University Business Mathematics — Counting and Probability Worksheet

All 11 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1The Addition and Multiplication Principles
   1. A café's lunch combo is a main, a drink and an optional dessert. The main is one of $5$ sandwiches or one of $3$ salads; the drink is one of $4$; the dessert is one of $2$ pastries, or none. How many different combos are there?
   2. A hardware store labels each product with a code made of one letter ($26$ choices) followed by three digits, where the first digit may not be $0$. How many codes are possible?
   3. How many of the codes in (b) have three different digits?
2. Q2Permutations and Factorial Notation
   1. Evaluate ${\displaystyle \frac{10!}{7!}}$ and simplify ${\displaystyle \frac{(n+1)!}{(n-1)!}}$ for an integer $n\ge 1$.
   2. Solve ${\displaystyle \frac{n!}{(n-2)!}}=56$ for the integer $n\ge 2$.
   3. A sales representative must visit $6$ client offices tomorrow, one after another. In how many orders can she make the visits?
   4. The $9$ members of a co-operative's board elect a chair, a vice-chair and a treasurer, three different people. In how many ways can the three posts be filled?
3. Q3Permutations and Factorial Notation

   A bookstore sets out $7$ different books in a row in its window: $3$ business titles and $4$ novels.

   1. In how many arrangements are all $3$ business titles side by side, in one unbroken group?
   2. In how many arrangements do novels and business titles alternate?
   3. The shop next door lines up $9$ mugs on a shelf: $4$ identical red ones, $3$ identical blue ones and $2$ identical white ones. How many different-looking rows are possible?
4. Q4Combinations and Mixed Counting Problems

   A firm forms a project team of $4$ from $7$ accountants and $5$ analysts.

   1. How many teams are possible?
   2. How many teams have exactly $2$ accountants and $2$ analysts?
   3. How many teams include at least one analyst?
5. Q5Combinations and Mixed Counting Problems
   1. A pizzeria offers $12$ toppings, $3$ sizes and $2$ crusts. How many different pizzas with exactly $3$ different toppings can be ordered?
   2. A coffee chain will open $3$ new branches, chosen from $8$ possible sites. In how many ways can the $3$ sites be chosen? In how many ways if the chain also decides the order in which the three will open?
   3. In how many ways can the $3$ sites be chosen if one particular site, the downtown one, must be among them?
6. Q6Sample Spaces, Events and Equally Likely Outcomes

   A manager will invite $2$ of $5$ job applicants — Ana, Ben, Chloé, Dev and Emma — to a second interview, choosing the pair at random so that every pair is equally likely.

   1. List the sample space.
   2. Find the probability that Ana is invited.
   3. Find the probability that neither Ben nor Chloé is invited.
   4. Find the probability that Ana or Ben (or both) is invited.
7. Q7Union, Intersection and Complement of Events

   At a wholesaler, an invoice is chosen at random. Let $L$ be the event that it was paid late and $E$ the event that it contains a billing error. The records give $P(L)=0.35$, $P(E)=0.20$ and $P(L\cap E)=0.12$. Find the probability that the invoice

   1. was paid late or contains an error;
   2. was paid on time and contains no error;
   3. was paid late but contains no error;
   4. has exactly one of the two problems.
8. Q8Union, Intersection and Complement of Events

   An office has $10$ laptops, $3$ of which have a faulty battery. A technician takes $4$ of them at random for a training session. Find the probability that

   1. none of the $4$ has a faulty battery;
   2. at least one has a faulty battery;
   3. exactly one has a faulty battery;
   4. at least two have a faulty battery.
9. Q9Conditional Probability, Probability Trees and Independence

   A hotel keeps $12$ sealed gift cards in a drawer for a staff draw: $4$ are worth $50 and $8$ are worth $20. The manager draws $2$ cards at random, one after the other, without replacement.

   1. Draw a probability tree for the two draws, with the probability on every branch.
   2. Find the probability that the two cards are worth $70 in total.
   3. Find the probability that at least one card is worth $50.
   4. Given that the first card is worth $20, what is the probability that the second is worth $50?
10. Q10Conditional Probability, Probability Trees and Independence

    For a loan application chosen at random at a credit union, let $A$ be the event that it is approved and $S$ the event that the applicant is self-employed. Records give $P(A)=0.6$, $P(S)=0.25$ and $P(A\cap S)=0.1$.

    1. Find $P(A\mid S)$ and $P(S\mid A)$, and say in words what each means.
    2. Find the approval rate for applicants who are not self-employed.
    3. Are the events $A$ and $S$ independent? Justify.
11. Q11Synthesis — drawing on several topics in this unit

    Six sales representatives, including Lina and Omar, are given the six presentation slots at a trade show (slots $1$ to $6$, in order), with every assignment equally likely.

    1. Find the probability that Lina presents before Omar.
    2. Find the probability that Lina and Omar present in consecutive slots, in either order.
    3. Given that Lina has slot $1$, find the probability that Omar has slot $6$.
    4. Given that Lina presents before Omar, find the probability that Lina has slot $1$.

The 7 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The addition and multiplication principles

Every count on this sheet starts with one question: is the task done in stages, one after another, or does it split into separate cases, exactly one of which happens? Stages multiply; cases add.

