Probability and Venn Diagrams – A Level Maths Year 1

1 Events and Probabilities

Term	Meaning	Notation
Sample space	The set of all possible outcomes.	$S$ or $\xi$
Event	A set of outcomes.	$A$, $B$
Complement	"$A$ does not happen".	$A'$
Intersection	"$A$ and $B$ both happen".	$A \cap B$
Union	"$A$ or $B$ (or both) happen".	$A \cup B$

Basic rules

$0 \leqslant \mathrm{P}(A) \leqslant 1$ $\mathrm{P}(A') = 1 - \mathrm{P}(A)$
For equally likely outcomes: $\mathrm{P}(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of outcomes}}$

The addition rule

$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$

Adding $\mathrm{P}(A)$ and $\mathrm{P}(B)$ counts the overlap $A \cap B$ twice, so it is subtracted once. A Venn diagram makes this obvious.

2 Venn Diagrams

A Venn diagram shows the sample space as a rectangle and events as overlapping circles. Fill it in from the middle outwards: the intersection first, then the "only" regions, then the region outside all circles.

Worked Example 1 — Filling in a Venn diagram

In a group of $40$ students, $18$ study French, $15$ study Spanish and $6$ study both. A student is chosen at random. Find (a) P(French or Spanish), (b) P(neither), (c) P(exactly one language).

①Both: $6$. French only: $18 - 6 = 12$. Spanish only: $15 - 6 = 9$.

②Neither: $40 - (12 + 6 + 9) = 13$.

③(a) $\mathrm{P}(F \cup S) = \dfrac{27}{40}$ (check: $\tfrac{18}{40} + \tfrac{15}{40} - \tfrac{6}{40} = \tfrac{27}{40}$ ✓)

④(b) $\mathrm{P}(\text{neither}) = \dfrac{13}{40}$. (c) $\mathrm{P}(\text{exactly one}) = \dfrac{12 + 9}{40} = \dfrac{21}{40}$.

Worked Example 2 — Using the addition rule

$\mathrm{P}(A) = 0.5$, $\mathrm{P}(B) = 0.4$ and $\mathrm{P}(A \cup B) = 0.7$. Find $\mathrm{P}(A \cap B)$ and $\mathrm{P}(A \cap B')$.

①$0.7 = 0.5 + 0.4 - \mathrm{P}(A \cap B)$, so $\mathrm{P}(A \cap B) = 0.2$.

②$\mathrm{P}(A \cap B')$ is "$A$ only": $0.5 - 0.2 = 0.3$.

3 Mutually Exclusive and Independent Events

Mutually exclusive events

$A$ and $B$ cannot both happen: $\mathrm{P}(A \cap B) = 0$, so $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B)$
On a Venn diagram the circles do not overlap

Independent events

One happening does not change the probability of the other:
$\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B)$

These are different ideas. Mutually exclusive events with non-zero probabilities can never be independent: if one happens, the other's probability drops to zero. "Mutually exclusive" is about the overlap; "independent" is about whether one event affects the other.

Worked Example 3 — Testing for independence

Using the probabilities in Worked Example 2, are $A$ and $B$ independent?

①$\mathrm{P}(A) \times \mathrm{P}(B) = 0.5 \times 0.4 = 0.2$.

②$\mathrm{P}(A \cap B) = 0.2$.

③They are equal, so $A$ and $B$ are independent.

To show independence or dependence you must calculate both sides and compare them explicitly.

Worked Example 4 — Independence in a Venn diagram

Using the French/Spanish data in Worked Example 1, are "studies French" and "studies Spanish" independent?

①$\mathrm{P}(F) \times \mathrm{P}(S) = \dfrac{18}{40} \times \dfrac{15}{40} = \dfrac{270}{1600} = 0.16875$.

②$\mathrm{P}(F \cap S) = \dfrac{6}{40} = 0.15$.

③$0.15 \neq 0.16875$, so the events are not independent.

Worked Example 5 — Finding unknowns

In a Venn diagram, $\mathrm{P}(A \text{ only}) = 0.3$, $\mathrm{P}(A \cap B) = x$, $\mathrm{P}(B \text{ only}) = 0.2$ and $\mathrm{P}(\text{neither}) = 0.3$. Find $x$ and determine whether $A$ and $B$ are independent.

