6.3 Probability

Learning objectives

6.3A Probability terminologyUse probability language including outcomes, equal likelihood, events and randomness.6.3B The probability scaleunderstand and use the probability scale P(certainty) = 1 P(impossibility) = 0 6.3C Theoretical probabilityunderstand and use estimates or measures of probability from theoretical models 6.3D Probabilities from a Venn diagramfind probabilities from a Venn diagram 6.3E Sample spaces and eventsUnderstand sample spaces and events, and determine event probabilities from a sample space.6.3F Systematic outcome lists for successive eventsList outcomes systematically for single events and for two successive events.6.3G Probabilities from previously collected dataestimate probabilities from previously collected data 6.3H Complementary eventscalculate the probability of the complement of an event using P(A′) = 1 − P(A)6.3I The addition rule of probability for mutually exclusiveuse the addition rule of probability for mutually exclusive events P(Either A or B occurring) = P(A) + P(B) when A and B are mutually exclusive 6.3J Expected frequencyunderstand and use the term ‘expected frequency’ Determine an estimate of the number of times an event with a probability of 0.4 will happen over 300 tries

An outcome is one possible result of a trial. An event is a specified outcome or set of outcomes. A random trial has an uncertain result, even though its possible outcomes may be known.

Term	Meaning
impossible	cannot occur
unlikely	probability below $\tfrac12$
evens	probability $\tfrac12$
likely	probability above $\tfrac12$
certain	must occur

Equally likely outcomes have the same chance. Do not assume outcomes are equally likely merely because they can all occur.

Probability language compares likelihood, not frequency already observed in a small sample. 'Random' does not mean every result must appear equally often.

Every probability lies from 0 to 1 inclusive. Probability 0 means impossible, $\tfrac12$ means an even chance, and 1 means certain.

Probability	Interpretation
$0$	impossible
between $0$ and $\tfrac12$	unlikely
$\tfrac12$	evens
between $\tfrac12$ and $1$	likely
$1$	certain

Fractions, decimals and percentages can name the same position: $\tfrac34=0.75=75\%$ .

A value below 0 or above 1 cannot be a probability. When marking a scale, use its subdivisions rather than estimating from the page width.

When all elementary outcomes are equally likely, theoretical probability is the number of favourable outcomes divided by the total number of possible outcomes.

Quantity	Rule
probability of event $A$	$P(A)=\dfrac{\text{favourable outcomes}}{\text{all equally likely outcomes}}$
check	$0\le P(A)\le1$

A fair six-sided die has three even faces, so $P(\text{even})=\tfrac36=\tfrac12$ .

Count outcomes, not labels. If sectors, objects or mechanisms are not equally likely, favourable-count over total-count is not justified without weighting them.

A Venn diagram partitions the universal set into disjoint regions. Read the region named by the event, add its frequencies, then divide by the total frequency when one item is chosen at random.

Event	Regions included
$A\cap B$	overlap of $A$ and $B$
$A\cup B$	every region in $A$ or $B$ , overlap once
$A'$	every region outside $A$
neither $A$ nor $B$	outside both circles

For a conditional statement such as 'given $B$ ', restrict the denominator to the total inside $B$ before counting the favourable part.

Do not count an overlap twice when finding a union, and do not omit the region outside all circles from the universal total.

A sample space lists every possible outcome of an experiment. An event is a subset of that sample space.

First coin	Second coin	Ordered outcome
H	H	HH
H	T	HT
T	H	TH
T	T	TT

For fair independent coins, the event 'exactly one head' is $\{HT,TH\}$ , so its probability is $\tfrac24=\tfrac12$ .

Outcomes such as HT and TH are different when order records successive results. A sample space must be exhaustive and must not repeat an outcome.

For two successive choices or trials, use an ordered list, two-way table or tree so every first-stage option is paired with every permitted second-stage option.

Step	Action
1	fix one first-stage outcome
2	pair it with every allowed second-stage outcome
3	repeat for each first-stage outcome
4	check restrictions such as no repetition or order ignored

If there are $m$ first-stage choices and $n$ independent second-stage choices, there are $mn$ ordered outcomes.

Do not use $mn$ when a choice is removed, repeats are forbidden, or AB and BA represent the same unordered pair; adjust the list to the actual rules.

Experimental probability uses observed results: divide the frequency of the event by the total number of trials or observations.

Quantity	Calculation
experimental probability	$\dfrac{\text{event frequency}}{\text{number of trials}}$
estimated event count in $N$ future trials	$N\times\text{experimental probability}$

A larger relevant sample usually gives a more stable estimate, but it does not guarantee the next outcome or make the estimate exact.

