University Introductory Statistics — Probability Rules and Conditional Probability Worksheet

10 of the 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. 10 of the 11 questions are printed below. The other 1 is built on a diagram or a table of values that does not translate to the page, so it is in the free PDF — marked below where it would have come.

1. Q1Sample Spaces, Events and Venn Diagrams

   A trail camera is left in a forest for three nights. Each night it either records a fox (D) or does not (N), so an outcome is a string such as DND (fox on nights $1$ and $3$ only). Let $A$ be the event that a fox is recorded on at least two of the nights, and $B$ the event that a fox is recorded on the first night.

   1. List the sample space.
   2. List the outcomes in $A$, $B$, $A\cap B$, $A\cup B$ and $A^c$.
   3. Are $A$ and $B$ mutually exclusive? Justify.
   4. Assuming the eight outcomes are equally likely, find $P(A\cup B)$ and $P(A^c)$.
2. Q2The Addition Rule and Complements

   A seed mix for a wildflower garden produces red, yellow, white or purple flowers. For a seed chosen at random, $P(\text{red})=0.35$, $P(\text{yellow})=0.25$ and $P(\text{white})=0.15$; each seed produces exactly one colour.

   1. Find $P(\text{purple})$.
   2. Find the probability that the flower is red or yellow.
   3. Find the probability that it is not white.
   4. Find the probability that it is neither red nor white.
3. Q3The Addition Rule and Complements

   In a cycling club, a member chosen at random owns a road bike with probability $0.55$ and a mountain bike with probability $0.40$, and owns at least one of the two with probability $0.80$.

   1. Find the probability that the member owns both kinds of bike.
   2. Find the probability that the member owns a road bike but no mountain bike.
   3. Find the probability that the member owns neither kind, and the probability that the member owns exactly one kind.
   4. Are the events “owns a road bike” and “owns a mountain bike” mutually exclusive? Justify.
4. Q4Counting with Permutations and Combinations

   An ecology class has $11$ sampling plots along a river, $4$ of them in wetland and $7$ on dry ground. A team chooses $3$ plots at random to survey, every set of $3$ being equally likely.

   1. How many different sets of $3$ plots are possible?
   2. Find the probability that exactly $2$ of the chosen plots are wetland.
   3. Find the probability that at least one chosen plot is wetland.
   4. The team will survey its $3$ plots in a fixed order: morning, midday and afternoon. How many different schedules (which plot at which time) can be made from the $11$ plots?
5. Q5Conditional Probability and the Multiplication Rule

   This question is built around a diagram or a table of values. Open it in the PDF.

6. Q6Conditional Probability and the Multiplication Rule
   1. A literary magazine passes $40\%$ of submitted stories through its first editorial screen. Of the stories that pass, $35\%$ are accepted by the reviewers, and of the accepted stories, $50\%$ are printed in the next issue. Find the probability that a submitted story is printed in the next issue, and the probability that it passes the screen but is then turned down by the reviewers.
   2. A tray holds $10$ pastries that look identical, $3$ of which have a nut filling. A guest takes $3$ pastries at random, one after another. Use the multiplication rule to find the probability that none of the three has a nut filling, and the probability that the first two both do.
7. Q7Independent Events

   For an order chosen at random at an online shop, let $G$ be the event that the order is gift-wrapped and $X$ the event that it is sent by express shipping. The shop's records give $P(G)=0.20$, $P(X)=0.35$ and $P(G\cap X)=0.07$.

   1. Decide whether $G$ and $X$ are independent.
   2. Find $P(G\mid X)$ and say what it tells the shop.
   3. Find $P(G\cup X)$.
   4. Are $G$ and $X$ mutually exclusive? Justify.
8. Q8Independent Events

   A warehouse is protected at night by a motion sensor, a camera and a guard patrol, which detect an intruder independently of one another with probabilities $0.9$, $0.8$ and $0.5$. Find the probability that an intruder is detected

   1. by all three;
   2. by none of them;
   3. by at least one of them;
   4. by the motion sensor only.
9. Q9Tree Diagrams, Total Probability and Bayes' Theorem

   A language-learning app gets its new users from three channels: $50\%$ from search advertising, $30\%$ from referrals by friends and $20\%$ from social media. The probability that a new user is still active after $30$ days is $0.20$ for search users, $0.40$ for referred users and $0.10$ for social-media users.

