Worked example · Easy
Fair die
A fair six-sided die is rolled. Find the probability of a multiple of 3.
Multiples of 3 are 3 and 6, giving two favourable outcomes from six.
\[P(\text{multiple of 3})=\frac26=\boxed{\frac13}\]Chapter 44
Probability
Learn to calculate, interpret and estimate probabilities using outcomes, sample spaces, Venn diagrams and probability trees.
By the end of this chapter, you should be able to
- interpret the probability scale and use probability notation;
- calculate single-event probabilities and complements;
- estimate expected frequencies from probabilities and populations;
- use relative frequency as an experimental estimate of probability;
- list combined outcomes using sample-space and Venn diagrams;
- apply rules for independent events and mutually exclusive outcomes;
- use probability trees and conditional branches for dependent events.
1
Subtopic 1
The probability of a single event
Probability describes how likely an event is to occur. Its value is between 0 (impossible) and 1 (certain). A probability of \(\tfrac12\) means an even chance. Fractions, decimals and percentages are equivalent ways to describe a chance.
A random experiment has possible outcomes. An event is the outcome or group of outcomes you are interested in. Count favourable outcomes only when all individual outcomes are equally likely.
Single-event probability
\[\boxed{P(A)=\frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}}\]\[0\le P(A)\le1\]For unequal chances (a biased spinner), use the actual given probabilities instead of assuming equal likelihood. Simplify a fractional answer.
Worked example · Medium
Counters of three colours
A bag has 5 red, 3 blue and 2 yellow counters. Find the probabilities of red and of blue or yellow.
Total counters: \(5+3+2=10\).
\[P(R)=\frac5{10}=\boxed{\tfrac12}\]\[P(B\text{ or }Y)=\frac{3+2}{10}=\boxed{\tfrac12}\]Worked example · Exam
Read a two-way table
Of 30 students, 8 boys and 12 girls study art, while 5 boys and 5 girls study music. One student is chosen at random. Find (a) the probability of an art student, (b) the probability of a girl who studies music, and (c) the probability of a girl.
Totals: art \(8+12=20\); music \(5+5=10\); girls \(12+5=17\).
\[\text{(a) }\frac{20}{30}=\boxed{\tfrac23}\quad\text{(b) }\frac5{30}=\boxed{\tfrac16}\quad\text{(c) }\frac{17}{30}\]2
Subtopic 2
The probability of an event not occurring
The complement of event \(A\) is the event that \(A\) does not happen. If one of several possible outcomes has an unknown probability, subtract the total of the other probabilities from 1.
For a complete set of mutually exclusive outcomes, all their probabilities add up to 1. You may see the complement written \(A'\), \(A^c\) or “not A”.
Complement rule
\[\boxed{P(A')=1-P(A)}\]For a complete list of separate outcomes: \(P(A_1)+P(A_2)+\cdots=1\).
Worked example · Easy
Not raining
The chance of rain tomorrow is \(\tfrac3{10}\). Find the chance of no rain.
\[P(\text{no rain})=1-\frac3{10}=\boxed{\frac7{10}}\]
Worked example · Medium
Find a missing spinner probability
A spinner has outcomes A, B, C and D, with probabilities 0.23, 0.18, 0.37 and \(p\). Find \(p\).
\[p=1-(0.23+0.18+0.37)=\boxed{0.22}\]
Worked example · Exam
Win, draw or lose
A team has probability 0.35 of a win and 0.25 of a draw. Find (a) the probability of a loss and (b) the probability that the team does not lose.
\[P(\text{loss})=1-0.35-0.25=\boxed{0.40}\]\[P(\text{not lose})=1-0.40=\boxed{0.60}\]
Part (b) also equals \(P(\text{win})+P(\text{draw})\).
3
Subtopic 3
Estimating from a population
A probability can be used to estimate the expected number of occurrences in many trials or a whole population. This is called the expected frequency. An expectation is not a guarantee: actual results vary.
You can use a probability stated as a fraction, decimal or percentage, but convert it to a number between 0 and 1 before multiplying.
Expected frequency
\[\boxed{\text{Expected frequency}=\text{number of trials}\times P(A)}\]To estimate how many people or items fall into a category, multiply the population total by the appropriate estimated probability.
