Mini-Lesson

Probability

4 missions. Guess, learn, then check. Your place is saved.

Mission 1A coin lands heads 7 times in a row. Is tails now "due"?
Mission 2Roll two dice and add them. Which total is most likely, and why?
Mission 3Roll a dice four times. Is at least one six better than evens?
Mission 41 in 100 are ill; a 90% accurate test says you are. Chance you're ill?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A coin lands heads 7 times in a row. Is tails now "due"?

Make a guess: A coin lands heads 7 times in a row. Is tails now "due"?

Probability · basics and relative frequency

The probability scale

Probability measures how likely an event is, from 0 (impossible) to 1 (certain). Evens is 0.5.
Write probabilities as fractions, decimals or percentages, never as "1 in 4" or "1 : 4" in an answer.
Words along the scale: impossible, unlikely, evens, likely, certain.
A probability can never be below 0 or above 1. If you get 1.2, something has gone wrong.

Probability · basics and relative frequency

Equally likely outcomes

When outcomes are equally likely: $P (event) = \frac{number of outcomes in the event}{total number of outcomes}$ P(event) = (number of outcomes in the event)/(total number of outcomes)
A fair dice: $P (5 or 6) = \frac{2}{6} = \frac{1}{3}$ P(5 or 6) = 2/6 = 1/3.
$P (not A) = 1 - P (A)$ P(not A) = 1 − P(A). A fair dice: $P (not 6) = 1 - \frac{1}{6} = \frac{5}{6}$ P(not 6) = 1 − 1/6 = 5/6.
The probabilities of all the possible outcomes, when they cannot happen together (mutually exclusive), add up to 1.

Worked example Probability · basics and relative frequency

A missing probability

Question

A spinner lands on red, blue, green or yellow.

P (red) = 0.35

and

P (blue) = 0.2

. Yellow is twice as likely as green. Find

P (yellow)

.

1
The four outcomes are mutually exclusive, so their probabilities sum to 1: green and yellow share $1 - 0.35 - 0.2 = 0.45$ 1 − 0.35 − 0.2 = 0.45
2
Let $P (green) = x$ P(green) = x, so $P (yellow) = 2 x$ P(yellow) = 2x and $3 x = 0.45$ 3x = 0.45
3
Solve: $x = 0.15$ x = 0.15, so $P (yellow) = 2 \times 0.15$ P(yellow) = 2 × 0.15

Answer

P (yellow) = 0.3

⚠️ Watch out

halving 0.45 gives green and yellow equal chances; the question says yellow is twice as likely.

Calculate

?A bag holds 5 red, 7 blue and 8 green counters. One counter is taken at random. Find the probability that it is not blue. Give your answer as a decimal.

Probability · basics and relative frequency

Relative frequency and expected outcomes

From an experiment: relative frequency $= \frac{number of times the event happened}{number of trials}$ = (number of times the event happened)/(number of trials). Use it as an estimate of the probability.
The more trials, the closer relative frequency tends to get to the true probability. A few trials can wobble a lot. Each flip or spin is unaffected by earlier ones: a coin has no memory.
Expected number of outcomes $= probability \times number of trials$ = probability × number of trials. It is an estimate, not a promise.
A fair (unbiased) coin, dice or spinner has equally likely outcomes. A large experiment whose relative frequencies are far from these suggests bias.

Worked example Probability · basics and relative frequency

Estimating from an experiment

Question

A spinner numbered 1 to 5 is spun 50 times and lands on 5 eighteen times. Estimate how many times it will land on 5 in 400 spins.

1
Relative frequency of 5: $\frac{18}{50} = 0.36$ 18/50 = 0.36
2
Use it as the probability: expected number $= 0.36 \times 400$ = 0.36 × 400
3
Calculate: $0.36 \times 400 = 144$ 0.36 × 400 = 144

Answer

About 144 times.

⚠️ Watch out

assuming the spinner is fair (

\frac{1}{5} \times 400 = 80

) ignores the experiment, and the experiment suggests it is biased towards 5.

Calculate

?Exam-style. Ade thinks a dice is biased. He rolls it 250 times and gets a six 70 times. He rolls it 600 more times. Use his results to estimate how many sixes he will get in the 600 rolls.

Calculate

?A frequency tree records 120 students. 70 are boys, and 28 of the boys wear glasses. 20 of the girls wear glasses. A student is picked at random. Find the probability that the student is a girl who does not wear glasses. Give your answer as a decimal.

Quick check

?Two groups test the same spinner. Group A: 9 reds in 20 spins. Group B: 124 reds in 400 spins. Which is the most reliable estimate of

P (red)

?