   1. Draw a tree diagram with every branch labelled.
   2. Find the probability that a new user is still active after $30$ days.
   3. An active user is chosen at random. Find the probability that the user came from a referral.
   4. A user who is not active is chosen at random. Find the probability that the user came from search advertising.
10. Q10Tree Diagrams, Total Probability and Bayes' Theorem

    At an orchard's packing line, $8\%$ of apples are bruised. An optical scanner flags a bruised apple with probability $0.95$ and flags a sound apple with probability $0.05$.

    1. Find the probability that a randomly chosen apple is flagged.
    2. Find the probability that a flagged apple is bruised.
    3. Find the probability that an apple that is not flagged is bruised.
11. Q11Synthesis — drawing on several topics in this unit

    A laboratory keeps two racks of sample vials. Rack~1 holds $5$ vials, $2$ of them contaminated; rack~2 holds $6$ vials, $1$ of them contaminated. A technician rolls a fair die and uses rack~1 on a roll of $1$ or $2$ and rack~2 otherwise, then takes $2$ vials at random from that rack.

    1. For each rack, find the probability that both vials taken are clean.
    2. Find the probability that both vials taken are clean.
    3. Given that both vials are clean, find the probability that they came from rack~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.

Sample spaces, events and Venn diagrams

An event is a set of outcomes, so before any arithmetic there has to be a list. Q1 records three nights of a trail camera, and each outcome is a string of three letters, one per night, in order — which is why DND and NDD are different outcomes rather than the same one written twice.

Q1(b) turns each event into a sublist. Read the definitions literally: "at least two" includes the outcome with all three, which is the slip that costs the most marks in this part, and an event defined by the first night alone is settled by the first letter whatever the other two do. A ∩ B holds the outcomes appearing on both sublists; A ∪ B holds every outcome on either, written once; Aᶜ holds everything in the sample space that A does not.

Q1(d) is where equally likely outcomes turn counts into probabilities. With every outcome equally likely, P(E) is the number of outcomes in E over the number in the sample space — so the counts come straight off the sublists. Both parts can be checked a second way, by the addition rule and by the complement rule, and it is worth doing: agreement between a count and a rule is the fastest confirmation available in this chapter.

The addition rule, complements and translation

From here on, probabilities are given rather than counted, and the work is translation. Q2 gives three of four colours in a seed mix and says each seed produces exactly one colour — which is the whole content of Q2(a): the four colour events are mutually exclusive and between them cover everything, so their probabilities must total 1.

Q2(b) is an "or" between events that cannot happen together, so the subtraction in the addition rule contributes nothing and the two probabilities simply add. Q2(c) is a complement. Q2(d) can be done either way — as the complement of an "or", or by adding the colours that remain — and doing it both ways is a free check.

Once the overlap is known, fill a Venn diagram from the centre outward and every remaining part of Q3 is read off it. "Owns a road bike but no mountain bike" in Q3(b) is the first circle with the overlap removed. In Q3(c), "neither" is outside both circles — the complement of the union — and "exactly one kind" is the union without the overlap. Q3(d) asks the same kind of question as Q1(c): mutually exclusive means the intersection is empty, so compare the probability found in Q3(a) with 0 and say what the comparison shows.

Counting with permutations and combinations

Counting appears here for one reason: to build an equally likely sample space that is too big to list. Q4 chooses three river plots out of eleven, every set of three equally likely, so every probability in it is one count over another.

Q4(d) changes the question. Morning, midday and afternoon are distinguishable slots, so the same three plots in a different order make a different schedule, and the count is a permutation rather than a combination. Notice that Q4(d) is a pure count and not a probability — read the instruction, because this part is answered with a number of schedules, not a fraction.

Conditional probability and the multiplication rule

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 — which in a two-way table is exactly a change of denominator.

Q5 classifies surveyed riders by fare type and trip purpose, and every part of it is one count over another. The only decision is which total goes underneath. Q5(a) is a marginal probability, so the grand total goes underneath. Q5(b) conditions on fare type, so the total for that fare type does. Q5(c) reverses the condition, so the total for that trip purpose does — the cell on top is the same one as in Q5(b), and only the denominator moves. Q5(d) pairs a joint probability, which keeps the grand total underneath because nothing has been assumed, with one more conditional.

Q6(a) is a chain of stages, each probability conditional on the one before it: a story passes a screen, then is accepted, then is printed. Multiply along the chain. The second half of Q6(a) asks for a path that succeeds at the first stage and then fails at the second, so take the complement within that stage before multiplying — the failure probability is 1 minus the acceptance probability quoted for the stories that passed, not 1 minus anything about the whole pool.