Worked example · Easy
Expected sixes
A fair die is rolled 180 times. Estimate the number of sixes.
\[180\times\frac16=\boxed{30\text{ sixes}}\]
Worked example · Medium
Survey to population
42 of 140 people surveyed prefer cycling. Estimate how many of 900 people would prefer cycling if the survey is representative.
\[P(\text{cycling})\approx\frac{42}{140}=0.3\]\[900\times0.3=\boxed{270\text{ people}}\]
Worked example · Exam
Production quality
A factory has a 1.5% probability of producing a faulty component. Estimate (a) how many faulty components occur in 800, and (b) how many are not faulty.
\[P(\text{faulty})=\frac{1.5}{100}=0.015\]\[800\times0.015=\boxed{12}\]\[800-12=\boxed{788\text{ not faulty}}\]
4
Subtopic 4
Relative frequency and probability
When events are not known to be equally likely (for example a drawing pin landing point up), use observations. Relative frequency is the proportion of trials in which the event occurred and provides an experimental estimate of probability.
The estimate may change from sample to sample. Results from larger, representative samples are generally more stable. Comparing observed frequencies with equal chances can suggest whether a spinner is biased.
Experimental probability
\[\boxed{\text{Relative frequency of }A=\frac{\text{observed occurrences of }A}{\text{total number of trials}}}\]To predict future counts from an experiment, multiply this estimated probability by the number of future trials.
Worked example · Easy
Estimate from trials
A drawing pin points upwards on 75 of 200 throws. Estimate the probability of point up.
\[P(\text{up})\approx\frac{75}{200}=\boxed{0.375}\]
Worked example · Medium
Compare two samples
A biased spinner shows red 18 times in 60 trials and 142 times in 500 more trials. Estimate the probability of red using all the data.
Combine both sets of results before dividing.
\[\frac{18+142}{60+500}=\frac{160}{560}=\boxed{\frac27\approx0.286}\]Worked example · Exam
Use relative frequency to forecast
In a survey of 300 students, 90 travel by bus. Estimate (a) the probability a randomly chosen student travels by bus, and (b) the number travelling by bus in a school of 1200, assuming the sample is representative.
\[P(\text{bus})\approx\frac{90}{300}=\boxed{0.3}\]\[0.3\times1200=\boxed{360\text{ students}}\]
5
Subtopic 5
The probability of combined events
For two experiments, list every possible outcome in a sample space diagram. Rows and columns can represent the first and second spinners, dice or coins; each table cell is one outcome.
Events are mutually exclusive if they cannot both occur in a single trial. If they overlap, adding their probabilities directly would count the overlap twice. A Venn diagram helps identify the intersection and union of events.
Combining events
\[\boxed{P(A\text{ or }B)=P(A)+P(B)}\quad\text{when }A,B\text{ are mutually exclusive}\]\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]\(A\cap B\) means both conditions are met; \(A\cup B\) means at least one. In a sample space with equally likely cells, count the favourable cells.
Worked example · Easy
Two coins
Two fair coins are tossed. Find the probability of getting exactly one head.
Sample space: HH, HT, TH, TT. Exactly one head occurs in HT or TH.
\[P(\text{one head})=\frac24=\boxed{\tfrac12}\]Worked example · Medium
Two dice: total of 7
Two fair six-sided dice are rolled. What is the probability that their total is 7?
There are \(6\times6=36\) equally likely ordered pairs. The favourable pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
\[P(\text{total }7)=\frac6{36}=\boxed{\tfrac16}\]Worked example · Exam
Venn diagram with overlap
In a class of 29 students, 14 play football, 12 play tennis and 5 play both. One student is selected at random. Find the probability they (a) play at least one sport and (b) play neither.
\[n(F\cup T)=14+12-5=21\]\[P(\text{at least one})=\boxed{\tfrac{21}{29}}\]\[n(\text{neither})=29-21=8\implies P=\boxed{\tfrac8{29}}\]
6
Subtopic 6
Independent events
Two events are independent when knowing the result of one does not change the probability of the other. Examples are separate coin tosses and selecting a ball with replacement, returning it before the second selection.
The word and often signals multiplying along a sequence. For more than two independent events, multiply all their probabilities. For events joined by or, first decide which distinct outcomes are being combined.
Multiplication for independent events
\[\boxed{P(A\cap B)=P(A)\times P(B)}\]\[P(A\cap B\cap C)=P(A)P(B)P(C)\]Use this only if independence is justified. “With replacement” keeps the same proportions after each draw.
