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E-Math notes · Chapter 18

Probability: O-Level E-Math Notes

Probability puts a number on chance. In this chapter you learn to measure chance on a scale from impossible to certain, list outcomes, use possibility diagrams and tree diagrams, and use the two rules for combined events: add for 'or' when events cannot happen together, multiply for 'and' when events do not affect each other.

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Chapter
18
Figure 1 · Three fair coin tosses. Each path has probability one half times one half times one half.
Three coin tossesThree-stage tree diagram for three fair coin tossesFirstSecondThirdH12H12H12T12T12H12T12T12H12H12T12T12H12T12Probability1818181818181818
4 outcomes · 21 worked examples · every step shown ✦
Maths in real life

Probability tells us how likely something is, so we can plan, play fair games and make decisions. Three real situations come first. Each is solved in full, with a Your Turn that has its own numbers. Come back to them when the ideas are new.

Probability as a measure of chance

Probability =number of favourable outcomesnumber of possible outcomes=\frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}

Single events

List the outcomes in order.

By the end of this chapter
  • Use probability as a measure of chance, from impossible to certain, and write probabilities as fractions.
  • Find the probability of a single event, including by listing all the possible outcomes.
  • Find the probability of combined events using possibility diagrams and tree diagrams.
  • Use addition for mutually exclusive events and multiplication for independent events.
Supports S2.3 · S2.2 · S2.4 · S2.1

Maths in real life: where probability is used

Probability tells us how likely something is, so we can plan, play fair games and make decisions. Three real situations come first. Each is solved in full, with a Your Turn that has its own numbers. Come back to them when the ideas are new.

A lucky dip

When a ticket is not put back, the chances on the next draw change. A tree diagram keeps track of every path, and we multiply along the branches.

Example 1.1

A lucky dip

A bag holds 3 prize and 7 blank tickets. Two are taken at random, one after the other, and the first is not put back. Draw a tree diagram to help.

  • (i)Find the probability that both are prizes.
  • (ii)Find the probability that one prize and one blank are taken.
  • (iii)Find the probability that at least one is a prize.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
P(prize then prize)  =310×29\displaystyle P(\text{prize then prize})\;=\dfrac{3}{10}\times\dfrac{2}{9}
multiply along the branches
=690\displaystyle =\dfrac{6}{90}
=115\displaystyle =\frac{1}{15}
simplify
(ii)
P(prize then blank)  =310×79\displaystyle P(\text{prize then blank})\;=\dfrac{3}{10}\times\dfrac{7}{9}
multiply along the branches
=2190\displaystyle =\dfrac{21}{90}
=730\displaystyle =\frac{7}{30}
simplify
P(blank then prize)  =710×39\displaystyle P(\text{blank then prize})\;=\dfrac{7}{10}\times\dfrac{3}{9}
multiply along the branches
=2190\displaystyle =\dfrac{21}{90}
=730\displaystyle =\frac{7}{30}
simplify
P(p,b or b,p)  =P(p,b)+P(b,p)=730+730\displaystyle P(\text{p,b}\text{ or }\text{b,p})\;=P(\text{p,b})+P(\text{b,p})=\frac{7}{30}+\frac{7}{30}
the two orders cannot both happen, so add the branches
=1430\displaystyle =\dfrac{14}{30}
same denominator: add the numerators
=715\displaystyle =\frac{7}{15}
simplify
(iii)
P(blank then blank)  =710×69\displaystyle P(\text{blank then blank})\;=\dfrac{7}{10}\times\dfrac{6}{9}
multiply along the branches
=4290\displaystyle =\dfrac{42}{90}
=715\displaystyle =\frac{7}{15}
simplify
P(at least one prize)  =1−P(no prize)=1−715=815\displaystyle P(\text{at least one prize})\;=1-P(\text{no prize})=1-\frac{7}{15}=\frac{8}{15}
easier than adding three branches: use the opposite event
Lucky dip: two tickets, no replacementLucky dip: two tickets, no replacementFirst ticketSecond ticketP310P29B79B710P13B23Probability115730730715
Two tickets drawn from the barrel. The first ticket is not put back, so the second branches change.
Final answer
(i)P(both prize)=115P(\text{both prize})=\frac{1}{15}
(ii)P(one of each)=715P(\text{one of each})=\frac{7}{15}
(iii)815\frac{8}{15}

A dice game

A possibility diagram lists every pair of results. Count the cells that win, and divide by the total number of cells.