Q6(b) is the multiplication rule for draws without replacement, and the question says to use it rather than counting combinations. Each draw changes the tray: on a branch where a nut pastry has gone, both the nut count and the total drop by one for the next factor. The second part constrains only the first two draws, so it is a product of two factors — reading in a third condition that is not there is the usual error.

Independent events, and what independence is not

Independence means knowing one event happened does not change the probability of the other. There are two equivalent tests, and either settles Q7(a).

Q7 gives P(G), P(X) and P(G ∩ X) for gift-wrapped and express orders, which is everything both tests need. Q7(a) asks for a decision, so compute both sides and write them next to each other before saying anything; a verdict with no comparison shown is not an answer. Q7(b) then asks for P(G | X) and for what it tells the shop — the sentence that earns the second half of the mark compares the conditional probability with P(G) and says, in the shop's own terms, whether knowing an order is express changes what is expected about gift wrapping. Q7(c) is the addition rule again, with the intersection already given.

Q8 runs three detectors that work independently of one another, which is what licenses multiplying their probabilities. Q8(a) multiplies the three detection probabilities. Q8(b) multiplies the three complements. Q8(c) is an "at least one", and its complement is precisely Q8(b) — one subtraction, rather than a sum of seven cases. Q8(d) turns on the word "only": it means one detector succeeds and the other two fail, so it is an intersection of one event with two complements, not a conditional probability and not the same as Q8(a) restricted.

Tree diagrams, total probability and Bayes' theorem

A tree is the tool for a two-stage experiment where the first stage decides which second-stage probabilities apply. Q9 splits new users of an app across three acquisition channels and gives a different retention probability on each.

Q9(c) and Q9(d) reverse the condition, and that reversal is Bayes' theorem in the only form this course wants: one path over the sum of the paths ending in the same outcome. The denominator for Q9(c) is the total from Q9(b), which is why Q9(b) is asked first. Q9(d) conditions on the outcome that was not given directly, so its numerator and its denominator are both built from the complements labelled in step 2 — either from a fresh sum, or as 1 minus the total in Q9(b) for the denominator.

Q10(a) is the law of total probability: an apple is flagged along either of two paths. Q10(c) reverses a condition again, and this time both the event and the condition are complements of what the question states — take care to pick the right path for the numerator before dividing.

The synthesis question

Q11 is the chapter in one question: a die decides which of two racks of vials is used, and two vials are then drawn from that rack without replacement. Every idea above appears once.

Q11(a) works inside a single rack, where the two draws are dependent — either the multiplication rule with the second factor adjusted for the vial already taken, or a ratio of combinations, and the two agree. Q11(b) is total probability across the two racks, with the first-stage weights coming from the die: read the rule carefully, because the two racks are not equally likely and assuming they are is the one error that makes the rest of the question meaningless. Q11(c) reverses the condition once more, dividing one branch's contribution by the total from Q11(b).

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Getting the most out of it

Translate the sentence into symbols first

"At least one" is a union. "Neither" is the complement of a union. "Exactly one" is a union with the overlap removed. "Only" is an intersection with complements. "Given that", "among those who" and "of the ones that" all signal a conditional, and they tell you what goes in the denominator. Write the symbols in the margin before touching the numbers, and most of the difficulty in this chapter disappears.

Draw the picture the question implies

Two overlapping events with a known intersection want a Venn diagram. Stages that happen one after another want a tree. Two categorical variables want a two-way table. The structure of the information, not the wording of the question, picks the picture — and with the right one drawn, nearly every part is read off it rather than derived.

Make everything add to 1

The four regions of a Venn diagram, the branches leaving any node of a tree, the ends of all paths, the probabilities of a set of outcomes that covers everything: each collection adds to 1. It is the cheapest check in probability, it takes seconds, and it catches a missed region or a complement applied to the wrong stage.

Name the condition out loud before dividing

Say the sentence "among the ___, the proportion that are ___" for every conditional probability you write. The first blank is the denominator. Doing this once per question is what stops P(A | B) and P(B | A) from being swapped, which is the single most common lost mark in this chapter and the reason a reversed question is worth practising until it is dull.

Justify, do not just decide

Several parts here ask whether events are mutually exclusive or independent. Both are settled by a comparison of two numbers, so write both numbers down and then the verdict. The marks are in the comparison; an answer that reports only the verdict cannot show the comparison was ever made.

Common questions