Worked example · Easy
Head and 4
A fair coin is tossed and a fair die rolled. Find the probability of a head and a 4.
\[\frac12\times\frac16=\boxed{\frac1{12}}\]
Worked example · Medium
With replacement
A bag has 3 green and 2 black counters. A counter is chosen, replaced and then another chosen. Find the probability of green followed by black.
\[P(G\text{ then }B)=\frac35\times\frac25=\boxed{\frac6{25}}\]
Worked example · Exam
At least one event in three independent trials
A machine independently makes faulty items with probability 0.04. Three items are tested. Calculate the probability that at least one is faulty.
The complement of “at least one faulty” is “none faulty”.
\[P(\text{none})=(1-0.04)^3=0.96^3=0.884736\]\[P(\text{at least one})=1-0.884736=\boxed{0.115264}\]7
Subtopic 7
Tree diagrams for combined events
A probability tree records possible routes through successive experiments. Each branch is labelled with a probability. The probabilities leaving any one node must add to 1.
To find the probability of a particular route, multiply probabilities along its branches. If several routes satisfy the question, add their route probabilities. This works with two or more stages.
Tree diagram rules
- Branch probabilities from a single node add to 1.
- Multiply along a route (AND).
- Add distinct successful routes (OR).
- All terminal route probabilities together add to 1.
Worked example · Easy
Exactly one head from a tree
Two fair coins are tossed. Use the branches to find the probability of exactly one head.
\[P(HT)+P(TH)=\frac12\cdot\frac12+\frac12\cdot\frac12=\boxed{\frac12}\]
Worked example · Medium
Two colours with replacement
A bag contains 4 red and 2 blue counters. Two counters are chosen with replacement. Find the probability that both have the same colour.
\[P(RR)+P(BB)=\frac46\cdot\frac46+\frac26\cdot\frac26\]\[=\frac{16+4}{36}=\boxed{\frac59}\]
Worked example · Exam
One win in two games
A team independently wins any game with probability 0.7. Use a two-stage tree to calculate the probability of winning exactly one of two games.
\[P(WL)+P(LW)=0.7(0.3)+0.3(0.7)\]\[=0.21+0.21=\boxed{0.42}\]
8
Subtopic 8
The probability of dependent events
Events are dependent if what happens first changes the probabilities later. Selecting objects without replacement is the central example: one object leaves the collection, reducing the total, and perhaps the number of favourable objects as well.
In the probability tree, the second-stage branches must be calculated separately for each first-stage outcome. For a conditional question, use the information about the first outcome to select the appropriate branches.
Probability without replacement
If the first draw removes an item, the new total is one smaller. For a particular path, multiply the first probability by the probability of the second outcome given the first.
\[\boxed{P(A\text{ then }B)=P(A)\times P(B\mid A)}\]\(P(B\mid A)\) is the probability of B after A has occurred. Add the paths only after calculating the branch probabilities correctly.
Worked example · Easy
Two red counters, no replacement
There are 3 red and 2 blue counters. Two are chosen without replacement. Find the probability that both are red.
\[P(RR)=\frac35\times\frac24=\boxed{\frac3{10}}\]
After the first red is taken, 2 red remain from 4 counters.
Worked example · Medium
Same colour, no replacement
A bag contains 5 red and 3 blue counters. Two are drawn without replacement. Find the probability that they have the same colour.
\[P(RR)+P(BB)=\frac58\cdot\frac47+\frac38\cdot\frac27\]\[=\frac{20+6}{56}=\boxed{\frac{13}{28}}\]
Worked example · Exam
Weather-dependent delay
The probability of rain is 0.3. If it rains, the chance of a bus delay is 0.6; if it does not rain, the chance of delay is 0.1. Find (a) the overall probability of delay and (b) the probability that it does not rain and the bus is on time.
Separate the rainy and dry routes:
\[P(\text{delay})=0.3(0.6)+0.7(0.1)=\boxed{0.25}\]\[P(\text{dry and on time})=0.7(0.9)=\boxed{0.63}\]✓
Independent practice
Chapter 44 exercises
40 original practice questions begin with straightforward probability, progress to interpretation and calculations, and finish with multi-step exam-style questions. Attempt each question before opening its Show Answer section.
Build confidence — Questions 1–8
1Fair dieEasy · [2 marks]
A fair die is rolled. Find the probability of an odd number.