Mission check

?A fair coin has landed heads 7 times in a row. What is the probability that the next flip is tails?

Mission 2

Roll two dice and add them. Which total is most likely, and why?

Make a guess: Roll two dice and add them. Which total is most likely, and why?

Probability · sample spaces, two-way tables and Venn diagrams

Sample spaces

A sample space is the list of every possible outcome. List it systematically so you miss none and repeat none.
Two coins: HH, HT, TH, TT. HT and TH are different outcomes (first coin, second coin).
For two dice use a sample space diagram: a 6 by 6 grid with one cell for each of the 36 equally likely outcomes.
When outcomes are equally likely:

P(event) = \frac{number of outcomes in the event}{total number of outcomes}

Worked example Probability · sample spaces, two-way tables and Venn diagrams

Using a sample space diagram

Question

Two fair dice are rolled and the scores are added. Find the probability that the total is 10 or more.

1
Draw the 6 by 6 grid of totals: there are 36 equally likely outcomes
2
Count the cells with total 10, 11 or 12: $3 + 2 + 1 = 6$ 3 + 2 + 1 = 6 cells
3
Divide by the total number of outcomes: $\frac{6}{36}$ 6/36
4
Simplify: $\frac{6}{36} = \frac{1}{6}$ 6/36 = 1/6

Answer

\frac{1}{6}

⚠️ Watch out

there are 11 possible totals, but they are not equally likely, so the answer is not

\frac{3}{11}

.

Quick check

?Two fair coins are thrown. What is the probability of getting exactly one head and one tail?

Calculate

?A fair four-sided spinner numbered 1 to 4 is spun and a fair dice is rolled. The two scores are multiplied. Find the probability that the product is 12. Give your answer as a decimal.

Worked example Probability · sample spaces, two-way tables and Venn diagrams

Completing a two-way table

Question

80 students in Years 10 and 11 each chose school dinner, packed lunch or going home. In Year 10, 18 chose dinner and 12 chose packed lunch, out of 38 students. In Year 11, 15 chose packed lunch and 11 went home. 34 students chose dinner altogether. A student is picked at random. Find the probability that they are in Year 11 and chose school dinner.

1
Draw a two-way table: Year 10 and Year 11 as rows; dinner, packed lunch, home and total as columns
2
Use the dinner column total: Year 11 dinner $= 34 - 18 = 16$ = 34 − 18 = 16
3
Check the rows: Year 10 home $= 38 - 18 - 12 = 8$ = 38 − 18 − 12 = 8; Year 11 total $= 16 + 15 + 11 = 42 = 80 - 38$ = 16 + 15 + 11 = 42 = 80 − 38
4
"Picked at random from all 80", so divide by the grand total: $\frac{16}{80}$ 16/80

Answer

\frac{16}{80} = \frac{1}{5}

⚠️ Watch out

dividing by the Year 11 total (42) answers a different question; the student is chosen from all 80.

Calculate

?A cinema surveyed 200 people: 120 adults and 80 children. Each watched Film X, Y or Z. 45 adults watched X and 30 adults watched Z. 22 children watched Y and 24 children watched Z. One of the 200 people is chosen at random. Find the probability that the person is a child who watched Film X. Give your answer as a decimal.

Probability · sample spaces, two-way tables and Venn diagrams

Venn diagrams

A Venn diagram sorts items into overlapping circles inside a rectangle. The overlap is in both sets; outside the circles is in neither.
Fill in the overlap first, then the "only" parts, then outside. All regions sum to the total.
Probability = (number in your region) ÷ (total in the rectangle).
The universal set (rectangle) is $ξ$ xi or $E$ ℰ; $A \cap B$ A ∩ B is the overlap; $A \cup B$ A ∪ B is A or B or both; $A^{'}$ A' is not A.
$P (A or B) = P (A) + P (B) - P (A and B)$ P(A or B) = P(A) + P(B) − P(A and B) (overlap counted twice).

Calculate

?A sports club has 80 members. 45 swim, 38 cycle and 9 do neither. A member is chosen at random. Find the probability that the member both swims and cycles. Give your answer as a decimal.

Quick check

?A Venn diagram shows 50 people and the sets

T

(likes tea) and

C

(likes coffee): 20 in

T

only, 8 in

T \cap C

, 12 in

C

only and 10 outside both. A person is chosen at random. Find the probability that they are in

T^{'} \cap C

.

Mission check

?Roll two dice and add them. Which total is most likely, and why?

Mission 3

Roll a dice four times. Is at least one six better than evens?

Make a guess: Roll a dice four times. Is at least one six better than evens?