Example 1.2

A dice game at the fair

At a fair, a player rolls two fair dice and wins a prize if the total is nine or more. The possibility diagram shows every outcome.

  • (i)How many equally likely outcomes are there?
  • (ii)Find the probability that a player wins.
  • (iii)The game is played 90 times. How many prizes are expected?
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
6×6  =36\displaystyle 6\times6\;=36
each die has six faces
(ii)
cells that win  =10\displaystyle \text{cells that win}\;=10
count the shaded cells in the possibility diagram
P(win)  =favourable outcomesall possible outcomes=1036\displaystyle P(\text{win})\;=\dfrac{\text{favourable outcomes}}{\text{all possible outcomes}}=\dfrac{10}{36}
all outcomes are equally likely
=518\displaystyle =\frac{5}{18}
simplify the fraction
(iii)
expected prizes  =P(win)×90=518×90\displaystyle \text{expected prizes}\;=P(\text{win})\times90=\frac{5}{18}\times90
probability times the number of plays
=25\displaystyle =25
Fair game: winning totals shadedFair game: winning totals shadedSecond dieFirst die+123456123456723456783456789456789105678910116789101112
All the equally likely outcomes of two dice. The shaded cells are the winning totals.
Final answer
(i)3636
(ii)P(win)=518P(\text{win})=\frac{5}{18}
(iii)About 2525 prizes.

Two days of weather

When one event does not affect the other, they are independent, and the probability that both happen is the product of the probabilities.

Example 1.3

Weekend weather

Suppose the probability of a thunderstorm on Saturday is 14\frac{1}{4} and on Sunday is 13\frac{1}{3}, and the two days are independent. A is a storm on Saturday and B is a storm on Sunday.

  • (i)Find the probability that both A and B happen.
  • (ii)Find the probability that neither happens.
  • (iii)Find the probability that exactly one of A and B happens.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
P(A and B)  =P(A)×P(B)=14×13\displaystyle P(A\text{ and }B)\;=P(A)\times P(B)=\frac{1}{4}\times\frac{1}{3}
independent events: multiply
=112\displaystyle =\dfrac{1}{12}
multiply the numerators, multiply the denominators
(ii)
P(not A)  =34, P(not B)=23\displaystyle P(\text{not }A)\;=\frac{3}{4},\allowbreak\ \qquad P(\text{not }B)=\frac{2}{3}
opposites: one minus the probability
P(not A and not B)  =P(not A)×P(not B)=34×23\displaystyle P(\text{not }A\text{ and }\text{not }B)\;=P(\text{not }A)\times P(\text{not }B)=\frac{3}{4}\times\frac{2}{3}
independent events: multiply the probabilities of the opposites
=612\displaystyle =\dfrac{6}{12}
multiply the numerators, multiply the denominators
=12\displaystyle =\frac{1}{2}
simplify
(iii)
P(A and not B)  =14×23=16\displaystyle P(A\text{ and not }B)\;=\frac{1}{4}\times\frac{2}{3}=\frac{1}{6}
first way: A happens and B does not
P(not A and B)  =34×13=14\displaystyle P(\text{not }A\text{ and }B)\;=\frac{3}{4}\times\frac{1}{3}=\frac{1}{4}
second way: B happens and A does not
P(one way or other way)  =P(one way)+P(other way)=16+14\displaystyle P(\text{one way}\text{ or }\text{other way})\;=P(\text{one way})+P(\text{other way})=\frac{1}{6}+\frac{1}{4}
the two ways cannot both happen: add
=212+312\displaystyle =\dfrac{2}{12}+\dfrac{3}{12}
give both fractions the same denominator
=512\displaystyle =\dfrac{5}{12}
add the numerators
Final answer
(i)P(A and B)=112P(A\text{ and }B)=\frac{1}{12}
(ii)P(neither)=12P(\text{neither})=\frac{1}{2}
(iii)P(exactly one)=512P(\text{exactly one})=\frac{5}{12}
Think About It
In the lucky dip, why is the probability of a prize on the second ticket different from the probability on the first?
Show the worked answerHide the worked answer
Colour keyadd, subtract, multiply, divide or simplify
If the first ticket is a prize, 2 prizes are left among the 9 tickets.
one prize has gone
P  =29\displaystyle P\;=\dfrac{2}{9}
prizes left over tickets left
If the first ticket is blank, 3 prizes are still among the 9 tickets.
all the prizes are still there
P  =39=13\displaystyle P\;=\dfrac{3}{9}=\frac{1}{3}
simplify
The second chance depends on what happened first, so the two draws are not independent.
no replacement
Final answerThe first ticket is not put back, so the barrel has fewer tickets for the second draw, and the numbers of prizes and blanks have changed. After a prize the second-draw chance of a prize is 29\frac{2}{9}, but after a blank it is 13\frac{1}{3}.
Supports S2.1 · S2.2

