with probabilities to be determinedStartKing (A)Not KingKing (B)Not KingKing (B)Not King4/4036/40????

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

Solve
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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 a performance 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 a performance T-shirt. On the other hand, among those who do not buy running shoes, only $15\%$ purchase the performance T-shirt.

  1. a) If we choose a customer at random, what is the probability that they buy the performance T-shirt?
  2. b) Are the events "buying running shoes" and "buying a T-shirt" independent? Reason your answer by analyzing whether buying running shoes changes the probability of buying the T-shirt.
Solve
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A digital content creator has two channels: one for video games ($V$) and one for cooking ($C$). $60\%$ of their subscribers follow channel $V$, $40\%$ follow channel…

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

  1. a) If we randomly select a subscriber who is known to follow the cooking channel, what is the probability that they also follow the video games channel?
  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

Multi-step problems

When there are dependent stages, the tree diagram helps: you multiply along each branch and add across the branches that lead to the requested event. Tables and Venn diagrams are also useful.

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In a school, $45\%$ of the 2nd-year Bachillerato students are in the Social Sciences track ($CS$), $35\%$ in Science and Technology ($CT$), and the rest in Arts ($A$).…

In a school, $45\%$ of the 2nd-year Bachillerato 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 that represents this situation, indicating the probability on each branch.
  1. b) Is liking statistics an event independent of the Bachillerato track studied? Prove it mathematically for the case of the Social Sciences track.
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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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 are shown 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 are shown 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) Given 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 reasons whether the events "being under 25 years old" and "driving a moped" are independent.
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
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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 playoff ("repesca") opportunity which they pass with probability $0.3$. If they pass the playoff, 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 ultimately passes the tournament (passes the second stage).
  2. b) If the player has passed the tournament, what is the probability that they did so without needing to go through the playoff?
Solve

Back to the beginning: the coin and the die are independent because $P(A\cap B)=P(A)P(B)$; the cards that are not put back are dependent because the second probability changes with the first.

Summary

See all the exercises in this section