2.03StatisticsCore

Probability Laws

Q: What does the vertical bar mean?

"Given that". P(A B) is the probability of A when you already know B has happened.

If 30% of students study Physics, 40% study Chemistry and 15% study both, what is the chance a randomly chosen student studies neither? Or, given that someone studies Chemistry, how likely is it they also study Physics? Questions like these need precise rules — the addition rule, the definition of independence, and conditional probability — and a diagram to keep track of which region is which.

45 min Probability

The big picture

Almost every probability problem at A-Level is solved by drawing the right diagram and applying three or four laws. Venn diagrams handle overlapping events; tree diagrams handle events in sequence; two-way tables handle counts. The addition rule corrects for double-counting the overlap. Independence has a strict mathematical meaning, $\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$ , which you test rather than assume. And conditional probability, $\mathrm{P}(A \mid B)$ , restricts attention to the outcomes where $B$ has happened — an idea that is simple to state and famously easy to get backwards. These laws are the foundation for probability distributions and hypothesis testing in the rest of the chapter.

What you'll be able to do

Use set notation and Venn diagrams to represent events
Apply $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$
Understand and test for mutually exclusive and independent events
Calculate conditional probabilities using diagrams and $\mathrm{P}(A \mid B) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$

Notation and Venn diagrams

$A \cap B$ (" $A$ and $B$ ") is the overlap; $A \cup B$ (" $A$ or $B$ or both") is everything in either; $A^{\prime}$ ("not $A$ ") is everything outside $A$ , with $\mathrm{P}(A^{\prime}) = 1 - \mathrm{P}(A)$ .

In a Venn diagram, fill in the intersection first, then the "only" regions, then the outside, so that the total is 1 (or the total count).

Example 1

\mathrm{P}(P) = 0.3

\mathrm{P}(C) = 0.4

and

\mathrm{P}(P \cap C) = 0.15

. Find the probability a student studies neither subject.

1Intersection

0.15

. Physics only:

0.3 - 0.15 = 0.15

. Chemistry only:

0.4 - 0.15 = 0.25

2Total inside the circles:

0.15 + 0.15 + 0.25 = 0.55

Answer

1 - 0.55 = 0.45

Tip — Always start a Venn diagram from the intersection. Starting from $\mathrm{P}(A)$ and $\mathrm{P}(B)$ as "only" regions double-counts the overlap.

The addition rule and mutually exclusive events

Adding $\mathrm{P}(A)$ and $\mathrm{P}(B)$ counts the overlap twice, so subtract it once: $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$ .

Events are $\textbf{mutually exclusive}$ if they cannot both happen: $\mathrm{P}(A \cap B) = 0$ . Then $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B)$ , and their Venn circles do not overlap.

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

The addition rule.

Independent events

Events are $\textbf{independent}$ if the occurrence of one does not change the probability of the other. Mathematically: $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B)$ .

To test for independence, calculate both sides and compare. Do not assume events are independent because they "seem unrelated" — check the numbers.

Mutually exclusive and independent are very different: mutually exclusive events with non-zero probabilities are never independent, since knowing one happened tells you the other did not.

A, B \text{ independent} \iff \mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B)

Definition of independence.

Example 1Are studying Physics and studying Chemistry independent in the example above?

\mathrm{P}(P)\mathrm{P}(C) = 0.3 \times 0.4 = 0.12

\mathrm{P}(P \cap C) = 0.15 \neq 0.12

AnswerNot independent — students taking one are more likely to take the other.

Conditional probability

$\mathrm{P}(A \mid B)$ means the probability of $A$ given that $B$ has occurred. You restrict the sample space to $B$ , so $\mathrm{P}(A \mid B) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$ .

Rearranged, $\mathrm{P}(A \cap B) = \mathrm{P}(B)\,\mathrm{P}(A \mid B)$ — this is exactly what you do when you multiply along the branches of a tree diagram.

If $A$ and $B$ are independent, then $\mathrm{P}(A \mid B) = \mathrm{P}(A)$ : knowing $B$ happened changes nothing.

