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BAC2 ss · Mathematics
Chance and probability · Probability

Dependent and independent events

Toss a coin and then roll a die: what comes up on the coin changes nothing about what can come up on the die. But draw a card and, without putting it back, draw another: now the first draw does matter, because the pack is already different. The difference between these two cases is independence, and deciding correctly which is which is the basis of almost all calculations with compound probability.

Before you start, make sure you know: Conditional probability (1st Bachillerato)
Before you start, make sure you know: Events (1st Bachillerato)

What it means for two events to be independent

Tree diagram showing that the die keeps a probability of one sixth for each face regardless of whether the coin lands heads or tails. coin: Hcoin: T die: 1/6 each face, always the same die: 1/6 each 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

Independence in draws: with and without replacement

Tree diagram of consecutive draws from an urn without replacement showing how the probabilities in the second draw vary depending on the previous result. 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

Checking whether two events are independent

Decision diagram comparing the product P(A) times P(B) with the intersection P(A and B) to decide whether the events are independent or dependent. P(A)·P(B) =? P(A∩B) they match →independent they do not match →dependent
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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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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

Tree diagram with conditional probabilities comparing the probability of an event depending on whether the previous event occurs or not, to check dependence. runningtechnical T-shirt they overlap differently in each group → dependent
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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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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

Summary

See all the exercises in this section