Use the total number represented by the data as the denominator. Do not confuse a cumulative frequency, a subgroup total or a graph reading with the whole sample.

An event and its complement cover every outcome without overlap, so their probabilities add to 1.

Required event	Complement method
not $A$	$P(A')=1-P(A)$
at least one success	$1-P(\text{no successes})$
any unlisted category	$1-\text{sum of listed category probabilities}$

If $P(\text{packed lunch})=0.79$ , then $P(\text{no packed lunch})=1-0.79=0.21$ .

Subtract from 1 only when the event being subtracted is exactly the complement of the required event. 'At least one' is complemented by 'none', not by 'exactly one'.

Mutually exclusive events cannot happen on the same trial. For such events, the probability of one or the other is the sum of their probabilities.

Condition	Addition rule
$A$ and $B$ mutually exclusive	$P(A\cup B)=P(A)+P(B)$
$A$ and $B$ may overlap	$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

If a spinner lands on exactly one colour, red and yellow are mutually exclusive, so $P(\text{red or yellow})=P(\text{red})+P(\text{yellow})$ .

The word 'or' does not by itself prove mutual exclusivity. Check whether both events can occur together before adding without an overlap correction.

Expected frequency is the long-run estimate of how often an event occurs: multiply its probability by the number of trials.

Known information	Calculation
probability $p$ , trials $n$	expected frequency $=np$
event frequency $f$ , probability $p$	estimated trials $=f/p$
expected frequency $f$ , trials $n$	probability $=f/n$

If $P(\text{blue})=0.4$ over 280 spins, the expected frequency is $280\times0.4=112$ .

Expected frequency is an estimate, not a guaranteed result. First find any missing probability, and keep trials and event counts in consistent units.

A probability tree shows successive stages. Branches leaving one node list all possible next outcomes and must sum to 1.

Step	Action
1	label every stage and branch outcome
2	complete each sibling pair or group so it sums to 1
3	multiply probabilities along one path
4	add probabilities of the distinct paths that satisfy the event

Decide whether later branch probabilities stay the same or change according to replacement, dependence and earlier outcomes.

Do not add probabilities along a path or multiply alternative paths. Multiplication means successive events on one path; addition combines disjoint completed paths.

Events are independent when knowing one occurred does not change the probability of the other. Then the probability of both is the product of their probabilities.

Relationship	Rule
independent $A$ and $B$	$P(A\cap B)=P(A)P(B)$
repeated independent event	use the same branch probabilities at each repetition
exactly one of two events	add the two orders $A B'$ and $A' B$

For at least one success in independent repetitions, the complement is often shorter: $1-P(\text{all failures})$ .

Repeated-looking trials are not automatically independent. Replacement or a separate mechanism may preserve probabilities; removing an item usually changes them.

Without replacement, the first selection changes both the total number of objects and possibly the favourable number. Later probabilities are conditional on the earlier path.

After one object is removed	Update
denominator	subtract 1 from the total
favourable numerator	subtract 1 only if the removed object was favourable
one order	multiply its successive conditional probabilities
several valid orders	calculate each order and add

From 7 red and 5 blue counters, $P(\text{two red without replacement})=\tfrac7{12}\times\tfrac6{11}$ .

Do not reuse the original fraction on the second draw. For 'one of each', include both red-then-blue and blue-then-red unless the order is fixed.

A multi-step probability problem is solved by defining the required event, choosing a representation, applying the correct rule at each stage, and checking that the answer is plausible.

Feature in the problem	Useful move
all outcomes can be listed	sample space or systematic table
successive stages	probability tree
'not' or 'at least one'	test a complement
disjoint alternatives	add path probabilities
without replacement	update conditional branches
repeated trials estimate	expected frequency $=np$

State assumptions, keep exact fractions until the end when practical, and verify $0\le P\le1$ and that sibling branches sum to 1.

Do not select a rule from a keyword alone. Translate the event first, then check for overlap, independence, replacement and whether the question asks for probability or expected count.

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—Pearson Edexcel International GCSE Mathematics (Specification A) (4MA1)Read

6.3 Probability

Learning objectives

Use precise probability language

Read and use the probability scale

Calculate theoretical probability

Find probabilities from a Venn diagram

Build a sample space and identify events

List successive outcomes systematically

Estimate probability from collected data

Use complementary events

Add mutually exclusive probabilities

Calculate expected frequency

Construct and use probability trees

Calculate with independent events

Calculate conditional probability without replacement

Model and solve mixed probability problems

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