Properties of probability
Met Éireann's forecast for St Patrick's Day gives a 70% chance of rain in Dublin and a 50% chance of rain in Cork. Could the chance of rain in at least one of the two cities be 120%? Here there are no equally likely outcomes to count: you are given probabilities and must work out others using a few basic rules, usually credited to Kolmogorov. These rules, and what follows from them, are the foundation for conditional probability and the binomial distribution later in the course.
Kolmogorov's axioms
An urn contains numbered balls. Let the events be $A$ = "drawing an even number" and $B$ = "drawing an odd number less than 5". If we know that $P(A) = 0.5$,…
An urn contains numbered balls. Let the events be $A$ = "drawing an even number" and $B$ = "drawing an odd number less than 5". If we know that $P(A) = 0.5$, $P(B) = 0.2$ and that the events are mutually exclusive, calculate the probability of the union $P(A \cup B)$.
In a Leaving Certificate Mathematics exam, two events $A$ and $B$ from the same sample space are presented with probabilities $P(A) = 0.7$ and $P(B) = 0.5$. Cian writes…
In a Leaving Certificate Mathematics exam, two events $A$ and $B$ from the same sample space are presented with probabilities $P(A) = 0.7$ and $P(B) = 0.5$. Cian writes in his working: "Since $P(A \cup B) = P(A) + P(B) = 0.7 + 0.5 = 1.2$, and by Kolmogorov's axioms no probability can exceed $1$, the problem statement contains an error and these two events cannot occur in the same experiment".\na) Identify and explain with reasoning the two conceptual errors made by Cian (one concerning the formula used and the other concerning the compatibility of the events).\nb) Using the properties derived from the axioms, calculate the minimum bound that $P(A \cap B)$ must take for the situation to be mathematically consistent, and determine the interval in which $P(A \cup B)$ can vary.
Two players, Ana and Blas, alternately toss a fair coin. The first player to obtain "Heads" wins. Ana tosses first. Define the infinite sample space of this experiment…
Two players, Ana and Blas, alternately toss a fair coin. The first player to obtain "Heads" wins. Ana tosses first. Define the infinite sample space of this experiment and, using the sum of a geometric progression, calculate the probability that Blas wins the game (that is, that the first "Heads" occurs on an even toss).
Probability of the complement and of the impossible event
The same works for three events: $(A\cup B\cup C)' = A'\cap B'\cap C'$, so $P(A'\cap B'\cap C') = 1 - P(A\cup B\cup C)$.
Events $A_1, A_2, \dots$ form a partition of the sample space if they are mutually exclusive in pairs and their union is the whole sample space, so their probabilities add up to $1$.
A study on digital habits reveals that the probability of a teenager using a microblogging social network is $P(S) = 0.72$. Calculate the probability of the…
A study on digital habits reveals that the probability of a teenager using a microblogging social network is $P(S) = 0.72$.
Calculate the probability of the complementary event, that is, the probability that a teenager chosen at random does not use that social network.
Using De Morgan's Laws, calculate the probability that neither event $A$ nor event $B$ occurs (that is, $P(\bar{A} \cap \bar{B})$), knowing that the probability of the…
Using De Morgan's Laws, calculate the probability that neither event $A$ nor event $B$ occurs (that is, $P(\bar{A} \cap \bar{B})$), knowing that the probability of the union is $P(A \cup B) = 0.84$.
Let $A$, $B$, and $C$ be three events from the same random experiment such that they form a complete system of events. The following relationships between their…
Let $A$, $B$, and $C$ be three events from the same random experiment such that they form a complete system of events. The following relationships between their probabilities are known:
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The probability of $A$ is twice that of $B$.
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The probability of the union of $B$ and $C$ is three times the probability of $B$.
Prove with reasoned arguments, using Kolmogorov's axioms and the properties of probability, whether it is possible that $P(C) = 0.4$. If not, determine the exact value of the probabilities of each event.
The probability of the union
For three events:
$$P(A \cup B \cup C) = P(A)+P(B)+P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)$$Each double intersection is counted twice and must be subtracted; but the central region common to all three was added three times and subtracted three times, so it has to be added back in.
On a streaming platform, $65\ \%$ of young users are subscribed to the Music Plan ($M$) and $40\ \%$ are subscribed to the Video Plan ($V$). It is also known that…
On a streaming platform, $65\ \%$ of young users are subscribed to the Music Plan ($M$) and $40\ \%$ are subscribed to the Video Plan ($V$). It is also known that $25\ \%$ of users enjoy both plans simultaneously.
Calculate the probability that a user chosen at random is subscribed to at least one of the two plans (music or video).
Let $A$ and $B$ be two events such that $P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cup B) = 0.7$. Determine whether the events $A$ and $B$ are compatible or mutually…
Let $A$ and $B$ be two events such that $P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cup B) = 0.7$. Determine whether the events $A$ and $B$ are compatible or mutually exclusive, justifying your answer by calculating $P(A \cap B)$.
