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BAC1 ss · Mathematics
Probability · Probability

Dependent and independent events

Whether it rains on Saturday and whether you pass Monday's exam have nothing to do with each other: knowing one does not change the probability of the other. But if the first card drawn from a pack is a king, that does change the probability that the second one is also a king. This section makes precise when two events are independent (they do not influence each other) and when they are dependent (they do), and how this simplifies — or not — the calculation of $P(A\cap B)$.

Before you start, make sure you know: Compound events (4th ESO)
Before you start, make sure you know: Dependence (4th ESO)

Independence of events

Visual diagram comparing independent events, where the condition does not change the probability, with dependent events, where it does. independent P(A|B) = P(A) knowing B changes nothing dependent P(A|B) ≠ P(A) knowing B does changethe probability
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A 1st Baccalaureate student has two exams on the same day: Mathematics and English. The probability of passing Mathematics is $P(M) = 0,7$ and that of passing English is…

A 1st Baccalaureate student has two exams on the same day: Mathematics and English. The probability of passing Mathematics is $P(M) = 0,7$ and that of passing English is $P(I) = 0,8$. Assuming that passing one exam does not affect the result of the other:

  1. a) Calculate the probability of passing both exams.
  2. b) Calculate the probability of passing neither exam.
Solve

The product rule for independent events

When $A$ and $B$ are independent, the general product rule $P(A\cap B)=P(B)\cdot P(A|B)$ simplifies, because $P(A|B)=P(A)$.

Diagram illustrating the product rule for two independent events, showing that the probability of their intersection is found by multiplying the probabilities of each event. P(M) = 0,7 · P(I) = 0,8 → independent P(M ∩ I) = 0,7 · 0,8 = 0,56
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Let $A$ and $B$ be two events associated with a random experiment. We know that the probability of event $A$ occurring is $P(A) = 0,6$ and the probability of event $B$…

Let $A$ and $B$ be two events associated with a random experiment. We know that the probability of event $A$ occurring is $P(A) = 0,6$ and the probability of event $B$ occurring is $P(B) = 0,3$. If the events are known to be independent, calculate the probability of both occurring at the same time, that is, $P(A \cap B)$.

Solve
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An urn contains 5 yellow balls and 3 purple balls. We draw two balls from the urn consecutively and without replacement. a) Draw the tree diagram corresponding to this…

An urn contains 5 yellow balls and 3 purple balls. We draw two balls from the urn consecutively and without replacement.

  1. a) Draw the tree diagram corresponding to this experiment, indicating the probability of each branch.
  1. b) Calculate the probability that the two drawn balls are the same colour.
URN 5 3 Total: 8 DRAW Without replacement 1ª 2ª

You can do it on paper and send a photo with your answer.

Solve

With replacement and independence; without replacement and dependence

Successive draws from an urn are the clearest example of when there is independence and when there is not: it depends on whether what was drawn is put back.

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In a gumball machine there are balls of three colours: 10 red, 6 blue and 4 green. A child draws a gumball, notes its colour, puts it back into the machine and draws a…

In a gumball machine there are balls of three colours: 10 red, 6 blue and 4 green. A child draws a gumball, notes its colour, puts it back into the machine and draws a second gumball.

  1. a) Justify whether the events "drawing a red ball on the first" and "drawing a blue ball on the second" are dependent or independent.
  2. b) Calculate the probability that the first gumball is blue and the second is green.
Returned (Replacement) 10 Rojas 6 Azules 4 Verdes Total: 20 Draws 1ª 2ª
Solve
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We draw two cards from a Spanish 40-card deck, one after the other and without returning the first to the deck. a) What is the probability that the first card is a King?…

We draw two cards from a Spanish 40-card deck, one after the other and without returning the first to the deck.

  1. a) What is the probability that the first card is a King?
  2. b) Given that the first card was a King, what is the probability that the second is also a King?
  3. c) Are these events dependent or independent?
Solve

Independence of several events

With three or more events, it is not enough to check independence in pairs: the product rule must also work for all of them together.

Diagram of the product rule for three independent coins, multiplying three factors of one half to obtain one eighth. heads1/2 × heads1/2 × heads1/2 3 coins tossed: P(heads, heads, heads) = 1/8 because they are mutually independent
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Two electronic components of a drone, $A$ and $B$, function independently. The probability of component $A$ failing is $0,05$ and the probability of component $B$…

Two electronic components of a drone, $A$ and $B$, function independently. The probability of component $A$ failing is $0,05$ and the probability of component $B$ failing is $0,08$.

  1. a) Calculate the probability that the drone can keep flying if at least one of the two components must function correctly for this to happen.
  1. b) Calculate the probability that exactly one of the two components fails.
Solve
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In a popular strategy video game, players open "legendary chests" to obtain rewards. The probability of getting an S-level hero from the first chest is only $0,05$.…

In a popular strategy video game, players open "legendary chests" to obtain rewards. The probability of getting an S-level hero from the first chest is only $0,05$. However, the game has an "accumulated luck" system: if you do not get the hero from a chest, the probability of getting it from the next one increases by $0,02$ compared with the probability for the previous chest. If you get the hero, the probability goes back to $0,05$ for the next attempt.

The figure shows the situation you must study: the first two chests do not give the hero and the hero appears in the third.

Three chests in a row: the first two fail and the third gives the S hero; after each failure the probability increases by 0,02 Accumulated luck system P(S) = 0,05 P(S) = ? P(S) = ? Chest 1 no hero Chest 2 no hero Chest 3 S hero +0,02 +0,02 First S-level hero in the third chest

Calculate the probability that a player needs to open exactly three chests to get their first S-level hero.

Solve

Summary

See all the exercises in this section