Probability as a measure of chance

Probability is a number that measures how likely an event is. It runs from 00 (the event is impossible) to 11 (the event is certain), and an even chance is 12\frac{1}{2}.

The probability scaleNumber line from zero to one with impossible, even chance and certain marked00.250.50.751impossibleeven chancecertain
The probability scale. Every probability lies between zero (impossible) and one (certain).
The Core Idea

Probability =number of favourable outcomesnumber of possible outcomes=\frac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}

This works when every outcome is equally likely. The answer is always between zero and one, and is best given as a fraction in its simplest form.

Example 2.1

Placing events on the probability scale

A fair six-sided die is rolled. For each event, find its probability and say where it lies on the scale from impossible to certain.

  • (i)Find the probability of rolling a seven and say how likely it is.
  • (ii)Find the probability of rolling a number less than seven and say how likely it is.
  • (iii)Find the probability of rolling an even number and say how likely it is.
  • (iv)Find the probability of rolling a six and say how likely it is.
  • (v)Find the probability of rolling a number greater than one and say how likely it is.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
P(rolling a seven) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =06\displaystyle P\;=\dfrac{0}{6}
count the favourable outcomes and all the possible outcomes
=0\displaystyle =0
simplify the fraction
A probability of this size is impossible.
place it on the scale: zero is impossible, one is certain
(ii)
P(rolling a number less than seven) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =66\displaystyle P\;=\dfrac{6}{6}
count the favourable outcomes and all the possible outcomes
=1\displaystyle =1
simplify the fraction
A probability of this size is certain.
place it on the scale: zero is impossible, one is certain
(iii)
P(rolling an even number) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =36\displaystyle P\;=\dfrac{3}{6}
count the favourable outcomes and all the possible outcomes
=12\displaystyle =\frac{1}{2}
simplify the fraction
A probability of this size is an even chance.
place it on the scale: zero is impossible, one is certain
(iv)
P(rolling a six) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =16\displaystyle P\;=\dfrac{1}{6}
count the favourable outcomes and all the possible outcomes
A probability of this size is unlikely.
place it on the scale: zero is impossible, one is certain
(v)
P(rolling a number greater than one) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =56\displaystyle P\;=\dfrac{5}{6}
count the favourable outcomes and all the possible outcomes
A probability of this size is likely.
place it on the scale: zero is impossible, one is certain
Final answer
(i)The probability is 00. It is impossible.
(ii)The probability is 11. It is certain.
(iii)The probability is 12\frac{1}{2}. It is an even chance.
(iv)P(rolling a six)=16.P(\text{rolling a six})=\frac{1}{6}. It is unlikely.
(v)The probability is 56\frac{5}{6}. It is likely.

The opposite event

Either an event happens or it does not, so the two probabilities add up to one: P(not E)=1−P(E)P(\text{not }E)=1-P(E).