Q1(a) has both principles in one line: the main is a choice between two kinds of item, so its blank is a sum, and the dessert is optional. "None" is a legitimate outcome of that stage and counts as one more choice — leaving it out is the usual slip. Q1(b) is the same shape with a restriction on one position: fill the restricted blank first, from what it is allowed to be.

Permutations and factorial notation

A permutation is an ordered arrangement. n! counts the orders of n different objects; P(n, r) = n!/(n − r)! counts the ordered selections of r of them.

Q2(a) and (b) are about the notation itself. Never expand a large factorial: write the bigger one as a product that stops at the smaller one, and let the smaller one cancel. The same move turns the quotient in Q2(b) into a short product in n, which is then an ordinary equation — keep only the root the condition on n allows. Q2(c) orders every object; Q2(d) fills three distinct posts from a larger group, which is a P(n, r) because a chair and a treasurer are different jobs.

Q3(c) is the arrangement of items that are not all different. Arrange them as if every item were distinguishable, then divide by the orders of each group of identical items, because swapping two identical mugs produces the same-looking row.

Combinations and mixed counting problems

A combination is an unordered selection: C(n, r) = n!/(r!(n − r)!). The whole decision is one question — would rearranging the chosen items give a different result? A team of four is the same team in any order, so Q4 is combinations throughout.

Q5 mixes the two kinds of count. In Q5(a) the toppings are a combination, since a pizza with the same three toppings is the same pizza whatever order they were named in, while size and crust are further stages of the multiplication principle. Q5(b) asks the same selection twice, once without order and once with an opening order, and the comparison between the two answers is the point: each unordered selection corresponds to r! ordered ones.

Sample spaces and equally likely outcomes

When every outcome is equally likely, a probability is a ratio of two counts:

Q6 asks you to list the sample space before using it. A pair of applicants is unordered, so list each pair once — fix the first name and run through the later ones — and check the length of the list against C(5, 2). Then Q6(b), (c) and (d) are counts on that list. "Neither Ben nor Chloé" means the pair avoids both names; "Ana or Ben (or both)" is an inclusive or, so a pair holding both names is counted once, not twice.

Union, intersection and complement

Two events split the sample space into four regions: in both, in the first only, in the second only, in neither. Q7 gives P(L), P(E) and P(L ∩ E); fill the four regions of a Venn diagram from the centre outward, and every part is read off it.

Translate each phrase before computing. "Late or contains an error" is the union. "On time and no error" is the region outside both circles, which is the complement of the union. "Late but no error" is the first circle with the overlap removed. "Exactly one of the two" is the union without the overlap. Q7(a) to (d) are these four readings in turn.

Q8 brings counting back into probability. Choosing 4 laptops from 10 is a combination, and each event is a count of selections: choose the faulty ones from the faulty and the good ones from the good, multiply, and divide by the total number of selections. Q8(a) and (c) are direct counts; Q8(b) and (d) are "at least" questions, where the complement of an event you have already found is the short route.

Conditional probability and probability trees

P(A | B) is the probability of A once B is known to have happened: the sample space shrinks to B, and A is measured inside it.

For Q9(b), translate the total into the combinations of card values that produce it, then find every path that gives one of them — order matters on a tree, so check whether more than one path leads to the same pair of values. Q9(c) is an "at least one", and the complement is a single path. Q9(d) is read straight off the second level of the tree: the condition names the branch you stand on.

Reading conditional probability from given values

Q10 gives the same information as Q7 — two probabilities and their intersection — and asks for conditional ones. P(A | S) divides by P(S); P(S | A) divides by P(A). The numerator is the same for both, and only the condition changes, which is exactly why the two are different numbers with different meanings. Say each in words starting from the condition: "among self-employed applicants, the proportion approved" is not "among approved applications, the proportion that are self-employed".

Q10(b) conditions on the complement. Find P(A ∩ Sᶜ) first — the approved applications with the self-employed ones removed — then divide by P(Sᶜ).

The synthesis question

Q11 puts all of it together: the sample space is the set of all orders of six people, so it is counted with a factorial, and each event is a count of orders. For Q11(a), look for a way to pair every order where Lina is first of the two with an order where she is second — a pairing between two events of the same size settles a probability without a long count. Q11(b) is Q3(a)'s block method: glue the two into one object and remember the two orders inside it.

Q11(c) and (d) are conditional. A condition that fixes someone's slot shrinks the sample space to the orders consistent with it; count inside that smaller space. In Q11(d), before dividing, ask whether one event contains the other: when it does, the intersection is simply the smaller event, and the definition of conditional probability becomes one division.

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Getting the most out of it

Decide "order or no order" out loud

Before writing P or C, say whether rearranging the chosen items gives a different outcome. Posts, slots and codes: yes. Teams, pairs and sets of toppings: no. Nearly every counting error is this decision made silently and wrongly.

Reach for the complement

"At least one", "at least two", "not both": write the complement next to the event and ask which of the two is quicker to count. It is usually the complement, and the subtraction from 1 or from the total is one line.

Check a small count by listing

When a count is small enough, list it, as Q6(a) asks. A list that disagrees with the formula shows which assumption about order or repetition was wrong, and that habit carries into the counts too large to list.

Make every table, diagram and tree add up

The four regions of a Venn diagram, the branches out of a node, the ends of all paths of a tree: each set adds to 1. It is the cheapest check in probability and it catches a missed region or a replacement handled wrongly.

Common questions