①Probabilities total $1$: $0.3 + x + 0.2 + 0.3 = 1$, so $x = 0.2$.

②$\mathrm{P}(A) = 0.5$, $\mathrm{P}(B) = 0.4$, and $0.5 \times 0.4 = 0.2 = \mathrm{P}(A \cap B)$.

③So $A$ and $B$ are independent.

4 Tree Diagrams

A tree diagram shows events happening in stages. Multiply along the branches to find the probability of a path; add the probabilities of different paths that give the outcome you want. The branches from any one point must add to $1$.

Worked Example 6 — A two-stage tree

The probability that it rains on a given morning is $0.3$. If it rains, the probability that Sam's bus is late is $0.6$; if it does not rain, it is $0.2$. Find the probability that the bus is late.

①Rain and late: $0.3 \times 0.6 = 0.18$.

②Dry and late: $0.7 \times 0.2 = 0.14$.

③$\mathrm{P}(\text{late}) = 0.18 + 0.14 = 0.32$.

Rain and lateness are not independent here — the second-stage probabilities depend on the first. Tree diagrams handle this naturally; formal conditional probability notation, $\mathrm{P}(B \mid A)$, comes in Year 2.

Worked Example 7 — Without replacement

A bag holds $5$ red and $3$ blue counters. Two are taken at random without replacement. Find the probability that they are the same colour.

①Both red: $\dfrac58 \times \dfrac47 = \dfrac{20}{56}$.

②Both blue: $\dfrac38 \times \dfrac27 = \dfrac{6}{56}$.

③$\mathrm{P}(\text{same}) = \dfrac{26}{56} = \dfrac{13}{28}$.

Without replacement, the second-stage denominators drop by one and the numerators change too.

Worked Example 8 — "At least one"

A fair die is rolled four times. Find the probability of at least one six.

①The complement of "at least one six" is "no sixes".

②Rolls are independent, so $\mathrm{P}(\text{no sixes}) = \left(\tfrac56\right)^4 = \tfrac{625}{1296}$.

③$\mathrm{P}(\text{at least one six}) = 1 - \tfrac{625}{1296} = \tfrac{671}{1296} \approx 0.518$.

5 Quick Reference

Complement

$\mathrm{P}(A') = 1 - \mathrm{P}(A)$

Addition rule

$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$

Mutually exclusive

$\mathrm{P}(A \cap B) = 0$

Independent

$\mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B)$

Venn diagram

Fill from the middle outwards.

Tree diagram

Multiply along, add between paths.

Without replacement

Second-stage probabilities change.

At least one

$1 - \mathrm{P}(\text{none})$

Check

All probabilities in a diagram total $1$.

6 Practice Questions

Question 1

$\mathrm{P}(A) = 0.35$. Find $\mathrm{P}(A')$.

▶ Show solution

$1 - 0.35 = 0.65$

Question 2

$A$ and $B$ are mutually exclusive, with $\mathrm{P}(A) = 0.25$ and $\mathrm{P}(B) = 0.4$. Find $\mathrm{P}(A \cup B)$ and $\mathrm{P}\big((A \cup B)'\big)$.

▶ Show solution

$\mathrm{P}(A \cup B) = 0.25 + 0.4 = 0.65$; $\mathrm{P}\big((A \cup B)'\big) = 0.35$.

Question 3

$A$ and $B$ are independent, with $\mathrm{P}(A) = 0.6$ and $\mathrm{P}(B) = 0.3$. Find $\mathrm{P}(A \cap B)$ and $\mathrm{P}(A \cup B)$.

▶ Show solution

$\mathrm{P}(A \cap B) = 0.6 \times 0.3 = 0.18$

$\mathrm{P}(A \cup B) = 0.6 + 0.3 - 0.18 = 0.72$

Question 4

Of $50$ people, $28$ drink tea, $20$ drink coffee and $7$ drink neither. Find the probability that a person chosen at random drinks exactly one of the two.