Example 2.2

The probability of an event not happening

The probability that a bus is late is 0.15.

  • (i)Find the probability that this does not happen.
  • (ii)Check that the two probabilities add up to one.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
P(not late)  =1−P(late)=1−0.15\displaystyle P(\text{not late})\;=1-P(\text{late})=1-0.15
an event and its opposite add up to one
=0.85\displaystyle =0.85
(ii)
0.15+0.85  =1\displaystyle 0.15+0.85\;=1
an event and its opposite cover every possibility
Final answer
(i)P(not late)=0.85P(\text{not late})=0.85
(ii)They add up to 11.

Expected number and trials

If an event has probability pp and the situation is repeated nn times, you expect about n×pn\times p successes. A real trial gives a relative frequency (successes divided by attempts), which is an estimate of the probability and gets better as the number of attempts grows.

Example 2.3

Expected number and relative frequency

A fair die is rolled 120 times, and a six is the event of interest.

  • (i)Estimate how many times you expect to get a six in 120 attempts.
  • (ii)In a trial, a six came up 62 times in 200 attempts. Estimate the probability from the trial.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare
(i)
expected number  =P×attempts=16×120\displaystyle \text{expected number}\;=P\times\text{attempts}=\frac{1}{6}\times120
probability times the number of attempts
=20\displaystyle =20
(ii)
relative frequency  =62200\displaystyle \text{relative frequency}\;=\dfrac{62}{200}
a trial gives an estimate, called the relative frequency
=31100\displaystyle =\frac{31}{100}
simplify
Final answer
(i)About 2020 times.
(ii)31100\frac{31}{100}
SGSchoolKaki Tip

An expected number is an average over many tries, not a promise. Rolling a die six times does not guarantee one six.

Think About It
Can a probability be 1.2 or -0.3? Explain.
Show the worked answerHide the worked answer
1.2  =65>1\displaystyle 1.2\;=\dfrac{6}{5}>1
six favourable outcomes out of five possible cannot happen
−0.3<0\displaystyle -0.3<0
a count of outcomes is never below zero
Final answerNo. The number of favourable outcomes can never be more than the number of possible outcomes, and it can never be negative, so a probability must be between zero and one.
Supports S2.2 · S2.3

Probability of single events and listing outcomes

A single event is something like 'the marble is red'. When the outcomes are equally likely, list them, count the ones that match the event, and divide by the total. Listing the possible outcomes in an orderly way is how you make sure none is missed.

Example 3.1

Marbles in a bag

A bag holds 3 red, 5 blue and 2 green marbles. One marble is taken at random.

  • (i)Find the probability that it is red.
  • (ii)Find the probability that it is not red.
  • (iii)Find the probability that it is blue or green.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
P(red)  =favourable outcomesall possible outcomes=310\displaystyle P(\text{red})\;=\dfrac{\text{favourable outcomes}}{\text{all possible outcomes}}=\dfrac{3}{10}
all outcomes are equally likely
(ii)
P(not red)  =1−P(red)=1−310\displaystyle P(\text{not red})\;=1-P(\text{red})=1-\frac{3}{10}
an event and its opposite add up to one
=710\displaystyle =\frac{7}{10}
(iii)
blue or green: 5+2  =7 marbles\displaystyle \text{blue or green: }5+2\;=7\ \text{marbles}
add the numbers of the two colours
P(blue or green) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =710\displaystyle P\;=\dfrac{7}{10}
count the favourable outcomes and all the possible outcomes
Final answer
(i)P(red)=310P(\text{red})=\frac{3}{10}
(ii)P(not red)=710P(\text{not red})=\frac{7}{10}
(iii)P(blue or green)=710P(\text{blue or green})=\frac{7}{10}
SGSchoolKaki Tip

Write the probability as a fraction first, then simplify. Do not turn it into a rounded decimal unless the question asks, because a fraction is exact.

Example 3.2

Choosing a letter

A letter is chosen at random from the word PROBABILITY. It has 11 letters in all.