Show Answer
\[P(\text{odd})=\frac36=\boxed{\frac12}\]
2Not successfulEasy · [2 marks]
The probability that a student catches a bus is 0.28. Find the probability that they miss it.
Show Answer
\[1-0.28=\boxed{0.72}\]
3Expected resultsEasy · [2 marks]
A fair four-section spinner is spun 80 times. How often would you expect it to show section 1?
Show Answer
\[80\times\frac14=\boxed{20\text{ times}}\]
4Experimental chanceEasy · [2 marks]
A player wins 18 out of 90 games. Estimate the probability that the player wins the next game.
Show Answer
\[\frac{18}{90}=\boxed{\frac15=0.2}\]
5Two fair coinsEasy · [2 marks]
Two fair coins are tossed. What is the probability of getting two tails?
Show Answer
Possible outcomes: HH, HT, TH, TT.
\[P(TT)=\boxed{\frac14}\]6Two independent headsEasy · [2 marks]
A fair coin is tossed twice. Calculate the probability of heads both times using multiplication.
Show Answer
\[\frac12\times\frac12=\boxed{\frac14}\]
7One A on two spinsEasy · [2 marks]
A fair spinner has two equal sectors, A and B. It is spun twice. Find the probability of getting exactly one A.
Show Answer
The two successful routes are AB and BA.
\[\frac14+\frac14=\boxed{\frac12}\]8Without replacementEasy · [2 marks]
A bag holds 4 green and 2 yellow balls. Two are drawn without replacement. Find the probability both are green.
Show Answer
\[\frac46\times\frac35=\boxed{\frac25}\]
Develop your methods — Questions 9–24
9Colours from a bagMedium · [2 marks]
A bag contains 8 red, 4 yellow and 3 black balls. Find the probability of not selecting black.
Show Answer
There are 15 balls. The 12 red or yellow balls are not black.
\[P(\text{not black})=\frac{12}{15}=\boxed{\frac45}\]10Repeated lettersMedium · [2 marks]
One letter is chosen at random from PROBABILITY. Find the probability of selecting B.
Show Answer
There are 11 letters and B appears twice.
\[P(B)=\boxed{\frac2{11}}\]11Unknown sectorMedium · [2 marks]
A spinner has four outcomes with probabilities 0.14, 0.26, 0.37 and \(p\). Find \(p\).
Show Answer
\[p=1-(0.14+0.26+0.37)=\boxed{0.23}\]
12Sport outcomesMedium · [2 marks]
The probability of winning a match is 0.40 and drawing it is 0.15. Calculate the probability of losing.
Show Answer
\[1-0.40-0.15=\boxed{0.45}\]
13Favourable productsMedium · [2 marks]
Each item has probability 0.35 of passing a quality check. Estimate how many of 240 items pass.
Show Answer
\[240\times0.35=\boxed{84}\]
14Faulty componentsMedium · [2 marks]
Three in every 100 manufactured parts are expected to be faulty. Estimate the number of faulty parts in a batch of 600.
Show Answer
\[600\times\frac3{100}=\boxed{18}\]
15Sample resultMedium · [2 marks]
A spinner shows blue 27 times in 150 spins. Estimate the probability of blue.
Show Answer
\[P(\text{blue})\approx\frac{27}{150}=\boxed{0.18}\]
16Forecast from observed dataMedium · [2 marks]
A bus is late on 60 of 250 observed journeys. Estimate how many delays occur in the next 500 journeys.
Show Answer
\[P(\text{late})\approx\frac{60}{250}=0.24\]\[500\times0.24=\boxed{120}\]
17Dice totalMedium · [2 marks]
Two fair dice are rolled. Find the probability that the total is 9.
Show Answer
Successful ordered pairs: (3,6), (4,5), (5,4), (6,3), out of 36.
\[P(9)=\frac4{36}=\boxed{\frac19}\]18Overlapping hobbiesMedium · [2 marks]
In a group of 28, 13 study music, 10 study drama and 4 study both. What is the probability that a randomly chosen member studies neither?
Show Answer
\[n(M\cup D)=13+10-4=19\]\[P(\text{neither})=\frac{28-19}{28}=\boxed{\frac9{28}}\]
19Coin and dieMedium · [2 marks]
Calculate the probability that a fair die shows an odd number and a fair coin shows tails.