Probability · combined events and tree diagrams

The AND rule and the OR rule

Events are independent if one happening does not change the probability of the other (two dice, a coin and a spinner, repeated spins).
AND for independent events: multiply. $P (A and B) = P (A) \times P (B)$ P(A and B) = P(A) × P(B)
OR for mutually exclusive events (they cannot both happen): add. $P (A or B) = P (A) + P (B)$ P(A or B) = P(A) + P(B)
$P (not A) = 1 - P (A)$ P(not A) = 1 − P(A), so P(at least one) $= 1 -$ = 1 − P(none).

Quick check

?Two fair dice are rolled. What is the probability that both show a six?

Probability · combined events and tree diagrams

Tree diagrams

A tree diagram shows each stage as a set of branches. Write the outcome at the end of each branch and its probability beside it.
The branches from any one point add to 1.
Multiply along a path to get the probability of that combined outcome.
Add the paths that give the event you want (the paths are mutually exclusive).
Check: all the end probabilities together add to 1.

Worked example Probability · combined events and tree diagrams

Exactly one of two events

Question

The probability that a train is late on any day is 0.2, independently of other days. Find the probability that it is late on exactly one of two days.

1
Draw a tree: Day 1 then Day 2, each with Late 0.2 and On time 0.8
2
Find the paths with exactly one late: (Late, On time) and (On time, Late)
3
Multiply along each path: $0.2 \times 0.8 = 0.16$ 0.2 × 0.8 = 0.16 and $0.8 \times 0.2 = 0.16$ 0.8 × 0.2 = 0.16
4
Add the two paths: $0.16 + 0.16 = 0.32$ 0.16 + 0.16 = 0.32

Answer

0.32

⚠️ Watch out

stopping at 0.16 misses the second path; the late day could be either day.

Calculate

?The probability that a seed germinates is 0.7, independently of other seeds. Three seeds are planted. Find the probability that at least one of them germinates. Give your answer as a decimal.

Sort it

?Add or multiply? Sort each calculation.

Tap a statement, then tap the group it belongs in. Then press Check.

Probability · combined events and tree diagrams

Without replacement

With replacement, the item goes back: the second-stage probabilities are the same as the first.
Without replacement, the events are dependent: the second branches change because there is one fewer item.
Take one away from the total on every second branch, and from the colour already taken on its own branch.
Then multiply along and add the paths as before.

Worked example Probability · combined events and tree diagrams

Two bulbs without replacement

Question

A box has 10 bulbs and 3 are faulty. Two are taken at random without replacement. Find the probability that exactly one is faulty.

1
First pick: P(faulty) $= \frac{3}{10}$ = 3/10, P(good) $= \frac{7}{10}$ = 7/10
2
Second pick, 9 bulbs left: after a faulty one, $\frac{2}{9}$ 2/9 faulty and $\frac{7}{9}$ 7/9 good; after a good one, $\frac{3}{9}$ 3/9 faulty and $\frac{6}{9}$ 6/9 good
3
Paths with exactly one faulty: $\frac{3}{10} \times \frac{7}{9} = \frac{21}{90}$ 3/10 × 7/9 = 21/90 and $\frac{7}{10} \times \frac{3}{9} = \frac{21}{90}$ 7/10 × 3/9 = 21/90
4
Add: $\frac{21}{90} + \frac{21}{90} = \frac{42}{90}$ 21/90 + 21/90 = 42/90

Answer

\frac{7}{15}

⚠️ Watch out

keeping

\frac{3}{10}

on the second pick treats it as with replacement.

Calculate

?A bag has 5 red and 3 green counters. Two counters are taken at random without replacement. Find the probability that they are the same colour. Give your answer to 3 significant figures.

Mission check

?Roll a dice four times. Is at least one six better than evens?

Mission 4

1 in 100 are ill; a 90% accurate test says you are. Chance you're ill?

Make a guess: 1 in 100 are ill; a 90% accurate test says you are. Chance you're ill?

Probability · conditional probability

"Given that" shrinks the total (Higher)

A conditional probability is a probability given that something has already happened.
"Given B" means you only look at the outcomes in B. B's total becomes the new denominator.

P (A given B) = \frac{number in both A and B}{number in B}

The formula $P (A and B) = P (A given B) \times P (B)$ P(A and B) = P(A given B) × P(B) must be known. Some questions write $P (A given B)$ P(A given B) as $P (A ∣ B)$ P(A | B).
Events are independent if $P (A given B) = P (A)$ P(A given B) = P(A).

Worked example Probability · conditional probability

Conditional probability from a two-way table (Higher)

Question

120 people at a gym chose a class. Yoga: 34 women and 14 men. Spin: 30 women and 42 men. A person is chosen at random from the spin class. Find the probability that the person is a woman.

