Probability, the two-way table and independence in Mathematik (CBK) at WU

Probability theory is the fifth of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna; the syllabus lists no subtopics for it.¹ The textbook by Birgit Rudloff, who leads the course, and Achim Zeileis has a fifth chapter with the same title, "Wahrscheinlichkeitsrechnung". Its first part explains what an event is and how probabilities combine; its second asks how a probability changes once you already know something about the outcome.

Most of the points lost here come from misreading what is given: an overlap counted twice, a condition read the wrong way round, independent events taken for events that exclude each other. The syllabus points you to the forum on Canvas for questions about exercises, and to the walk-in tutorial, where you work through problems with a tutor on hand.¹ German terms are given in brackets.

Random experiments, sample space and events: Mathematik (CBK)

A random experiment (Zufallsexperiment) is a process whose outcome you do not know in advance. Our example: a Viennese coffee house had 200 guests on a Saturday, and one of them is picked at random. The sample space (Ergebnismenge) $\Omega$ is the set of all possible outcomes, here the 200 guests. An event (Ereignis) is a subset of $\Omega$, such as $K$, "the guest ordered cake" (80 guests), or $Z$, "the guest paid by card" (125 guests). 50 guests did both.

Events combine like sets. The union $K\cup Z$, "$K$ or $Z$", happens when at least one of the two happens; the intersection $K\cap Z$, "$K$ and $Z$", when both do. The complement $\overline{K}$ (Gegenereignis) is everything outside $K$, here the 120 guests without cake. Two events that can never happen together are called mutually exclusive or disjoint (disjunkt), $A\cap B=\varnothing$. "Ordered cake" and "ordered only a drink" are such a pair.

Probabilities and the addition rule: Mathematik (CBK)

A probability assigns each event a number between 0 and 1, with $P(\Omega )=1$. For disjoint events, probabilities add, and the complement rule follows from that: $P(\overline{A})=1-P(A)$. When all outcomes are equally likely, as with a guest picked at random, the probability of an event is the number of its outcomes divided by the number of all outcomes: $P(K)=80/200=0.4$ and $P(Z)=125/200=0.625$. Outside this special case, the question has to give you the probabilities.

For "at least one of the two" it is tempting to add $P(K)$ and $P(Z)$. That gives $0.4+0.625=1.025$, more than 1 and so impossible: the 50 guests with cake who paid by card sit in both terms and are counted twice. The general addition rule (Additionsgesetz) takes them off once:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

In the coffee house, $P(K\cup Z)=0.4+0.625-0.25=0.775$, which is 155 of the 200 guests. For disjoint events $P(A\cap B)=0$, and the plain sum remains. "Neither of the two" is the complement: $1-0.775=0.225$, 45 guests.

Conditional probability and the two-way table: Mathematik (CBK)

Once you know that $B$ has happened, only the part of $\Omega$ inside $B$ still counts. The conditional probability (bedingte Wahrscheinlichkeit) of $A$ given $B$ is

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)},P(B)>0$$

Of the 125 guests who paid by card, 50 had cake: $P(K\mid Z)=0.25/0.625=0.4$. Of the 80 guests with cake, 50 paid by card: $P(Z\mid K)=0.25/0.4=0.625$. The numerator is the same both times, the denominator is not, which is why $P(A\mid B)$ is in general a different number from $P(B\mid A)$. Rearranged, the formula gives the multiplication rule $P(A\cap B)=P(A)\cdot P(B\mid A)$: it takes you from a share within one group to a share of everyone.

1. 200 guests, one dot per guest: 80 with cake (K), 120 without.
2. 125 pay by card (Z), 75 in cash.
3. 50 guests have both.
4. Only among the 125 paying by card: P(K | Z) = 50 / 125 = 0.4.
5. Only among the 80 with cake: P(Z | K) = 50 / 80 = 0.625.
6. Same numerator, different denominator.

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): 200 guests, 80 with cake (K), 125 paying by card (Z), 50 with both. Each column of dots is 8 guests, each row 25.

P(K | Z) = 50 / 125 = 0.4 and P(Z | K) = 50 / 80 = 0.625. The dots are sorted by group, not by the order of the day.

The two-way table (Vierfeldertafel) keeps this in order: the four intersections of two events inside, their totals in the margins; a tree diagram shows the same thing over several steps.