\mathrm{P}(A \mid B) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}

Conditional probability.

Example 1A test for a condition affecting 2% of people is 95% accurate for those with it and 90% accurate for those without. A person tests positive. Find the probability they have the condition.

1P(condition and positive)

= 0.02 \times 0.95 = 0.019

2P(no condition and positive)

= 0.98 \times 0.10 = 0.098

3P(positive)

= 0.019 + 0.098 = 0.117

4P(condition | positive)

= \dfrac{0.019}{0.117}

Answer

0.162

(3 s.f.)

Why this works

Despite a "95% accurate" test, a positive result means only about a 16% chance of having the condition, because the condition is rare and false positives from the large healthy group swamp the true positives. Confusing $\mathrm{P}(\text{positive} \mid \text{condition})$ with $\mathrm{P}(\text{condition} \mid \text{positive})$ is a real and serious error.

Think like an examiner

Use correct set notation in working and in answers.
Show a numerical test when asked whether events are independent, with a conclusion.
Draw a tree diagram or Venn diagram and label it clearly.
Identify the "given" event as the denominator in conditional probability.

Common misconceptions

Adding

\mathrm{P}(A)

and

\mathrm{P}(B)

for overlapping events.

Subtract

\mathrm{P}(A \cap B)

to avoid double-counting.

Assuming independent means mutually exclusive.

Independent:

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

. Mutually exclusive:

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

Dividing by

\mathrm{P}(A)

for

\mathrm{P}(A \mid B)

Divide by the probability of the given event,

\mathrm{P}(B)

Keeping probabilities the same on second branches when sampling without replacement.

Adjust the second-stage probabilities for the item removed.

Probability laws

\mathrm{P}(A^{\prime}) = 1 - \mathrm{P}(A)

Complement.

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

Addition rule.

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

Independence.

\mathrm{P}(A \mid B) = \tfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}

Conditional.

Stretch yourself

Events $A$ and $B$ are independent, with $\mathrm{P}(A) = 0.4$ and $\mathrm{P}(A \cup B) = 0.7$ . Find $\mathrm{P}(B)$ and $\mathrm{P}(A^{\prime} \mid B^{\prime})$ .

Hint — Write $\mathrm{P}(A \cap B) = 0.4\,\mathrm{P}(B)$ and substitute into the addition rule.

Reveal solution

SolutionLet

\mathrm{P}(B) = p

. Independence gives

\mathrm{P}(A \cap B) = 0.4p

. Addition rule:

0.7 = 0.4 + p - 0.4p

, so

0.6p = 0.3

and

p = 0.5

. For the second part,

A^{\prime}

and

B^{\prime}

are also independent (complements of independent events are independent), so

\mathrm{P}(A^{\prime} \mid B^{\prime}) = \mathrm{P}(A^{\prime}) = 0.6

. Check directly:

\mathrm{P}(A^{\prime} \cap B^{\prime}) = 1 - \mathrm{P}(A \cup B) = 0.3

and

\mathrm{P}(B^{\prime}) = 0.5

, so

\dfrac{0.3}{0.5} = 0.6

✓.

\blacksquare

Questions students ask

How do I know which diagram to use?

Venn diagrams for overlapping characteristics, tree diagrams for events in stages or sequences, two-way tables for counts in two categories.

Can two events be both independent and mutually exclusive?

Only if at least one has probability zero. Otherwise mutually exclusive events are dependent.

What does the vertical bar mean?

"Given that".

\mathrm{P}(A \mid B)

is the probability of

A

when you already know

B

has happened.

Why multiply along tree branches?

Each branch probability is conditional on the path so far, and

\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B \mid A)

Key takeaways

Use set notation: $\cap$ for and, $\cup$ for or, $^{\prime}$ for not.
Addition rule subtracts the overlap; mutually exclusive events have zero overlap.
Independence means $\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$ — test it numerically.
$\mathrm{P}(A \mid B) = \mathrm{P}(A \cap B)/\mathrm{P}(B)$ ; restrict to the given event.
Tree diagrams multiply conditional probabilities along branches.