In a study on digital content consumption in a Year 13 class, it is known that the probability of a student regularly using TikTok (event $A$) is $0.65$, the probability…
In a study on digital content consumption in a Year 13 class, it is known that the probability of a student regularly using TikTok (event $A$) is $0.65$, the probability of using Instagram (event $B$) is $0.70$ and the probability of using neither of the two social networks is $0.15$.
Using the properties of probability, calculate:
- a) The probability that a student, chosen at random, uses both social networks.
- b) The probability that they use only one of the two.
- c) Are the events $A$ and $B$ independent? Justify your answer.
Venn diagrams with proportional areas
Drawing a Venn diagram so that each region has an area proportional to its probability is a visual way of checking that the calculations are consistent: the whole of $E$ must have total area $1$, and no probability can ever be missing or left over.
To read a single region, subtract: $P(A\cap B') = P(A) - P(A\cap B)$, because $A$ is made of the two mutually exclusive pieces $A\cap B$ and $A\cap B'$.
Observe the following Venn diagram representing the events $A$ (students who play basketball) and $B$ (students who do swimming) at a sports centre: Knowing that the…
Observe the following Venn diagram representing the events $A$ (students who play basketball) and $B$ (students who do swimming) at a sports centre:
Knowing that the numbers represent the probability of each region, use the properties of probability to calculate:
- a) $P(A)$
- b) $P(A \cap B)$
- c) $P(\bar{A})$ (Probability of not playing basketball).
In an e-sports competition, the probability that team "Alfa" wins the first game is $P(A) = 0.6$. The probability that it wins the second is $P(B) = 0.5$. If the…
In an e-sports competition, the probability that team "Alfa" wins the first game is $P(A) = 0.6$. The probability that it wins the second is $P(B) = 0.5$. If the probability that it wins both is $P(A \cap B) = 0.3$, calculate:
- a) The probability that it wins the first game but not the second ($P(A \cap \bar{B})$).
- b) Illustrate the situation using an area diagram like the following one, in which the whole square represents probability 1 and each region has an area equal to its probability. Fill in the probability of the other three regions.
At a sports center, $60\%$ of the members do strength training (event $F$) and $45\%$ do yoga (event $Y$). It is also known that $20\%$ of those who do strength training…
At a sports center, $60\%$ of the members do strength training (event $F$) and $45\%$ do yoga (event $Y$). It is also known that $20\%$ of those who do strength training are also enrolled in yoga.
Construct a contingency table or a Venn diagram to answer the following questions:
- a) Probability that a randomly chosen member does neither strength training nor yoga.
- b) If we know that a member does yoga, what is the probability that they do not do strength training?
- c) Are the events "doing strength training" and "doing yoga" independent?
You can do it on paper and send a photo with your answer.
At least one, exactly one, at least two
Exam questions often describe an event in words. Translate the words into regions of the Venn diagram first.
An aeronautical engineer analyses the reliability of a critical component. Let $A$ and $B$ be two events such that $P(A) = \frac{1}{3}$ and $P(B) = p$. 1. If $A$ and $B$…
An aeronautical engineer analyses the reliability of a critical component. Let $A$ and $B$ be two events such that $P(A) = \frac{1}{3}$ and $P(B) = p$.
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If $A$ and $B$ are independent, express the probability that exactly one of the two events occurs in terms of $p$.
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If $A$ and $B$ are mutually exclusive, what is the maximum value that $p$ can take?
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Determine the value of $p$ so that the probability that neither $A$ nor $B$ occurs is equal to $\frac{1}{4}$, assuming that $A$ and $B$ are independent.
In a biotechnology laboratory, the reaction of a protein to two reagents, $\alpha$ and $\beta$, is studied. The probability of reaction to $\alpha$ is $P(\alpha) = 0.4$.…
In a biotechnology laboratory, the reaction of a protein to two reagents, $\alpha$ and $\beta$, is studied. The probability of reaction to $\alpha$ is $P(\alpha) = 0.4$. The probability of not reacting to either is $P(\alpha^c \cap \beta^c) = 0.3$.
If we denote by $R$ the event "the protein reacts to exactly one of the two reagents", and we know that $P(R) = 0.5$, determine:
- a) The probability of reaction to reagent $\beta$.
- b) Are the events of reacting to $\alpha$ and to $\beta$ independent? Justify your answer with calculations.
- c) Are they compatible?
In a redundant security system for a server infrastructure, three control nodes ($N_1, N_2, N_3$) are installed. The system is known to fail if at least two of the nodes…
In a redundant security system for a server infrastructure, three control nodes ($N_1, N_2, N_3$) are installed. The system is known to fail if at least two of the nodes go down simultaneously. The failure probabilities of each node are $P(N_1) = 0.05$, $P(N_2) = 0.08$ and $P(N_3) = 0.10$.
Because they share the same power grid, the failures are not independent. It is known that:
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The probability that the first and second nodes fail is $P(N_1 \cap N_2) = 0.02$.
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The probability that the second and third nodes fail is $P(N_2 \cap N_3) = 0.03$.
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The probability that the first and third nodes fail is $P(N_1 \cap N_3) = 0.01$.