	$Z$ (card)	$\overline{Z}$ (cash)	Total
$K$ (cake)	$0.25$	$0.15$	$0.4$
$\overline{K}$ (no cake)	$0.375$	$0.225$	$0.6$
Total	$0.625$	$0.375$	$1$

A conditional probability is an inner cell divided by the total of the row or column that belongs to the condition. Every row and column adds up to its margin, so a missing cell can be filled in from the others.

Independent events: Mathematik (CBK)

Two events are independent (unabhängig) when knowing about one does not change the probability of the other, $P(A\mid B)=P(A)$. Equivalent, and handier for calculating, is

$$P(A\cap B)=P(A)\cdot P(B)$$

That holds in the coffee house: $0.4\cdot 0.625=0.25$, and $P(K\mid Z)=0.4$ is exactly $P(K)$. In both rows of the table, 62.5% paid by card: 0.25 out of 0.4 and $0.375$ out of 0.6.

Our calculation

Numbers from the article’s example: 200 guests, 80 with cake, 125 by card, 50 with both; every number in the table is a share of all 200.

0.25 / 0.4 = 0.625 and 0.375 / 0.6 = 0.625, both equal to P(Z) = 0.625. So P(K) · P(Z) = 0.4 · 0.625 = 0.25 = P(K ∩ Z).

The most common mix-up is between independent and disjoint. Disjoint events rule each other out: a guest who ordered cake did not order only a drink. Knowing one therefore changes the other as much as anything can, and $P(A\cap B)=0$ is smaller than $P(A)\cdot P(B)$ as soon as both are positive. Two disjoint events with positive probability are therefore never independent. Independent events, on the other hand, do happen together, and simply adding their probabilities counts the intersection twice.

One task with a two-way table: Mathematik (CBK)

At an online fashion retailer, 25% of orders are paid by invoice, the event $R$ (Rechnung). Of those orders, 60% are sent back; of the others, 30%. $S$ stands for a return. A return arrives. What you want is the probability that the order was paid by invoice, $P(R\mid S)$.

You are given $P(R)=0.25$, $P(S\mid R)=0.6$ and $P(S\mid \overline{R})=0.3$, and asked for the condition the other way round. Fill in the two-way table row by row with the multiplication rule:

	$S$ (returned)	$\overline{S}$ (kept)	Total
$R$	$0.25\cdot 0.6=0.15$	$0.1$	$0.25$
$\overline{R}$	$0.75\cdot 0.3=0.225$	$0.525$	$0.75$
Total	$0.375$	$0.625$	$1$

The column total $P(S)=0.15+0.225=0.375$ collects both routes to a return, through invoice orders and through all the others. So

$$P(R\mid S)=\frac{P(R\cap S)}{P(S)}=0.4$$

Our calculation

Numbers from the article’s task (our own numbers, not an exam question): P(R) = 0.25, P(S | R) = 0.6, P(S | R̄) = 0.3.

P(S) = 0.25 · 0.6 + 0.75 · 0.3 = 0.15 + 0.225 = 0.375, P(R | S) = 0.15 / 0.375 = 0.4.

40% of returns come from invoice orders, although only 25% of orders are paid that way. Thought of as 400 orders: 100 on invoice, 60 of them returned; 300 others, 90 of them returned. Of the 150 returns, 60 were paid by invoice. This kind of reversal is also known as Bayes' theorem.

1. 400 orders.
2. 100 on invoice, 300 other.
3. Of the 100, 60 are returned.
4. Of the 300, 90 are returned.
5. Together 150 returns.
6. Invoice: 100 of 400 orders (25%), but 60 of 150 returns (40%).

Our calculation

The article’s task thought of as 400 orders: 0.25 · 400 = 100 on invoice, 0.6 · 100 = 60 of them returned; 0.3 · 300 = 90 of the others returned.

60 / 150 = 0.4, the same as P(R | S) = 0.4.

Four wrong results are close at hand. The 0.6 from the question is $P(S\mid R)$, the condition read the wrong way round. The 0.15 is the intersection $P(R\cap S)$, a share of all orders, not divided by $P(S)$. The 0.25 ignores the information "returned" and would only be right if $R$ and $S$ were independent. Leave the second route out of the denominator, and $P(S)=0.15$ and the result is 1, as if every return came from an invoice order.