Show Answer
\[\frac36\times\frac12=\boxed{\frac14}\]
20Repeated independent spinsMedium · [2 marks]
A spinner shows red with probability 0.6. Find the probability of red on all three independent spins.
Show Answer
\[0.6^3=\boxed{0.216}\]
21Same colour with replacementMedium · [2 marks]
A bag contains 2 red and 3 blue balls. Two are selected with replacement. Find the probability of the same colour twice.
Show Answer
\[P(RR)+P(BB)=\frac25\cdot\frac25+\frac35\cdot\frac35=\boxed{\frac{13}{25}}\]
22Exactly one failureMedium · [2 marks]
A player independently succeeds on each attempt with probability 0.8. Find the probability of exactly one failure in two attempts.
Show Answer
\[P(SF)+P(FS)=0.8(0.2)+0.2(0.8)=\boxed{0.32}\]
23Red then blue, no replacementMedium · [2 marks]
A bag has 6 red and 4 blue balls. Two are drawn without replacement. Find the probability of red then blue.
Show Answer
\[\frac6{10}\cdot\frac49=\frac{24}{90}=\boxed{\frac4{15}}\]
24Given the first ballMedium · [2 marks]
There are 7 green and 3 blue balls. One blue ball is taken and not replaced. What is the probability the next ball is green?
Show Answer
There are still 7 green, but only 9 balls remain.
\[P(G\mid\text{first blue})=\boxed{\frac79}\]Exam-style challenge — Questions 25–40
25Two-way table probabilityExam-style · [5 marks]
A class of 42 pupils contains 12 left-handed boys, 9 left-handed girls, 8 right-handed boys and 13 right-handed girls. One student is selected at random. Calculate (a) the probability of a left-handed student, (b) of a right-handed girl, and (c) of a boy.
Show Answer
\[P(L)=\frac{12+9}{42}=\boxed{\tfrac12}\]\[P(\text{right-handed girl})=\boxed{\tfrac{13}{42}}\]\[P(\text{boy})=\frac{12+8}{42}=\boxed{\tfrac{10}{21}}\]
26Interpret the textbook bar chartExam-style · [4 marks]
The Chapter 44 car-colour bar chart shows 10 black, 6 white, 7 red, 8 silver and 4 blue cars. If a car is selected at random, find the probability that it is red or blue and explain your denominator.
Show Answer
There are \(10+6+7+8+4=35\) cars. Red and blue are separate possible colours.
\[P(R\text{ or }B)=\frac{7+4}{35}=\boxed{\frac{11}{35}}\]27Missing outcome and complementExam-style · [4 marks]
A bag contains red, blue and green counters. The probability of red is \(\tfrac15\) and the probability of blue is \(\tfrac3{10}\). Find (a) the probability of green and (b) the probability of not selecting blue.
Show Answer
\[P(G)=1-\frac15-\frac3{10}=\boxed{\frac12}\]\[P(\text{not blue})=1-\frac3{10}=\boxed{\frac7{10}}\]
28Two linked complement calculationsExam-style · [4 marks]
On a journey the probabilities of arriving early and on time are 0.18 and 0.57, respectively. The other possibility is arriving late. Find (a) the probability of late arrival and (b) the probability of not arriving early.
Show Answer
\[P(\text{late})=1-0.18-0.57=\boxed{0.25}\]\[P(\text{not early})=1-0.18=\boxed{0.82}\]
29Expected frequency in a schoolExam-style · [5 marks]
A survey estimates that \(\tfrac3{14}\) of students cycle to school. There are 420 students. Estimate (a) the number who cycle and (b) the number who do not cycle. Explain why your prediction need not be exact.
Show Answer
\[420\cdot\frac3{14}=\boxed{90}\]\[420-90=\boxed{330}\]
These are estimates based on a probability, so actual observed counts may differ.
30Two production categoriesExam-style · [4 marks]
The probability that a product is faulty is 0.025. A company makes 1200 products. Estimate how many are (a) faulty and (b) not faulty, showing your workings.
Show Answer
\[1200\times0.025=\boxed{30\text{ faulty}}\]\[1200-30=\boxed{1170\text{ not faulty}}\]
31Biased-spinner estimateExam-style · [6 marks]
In 400 spins a spinner lands on red 96 times. Use the results to estimate (a) the probability of red, (b) the probability of not red and (c) the number of reds expected in the next 750 spins.