  • (i)Find the probability that the letter is B.
  • (ii)Find the probability that the letter is a vowel.
  • (iii)Find the probability that the letter is not a vowel.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
P(letter is B)  =favourable outcomesall possible outcomes=211\displaystyle P(\text{letter is B})\;=\dfrac{\text{favourable outcomes}}{\text{all possible outcomes}}=\dfrac{2}{11}
all outcomes are equally likely
(ii)
P(vowel)  =favourable outcomesall possible outcomes=411\displaystyle P(\text{vowel})\;=\dfrac{\text{favourable outcomes}}{\text{all possible outcomes}}=\dfrac{4}{11}
count the vowels in the word
(iii)
P(not vowel)  =1−P(vowel)=1−411\displaystyle P(\text{not vowel})\;=1-P(\text{vowel})=1-\frac{4}{11}
an event and its opposite add up to one
=711\displaystyle =\frac{7}{11}
Final answer
(i)P(letter is B)=211P(\text{letter is B})=\frac{2}{11}
(ii)P(vowel)=411P(\text{vowel})=\frac{4}{11}
(iii)P(not vowel)=711P(\text{not vowel})=\frac{7}{11}
The Core Idea

List the outcomes in order.

With two stages, write every outcome of the first stage, paired in turn with every outcome of the second. The number of outcomes is the product of the two counts.

Example 3.3

Listing all the outcomes

Two fair coins are tossed together.

  • (i)List all the possible outcomes.
  • (ii)Find the probability of at least one head.
  • (iii)Find the probability of two tails.
Show solutionHide solution
(i)
outcomes: HH, HT, TH, TT\displaystyle \text{outcomes: }\text{HH},\allowbreak\ \text{HT},\allowbreak\ \text{TH},\allowbreak\ \text{TT}
write every combination once, in an orderly way
number of outcomes  =2×2=4\displaystyle \text{number of outcomes}\;=2\times2=4
the first choice times the second choice
(ii)
favourable: HH, HT, TH\displaystyle \text{favourable: }\text{HH},\allowbreak\ \text{HT},\allowbreak\ \text{TH}
pick out the outcomes that match
P(at least one head) is the number of favourable outcomes over all the possible outcomes.
all outcomes are equally likely
P  =34\displaystyle P\;=\dfrac{3}{4}
count the favourable outcomes and all the possible outcomes
(iii)
favourable: TT\displaystyle \text{favourable: }\text{TT}
pick out the outcomes that match
P(two tails)  =favourable outcomesall possible outcomes=14\displaystyle P(\text{two tails})\;=\dfrac{\text{favourable outcomes}}{\text{all possible outcomes}}=\dfrac{1}{4}
all outcomes are equally likely
Final answer
(i)HH, HT, TH, TT\text{HH},\allowbreak\ \text{HT},\allowbreak\ \text{TH},\allowbreak\ \text{TT}
(ii)The probability is 34\frac{3}{4}.
(iii)P(two tails)=14P(\text{two tails})=\frac{1}{4}
SGSchoolKaki Tip

All the outcomes must be equally likely before you use favourable over possible. A spinner with sectors of different sizes, or a biased coin, needs the probabilities of the outcomes given to you.

Think About It
When two fair coins are tossed, why is the probability of one head and one tail 12\frac{1}{2} and not 13\frac{1}{3}?
Show the worked answerHide the worked answer
outcomes: HH, HT, TH, TT\displaystyle \text{outcomes: }\text{HH},\allowbreak\ \text{HT},\allowbreak\ \text{TH},\allowbreak\ \text{TT}
list the outcomes in order
One head and one tail: HT and TH.
two of the four outcomes
P  =24=12\displaystyle P\;=\dfrac{2}{4}=\frac{1}{2}
favourable over possible
Writing only HH, HT and TT would treat three unequal outcomes as equally likely.
the common mistake
Final answerBecause HT and TH are different outcomes. The four outcomes HH, HT, TH and TT are equally likely, and two of them have one head and one tail, so the probability is two out of four, which is 12\frac{1}{2}.
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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 18 of 19