Practice questions on probability: Mathematik (CBK)

The exam consists of 20 multiple-choice questions, each with 5 options of which exactly one is correct. Each correct answer is worth 1 point; of the wrong ones, the first 3 are free and each further one costs 0.25 points. You need 11 points to pass.¹

We wrote the three questions for this article in that format. The stories, numbers and wording are our own; none of them comes from a WU exam or from the textbook. Every wrong option is the result of a typical mistake, most of them the ones described above. Each solution shows the working, the typical mistake and how to rule options out.

Practice question 1Probability

Among the first-year students of a degree programme in Vienna, 40% have a Klimaticket (the annual public-transport pass) and 25% have a bike of their own. Of those with a Klimaticket, 20% also have a bike of their own. What is the probability that a first-year student picked at random has at least one of the two?

1. a)0.08
2. b)0.45
3. c)0.55
4. d)0.57
5. e)0.65

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. K stands for Klimaticket, F for bike. Given are P(K) = 0.40, P(F) = 0.25 and the conditional probability P(F | K) = 0.20.
2. Intersection by the multiplication rule: P(K ∩ F) = P(K) · P(F | K) = 0.40 · 0.20 = 0.08.
3. Addition rule: P(K ∪ F) = P(K) + P(F) − P(K ∩ F) = 0.40 + 0.25 − 0.08 = 0.57.

a)

Wrong. The probability of both, P(K ∩ F) = 0.40 · 0.20, instead of at least one.

0 or −0.25

b)

Wrong. The 20% subtracted as the intersection: 0.40 + 0.25 − 0.20. But the 20% is a share within the Klimaticket group, P(F | K), not a share of everyone.

0 or −0.25

c)

Wrong. Independence assumed: 0.40 + 0.25 − 0.40 · 0.25. But since P(F | K) = 0.20 ≠ P(F) = 0.25, K and F are not independent.

0 or −0.25

d)

Right. The addition rule, with the intersection from the multiplication rule: P(K ∩ F) = 0.40 · 0.20 = 0.08, so P(K ∪ F) = 0.40 + 0.25 − 0.08 = 0.57.

+1

e)

Wrong. Simply added, 0.40 + 0.25: those who have both are counted twice.

0 or −0.25

–

Left blank.

0

Typical mistake. Reading the 20% as the intersection. "Of those with a Klimaticket, 20% …" is a conditional probability, the share within group K. Subtracted as P(K ∩ F), it gives 0.45; the intersection is 0.40 · 0.20 = 0.08, and the result 0.57.

Ruling out before you finish. At least one of the two is at least as likely as the more common one alone, 0.40, and less likely than the sum, 0.65, because some students have both. That rules out a and e and leaves three.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 2Probability

A company is filling a trainee programme. 30% of applications come through a referral from staff. Of those applications, 50% lead to an interview invitation; of the others, 10%. A person is invited. What is the probability that their application came through a referral (rounded)?

1. a)0.15
2. b)0.22
3. c)0.30
4. d)0.50
5. e)0.68

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. R stands for referral, I for the interview invitation. Two-way table, row R: P(R ∩ I) = 0.30 · 0.50 = 0.15.
2. Row without referral: P(R̄ ∩ I) = 0.70 · 0.10 = 0.07.
3. Column I, both routes together: P(I) = 0.15 + 0.07 = 0.22.
4. Backwards: P(R | I) = 0.15 / 0.22 ≈ 0.682, rounded 0.68.

a)

Wrong. The intersection P(R ∩ I) = 0.30 · 0.50, “referred and invited” as a share of all applications, not divided by P(I).

0 or −0.25

b)

Wrong. The probability of being invited, P(I) = 0.15 + 0.07: the denominator, not the result.

0 or −0.25

c)

Wrong. The share of referrals before knowing “invited”; the condition ignored.

0 or −0.25

d)

Wrong. The condition reversed: P(I | R), the share invited among the referred, instead of P(R | I).

0 or −0.25

e)

Right. P(R | I) = P(R ∩ I) / P(I) = 0.15 / 0.22 ≈ 0.682. Of all invitations (0.22), 0.15 go to referred applicants.

+1

–

Left blank.

0

Typical mistake. Turning the condition round. Given is P(I | R) = 0.50, asked is P(R | I). Copying the 0.50 answers how many referred applicants are invited, not how many invited applicants were referred. The direction asked for also needs the second route to an invitation, 0.07 through applications without a referral, in the denominator.