Show Answer
\[P(R)\approx\frac{96}{400}=\boxed{0.24}\]\[P(\text{not }R)\approx1-0.24=\boxed{0.76}\]\[750(0.24)=\boxed{180}\]
32Combine two experimentsExam-style · [5 marks]
A factory inspects 50 components and finds 20 faulty. Later it inspects a further 250 and finds 82 faulty. Estimate (a) the overall faulty probability and (b) the faulty count in the next 200 components.
Show Answer
Combine both samples:
\[P(\text{faulty})\approx\frac{20+82}{50+250}=\frac{102}{300}=\boxed{0.34}\]\[200\times0.34=\boxed{68}\]33Two dice: high totalsExam-style · [5 marks]
Two fair dice are rolled. Find the probability that the sum is greater than 9, showing all the successful totals and their frequencies.
Show Answer
Sum 10: 3 pairs; sum 11: 2 pairs; sum 12: 1 pair. In all, 6 favourable outcomes out of 36.
\[P(\text{total}>9)=\frac6{36}=\boxed{\frac16}\]34Venn: at least one or neitherExam-style · [5 marks]
In a group of 40 pupils, 19 take music, 17 take drama and 8 take both. (a) How many take at least one? (b) Find the probability a randomly chosen pupil takes neither.
Show Answer
\[n(M\cup D)=19+17-8=\boxed{28}\]\[n(\text{neither})=40-28=12\]\[P(\text{neither})=\frac{12}{40}=\boxed{\frac3{10}}\]
35Three consecutive daysExam-style · [5 marks]
On any day a bus is late with probability 0.1. Assume days are independent. Over three days find (a) the probability it is never late and (b) the probability it is late at least once.
Show Answer
\[P(\text{never late})=0.9^3=\boxed{0.729}\]\[P(\text{late at least once})=1-0.729=\boxed{0.271}\]
36Exactly one blackExam-style · [5 marks]
A bag contains 7 white and 3 black balls. A ball is selected, replaced and then another ball is selected. Calculate the probability of exactly one black ball.
Show Answer
\[P(BW)+P(WB)=\frac3{10}\cdot\frac7{10}+\frac7{10}\cdot\frac3{10}\]\[=\frac{42}{100}=\boxed{0.42}\]
37Exactly two heads in three tossesExam-style · [5 marks]
Draw or use a tree diagram for three fair coin tosses. Find the probability of exactly two heads, naming the possible routes.
Show Answer
Routes: HHT, HTH and THH. Each has probability \(\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18\).
\[P(\text{exactly two heads})=3\times\frac18=\boxed{\frac38}\]38Biased coin on two tossesExam-style · [5 marks]
A biased coin shows heads with probability 0.3 on each independent toss. Use a tree or a complement to find the probability of at least one head in two tosses.
Show Answer
The complement is two tails, each with probability 0.7.
\[P(\text{at least one H})=1-(0.7\times0.7)=\boxed{0.51}\]39Two colours, any order without replacementExam-style · [6 marks]
A bag contains 5 green, 3 yellow and 2 red counters. Two counters are drawn without replacement. Calculate the probability that exactly one is green and one is yellow.
Show Answer
Both orders count: GY and YG.
\[\frac5{10}\cdot\frac39+\frac3{10}\cdot\frac59=\frac{30}{90}=\boxed{\frac13}\]40Dependent travel outcomesExam-style · [6 marks]
A pupil walks to school with probability 0.4, cycles with probability 0.3 and takes a bus otherwise. The probabilities of being late are 0.1 when walking, 0.2 when cycling and 0.5 when on a bus. Find the probability of being on time.
Show Answer
The bus probability is \(1-0.4-0.3=0.3\). Sum the separate late routes:
\[P(L)=0.4(0.1)+0.3(0.2)+0.3(0.5)=0.25\]\[P(\text{on time})=1-0.25=\boxed{0.75}\]Chapter 44 high-grade checklist
Scale
I know probability ranges from 0 (impossible) to 1 (certain).
Favourable outcomes
I count the correct equally likely outcomes and simplify the fraction.
Complements
I subtract from 1 when finding a missing or opposite probability.
Expected frequency
I multiply probability by the number of trials or the population.
Experiments
I calculate relative frequency using the observed results.
Combined events
I use sample-space tables and do not double-count overlaps.
Tree diagrams
I multiply along a path and add alternative paths.
Dependence
I recalculate branch probabilities after a draw without replacement.