▶ Show solution

Drink at least one: $50 - 7 = 43$. Both: $28 + 20 - 43 = 5$.

Tea only $23$, coffee only $15$, so exactly one: $38$.

$\mathrm{P} = \dfrac{38}{50} = 0.76$

Question 5

$\mathrm{P}(A) = 0.4$, $\mathrm{P}(B) = 0.25$ and $\mathrm{P}(A \cap B) = 0.1$. Are $A$ and $B$ independent? Are they mutually exclusive?

▶ Show solution

$0.4 \times 0.25 = 0.1 = \mathrm{P}(A \cap B)$, so they are independent.

$\mathrm{P}(A \cap B) \neq 0$, so they are not mutually exclusive.

Question 6

A darts player hits the target with probability $0.7$ on each throw, independently. In three throws, find the probability that she hits the target (a) exactly twice, (b) at least once.

▶ Show solution

(a) The miss can be on any of the $3$ throws: $3 \times 0.7^2 \times 0.3 = 0.441$.

(b) $1 - 0.3^3 = 1 - 0.027 = 0.973$.

Question 7

A box contains $4$ green and $6$ yellow balls. Two are drawn without replacement. Find the probability that they are different colours.

▶ Show solution

GY: $\tfrac{4}{10} \times \tfrac69 = \tfrac{24}{90}$. YG: $\tfrac{6}{10} \times \tfrac49 = \tfrac{24}{90}$.

Total: $\dfrac{48}{90} = \dfrac{8}{15}$.

Question 8

$\mathrm{P}(A) = 0.5$, $\mathrm{P}(B) = 0.3$ and $\mathrm{P}(A \cup B) = 0.65$. Draw a Venn diagram showing the probability of each region, and determine whether $A$ and $B$ are independent.

▶ Show solution

$\mathrm{P}(A \cap B) = 0.5 + 0.3 - 0.65 = 0.15$.

Regions: $A$ only $0.35$; both $0.15$; $B$ only $0.15$; neither $0.35$.

$0.5 \times 0.3 = 0.15 = \mathrm{P}(A \cap B)$, so independent.

Question 9

Of $100$ students: $4$ play football, hockey and swim; $6$ play football and hockey only; $5$ play football and swim only; $3$ play hockey and swim only; $20$ play football only; $12$ hockey only; $15$ swim only. (a) How many do none? (b) Find P(exactly two activities). (c) Are "plays football" and "swims" independent?

▶ Show solution

(a) $4 + 6 + 5 + 3 + 20 + 12 + 15 = 65$, so $35$ do none.

(b) $\dfrac{6 + 5 + 3}{100} = 0.14$

(c) $\mathrm{P}(F) = \dfrac{20 + 6 + 5 + 4}{100} = 0.35$, $\mathrm{P}(S) = \dfrac{15 + 5 + 3 + 4}{100} = 0.27$, $\mathrm{P}(F \cap S) = \dfrac{5 + 4}{100} = 0.09$.

$0.35 \times 0.27 = 0.0945 \neq 0.09$, so not independent.

Question 10

A factory has two machines. On any day, machine A breaks down with probability $0.05$ and machine B with probability $0.08$, independently.

(a) Find the probability that on a given day both break down. (b) Find the probability that at least one breaks down. (c) Find the probability that exactly one breaks down. (d) Find the probability that neither breaks down on any of $5$ consecutive days. (e) The machines share a power supply. Comment on the independence assumption.

▶ Show solution

(a) $0.05 \times 0.08 = 0.004$

(b) $\mathrm{P}(\text{neither}) = 0.95 \times 0.92 = 0.874$, so $1 - 0.874 = 0.126$.

(c) $0.05 \times 0.92 + 0.95 \times 0.08 = 0.046 + 0.076 = 0.122$ (check: $0.126 - 0.004 = 0.122$ ✓)

(d) Assuming days are independent: $0.874^5 = 0.510$ (3 s.f.).

(e) A power failure could stop both machines at once, so breakdowns would tend to happen together. Then $\mathrm{P}(\text{both})$ would be larger than $0.004$ and the events would not be independent.

Probability and Venn Diagrams · A Level Maths Year 1 · Created with MathJax