Ruling out before you finish. Referred applicants are invited five times as often as the others, so knowing “invited” makes a referral more likely than the 0.30 from before. That rules out a, b and c and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Practice question 3Probability

A café gets milk and pastries every morning from two different suppliers. The milk is late with probability 0.15, the pastries with 0.08, and the two delays are independent of each other. What is the probability that on a given morning at least one of the two deliveries is late?

1. a)0.012
2. b)0.218
3. c)0.230
4. d)0.782
5. e)0.988

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. The complement of “at least one delivery late” is “both on time”, with probabilities 0.85 for the milk and 0.92 for the pastries.
2. Because the delays are independent, multiply: P(both on time) = 0.85 · 0.92 = 0.782.
3. P(at least one late) = 1 − 0.782 = 0.218.

a)

Wrong. The probability that both are late, 0.15 · 0.08, instead of at least one.

0 or −0.25

b)

Right. Through the complement “both on time”: 1 − 0.85 · 0.92 = 1 − 0.782 = 0.218. The addition rule gives the same: 0.15 + 0.08 − 0.012 = 0.218.

+1

c)

Wrong. Simply added, 0.15 + 0.08, as if the events were disjoint. But independent delays also happen together, and that intersection is counted twice.

0 or −0.25

d)

Wrong. The probability that both are on time, 0.85 · 0.92: the complement worked out and not subtracted from 1.

0 or −0.25

e)

Wrong. The wrong complement: 1 − 0.15 · 0.08, which is “not both late”. But the opposite of “at least one late” is “none late”.

0 or −0.25

–

Left blank.

0

Typical mistake. Confusing independent with disjoint. You may simply add only when the two events never happen together. Independent delays coincide with probability 0.15 · 0.08 = 0.012, and the sum 0.230 counts those mornings twice.

Ruling out before you finish. At least one delay is at least as likely as the more common one alone, 0.15, and at most as likely as the sum 0.15 + 0.08 = 0.23. That rules out a, d and e and leaves b and c.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Probability in the Mathematik (CBK) exam

How the 20 questions spread over the chapters is something WU does not publish, so this article follows the structure of the textbook. Its fifth chapter opens with the section "Grundbegriffe" (basic concepts), containing "Zufallsexperimente und Ergebnismenge", "Ereignisse", "Verknüpfung von Ereignissen", "Wahrscheinlichkeiten" and "Eine Verallgemeinerung des Additionsgesetzes" (a generalisation of the addition rule). Later comes "Bedingte Wahrscheinlichkeiten" (conditional probabilities), with the subsections "Vierfeldertafeln" and "Unabhängige Ereignisse".²

A calculator is allowed as long as it has no extra functions for differential calculus, integration or matrices, no solver for linear systems and no text memory.¹ Products, sums and one fraction are all this chapter's calculations need.

Unless the syllabus provides otherwise, WU's exam guideline rules out formula sheets and watches of any kind.³ Whether a formula sheet or a table of the normal distribution is handed out in the exam, WU does not publish. Learn the addition rule, the formula for conditional probability with the multiplication rule, and the product rule for independent events by heart.

The exam plan reserves two hours for the whole exam; the actual writing time is not in the syllabus.⁴ With 20 questions, that leaves at most six minutes for each, enough to sketch a two-way table before you calculate.

Next step after conditional probability: Mathematik (CBK)

1. Do the three questions above before opening the solutions, and for each one first write down which probability is given and which is asked for, with the condition to the right of the bar.
2. If you got one wrong, find the mistake in the solution that matches your answer. If it was the reversed condition, draw the two-way table first from now on. For more practice, see the fifth chapter of the free textbook by Birgit Rudloff and Achim Zeileis.⁵
3. Random variables, expected value and the binomial and normal distributions are covered in the article on expected value, binomial and normal distribution. Exam dates and registration are in the overview of the Mathematik (CBK) exam.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften, Kapitel 5: Wahrscheinlichkeitsrechnungmathe4wiwi.org · undated page · checked on 27 September 2026
3. 3Richtlinie zur Abhaltung von Präsenzprüfungen, 2026 (PDF)WU Wien · document dated 2 March 2026 · checked on 27 September 2026
4. 4Prüfungsplan Semestermitte WiSe 2026/27, Prüfungswoche November 2026 (PDF)WU Wien, Prüfungsorganisation · document dated 30 March 2026 · checked on 27 September 2026
5. 5Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften (Online-Buch, Version 2023)mathe4wiwi.org · undated page · checked on 26 September 2026