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The probability that all three fail at the same time is $P(N_1 \cap N_2 \cap N_3) = 0.005$.
- a) Draw a Venn diagram with the three events $N_1$, $N_2$ and $N_3$ inside the sample space and calculate the probability of each of the eight regions (including the one outside the three circles).
- b) Calculate the probability that the overall system does not fail, that is, that at most one node fails. Use the diagram from the previous part to justify your answer.
Independent events and conditional probability
The probability of $A$ given that $B$ has occurred is
$$P(A\mid B) = \frac{P(A\cap B)}{P(B)}, \quad P(B)>0$$Rearranged, this is the multiplication rule: $P(A\cap B) = P(B)\,P(A\mid B)$.
Two esports teams, "Banshees" and "Shamrocks", participate in a tournament. The probability of the Banshees winning their match is $P(T) = 0.8$ and that of the Shamrocks…
Two esports teams, "Banshees" and "Shamrocks", participate in a tournament. The probability of the Banshees winning their match is $P(T) = 0.8$ and that of the Shamrocks winning theirs is $P(F) = 0.7$. Assuming that the results of both teams are independent, calculate the following probabilities using the algebra of events:
- a) That both teams win their respective matches.
- b) That at least one of the two wins.
- c) That only the Banshees team wins.
Given two events $A$ and $B$, it is known that $P(A) = 0.3$, $P(B) = 0.6$ and $P(A \cup B) = 0.72$. a) Calculate $P(A \cap B)$. b) Are the events $A$ and $B$…
Given two events $A$ and $B$, it is known that $P(A) = 0.3$, $P(B) = 0.6$ and $P(A \cup B) = 0.72$.
- a) Calculate $P(A \cap B)$.
- b) Are the events $A$ and $B$ independent? Reason your answer by comparing $P(A \cap B)$ with the product of their individual probabilities.
- c) Calculate $P(\bar{A} / B)$.
A streaming platform classifies its users as "Premium" ($P$) or "Standard" ($S$). It is estimated that $40\%$ are Premium. In addition, $70\%$ of Premium users watch…
A streaming platform classifies its users as "Premium" ($P$) or "Standard" ($S$). It is estimated that $40\%$ are Premium. In addition, $70\%$ of Premium users watch science fiction series ($C$), whereas only $30\%$ of Standard users watch them. If a user is chosen at random:
- a) Calculate the probability that they watch science fiction series.
- b) If the chosen user is watching a science fiction series, what is the probability that they are a Premium user?
- c) Are the events "Being a Premium user" and "Watching science fiction series" independent? Justify mathematically using properties of conditional probability.
Bounds on an unknown intersection
The axioms do not only give equalities: they also let us bound a probability when a piece of data is missing, using the fact that no probability can exceed $1$ or be negative.
For three events the same idea gives $P(A\cap B\cap C) \ge P(A)+P(B)+P(C)-2$. Indeed, $P(A\cap B\cap C) = 1-P(A'\cup B'\cup C') \ge 1-\big(P(A')+P(B')+P(C')\big)$. For example, with probabilities $0.6$, $0.7$ and $0.8$, the triple intersection is at least $2.1-2 = 0.1$.
In the election of student representatives at a school, let $A$ be the event "the chosen person is in the Science track" with $P(A) = 0.65$, and let $B$ be the event…
In the election of student representatives at a school, let $A$ be the event "the chosen person is in the Science track" with $P(A) = 0.65$, and let $B$ be the event "the chosen person participates in the debate club" with $P(B) = 0.45$.\na) Estimate, before performing any algebraic calculation, whether it is possible for both characteristics to be mutually exclusive ($A \cap B = \emptyset$). Justify your intuition based on the sum of their probabilities relative to the total sample space.\nb) Check your estimate by formally calculating the Fréchet bounds for the intersection; that is, the interval $[p_{\min}, p_{\max}]$ in which $P(A \cap B)$ must necessarily lie.\nc) Critically assess whether a prior estimate proposing $P(A \cap B) = 0.50$ would be admissible or if it contradicts the monotonicity property of probability ($A \cap B \subseteq B$).
Consider two events $A$ and $B$ of a random experiment. It is known that $P(A) = 0.6$ and $P(B) = 0.7$. Based on Kolmogorov's axioms, prove that: 1. The minimum possible…
Consider two events $A$ and $B$ of a random experiment. It is known that $P(A) = 0.6$ and $P(B) = 0.7$.
Based on Kolmogorov's axioms, prove that:
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The minimum possible value for $P(A \cap B)$ is $0.3$.
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The maximum possible value for $P(A \cap B)$ is $0.6$.
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If a third event $C$ is added such that $P(C) = 0.8$, what is the minimum value of the probability that all three events occur simultaneously, $P(A \cap B \cap C)$? Justify your answer using Bonferroni's inequality: $P(A \cap B) \geq P(A) + P(B) - 1$.
Back to the start: if rain in Dublin and rain in Cork were mutually exclusive, the chance of rain in at least one would be $0.7+0.5=1.2$, which is impossible. The lower bound gives $P(\text{both}) \ge 0.2$, so the 120% is ruled out.