1
"From the spin class" is the condition: total spin $= 30 + 42 = 72$ = 30 + 42 = 72
2
Count spin-class women: 30
3
Divide by the spin total, not by 120: $\frac{30}{72}$ 30/72
4
Simplify: $\frac{30}{72} = \frac{5}{12}$ 30/72 = 5/12

Answer

\frac{5}{12}

⚠️ Watch out

\frac{30}{120}

is P(woman and spin), and

\frac{30}{64}

is P(spin given woman); read which group the person comes from.

Quick check

?Use the gym table: yoga 34 women, 14 men; spin 30 women, 42 men. A man is chosen at random. What is the probability that he does spin?

Probability · conditional probability

Conditional probability from a Venn diagram (Higher)

"Given B": cover everything outside circle B. Circle B's total is the new denominator.
The numerator is the overlap (the part of B that is also in A).
If you are given set sizes and "neither", find the overlap first: overlap = A + B − (total − neither).

Calculate

?In a group of 50 students, 21 study history, 28 study geography and 8 study neither. A student who studies geography is chosen at random. Find the probability that this student also studies history. Give your answer as a decimal.

Probability · conditional probability

Reversing a tree with expected frequencies (Higher)

The second branches of a tree are already conditional: "late, given it rained".
To reverse the condition ("rained, given late"), imagine a round number, such as 100 or 1,000, going through the tree. These are expected frequencies.
Add up every path that ends in the given outcome: that is the new total.
Divide the path you want by that total.

Worked example Probability · conditional probability

Rain, given the bus was late (Higher)

Question

On 30% of days it rains. On a rainy day the probability the bus is late is 0.5; on a dry day it is 0.1. The bus was late. Find the probability that it was raining.

1
Imagine 100 days: 30 rainy, 70 dry
2
Late days: rainy $30 \times 0.5 = 15$ 30 × 0.5 = 15, dry $70 \times 0.1 = 7$ 70 × 0.1 = 7
3
The condition is "late": $15 + 7 = 22$ 15 + 7 = 22 late days in total
4
Of these, 15 were rainy: $\frac{15}{22}$ 15/22

Answer

\frac{15}{22} \approx 0.682

⚠️ Watch out

0.5 is P(late given rain), not P(rain given late); the condition is the other way round.

Calculate

?Machine A makes 60% of a factory's items and 5% of them are faulty. Machine B makes the rest and 10% of them are faulty. An item is chosen at random and is faulty. Find the probability that it was made by machine B. Give your answer to 3 significant figures.

Quick check

?100 students: 60 girls and 40 boys. 24 of the girls and 16 of the boys walk to school. Are the events "walks to school" and "is a girl" independent?

Mission check

?1 in 100 people are ill. A 90% accurate test says you are. What is the chance you are ill?

Recap

Big ideas

Probability runs from 0 (impossible) to 1 (certain); $P (not A) = 1 - P (A)$ P(not A) = 1 − P(A); all mutually exclusive outcomes sum to 1.
Relative frequency estimates probability, and gets more reliable with more trials. Expected number = probability × trials.
A fair coin has no memory: past results never make an outcome "due".
List outcomes systematically; for two events use a grid. P(event) = favourable ÷ total, only when the outcomes are equally likely.
In a two-way table or Venn diagram, find the missing values from the totals first, then divide by the total you are choosing from.
$A \cap B$ A ∩ B is the overlap, $A \cup B$ A ∪ B is either or both, and $A^{'}$ A' is everything outside $A$ A.
AND (independent events): multiply. OR (mutually exclusive events): add. P(at least one) $= 1 -$ = 1 − P(none).
On a tree diagram, multiply along the branches and add the paths you want; each set of branches adds to 1.
Without replacement, the second-stage probabilities change: one fewer in the total and in the group already taken.
"Given B" means choose only from B: B's total is the denominator.
From a two-way table or Venn diagram, divide the overlap by the total of the given group; to reverse a tree, use expected frequencies of a round number.
$P (A and B) = P (A given B) \times P (B)$ P(A and B) = P(A given B) × P(B); $A$ A and $B$ B are independent if $P (A given B) = P (A)$ P(A given B) = P(A).

🏆

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⭐⭐⭐

You've worked through Probability for Eduqas GCSE Maths. 🎉

Mission 1A coin lands heads 7 times in a row. Is tails now "due"?
Mission 2Roll two dice and add them. Which total is most likely, and why?
Mission 3Roll a dice four times. Is at least one six better than evens?
Mission 41 in 100 are ill; a 90% accurate test says you are. Chance you're ill?

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Specification points and sources for this lesson are listed at the bottom of the page.