Mini-Lesson

Probability

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

Mission 1Is a dice total of 7 linked to what the first die shows?
Mission 2A 95% accurate test says you have a rare disease. How likely is it?

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

Mission 1

Is a dice total of 7 linked to what the first die shows?

Make a guess: Is a dice total of 7 linked to what the first die shows?

Probability · Venn diagrams and independence

Events and Venn diagrams

$P (A^{'})$ P(A') is the probability that $A$ A does not happen: $P (A^{'}) = 1 - P (A)$ P(A') = 1 − P(A).
$P (A \cap B)$ P(A ∩ B) means " $A$ A and $B$ B"; $P (A \cup B)$ P(A ∪ B) means " $A$ A or $B$ B or both".
In a Venn diagram each region holds a frequency or a probability. Fill the overlap first, then the rest of each circle, then the outside. All the regions add up to 1 (or the total).

Worked example Probability · Venn diagrams and independence

Completing a Venn diagram

Question

P (A) = 0.5

,

P (B) = 0.35

and

P (A \cap B) = 0.15

. Find

P (A \cup B)

and the probability that exactly one of

A

and

B

happens.

1
Overlap first: $A \cap B$ A ∩ B region $= 0.15$ = 0.15
2
$A$ A only: $0.5 - 0.15 = 0.35$ 0.5 − 0.15 = 0.35; $B$ B only: $0.35 - 0.15 = 0.2$ 0.35 − 0.15 = 0.2
3
Outside both: $1 - (0.35 + 0.15 + 0.2) = 0.3$ 1 − (0.35 + 0.15 + 0.2) = 0.3
4
Read off: $P (A \cup B) = 0.35 + 0.15 + 0.2 = 0.7$ P(A ∪ B) = 0.35 + 0.15 + 0.2 = 0.7; exactly one $= 0.35 + 0.2 = 0.55$ = 0.35 + 0.2 = 0.55

Answer

P (A \cup B) = 0.7

and

P (exactly one) = 0.55

.

⚠️ Watch out

P (A) + P (B) = 0.85

counts the overlap twice. Subtract it once.

Calculate

?In a group of 60 students, 32 study Biology, 25 study Chemistry and 14 study neither. A student is chosen at random. Find the probability that they study exactly one of the two subjects. Give your answer to 3 decimal places.

Probability · Venn diagrams and independence

Mutually exclusive or independent?

Mutually exclusive: the events cannot both happen. $P (A \cap B) = 0$ P(A ∩ B) = 0, so $P (A \cup B) = P (A) + P (B)$ P(A ∪ B) = P(A) + P(B). The circles do not overlap.
Independent: one event happening does not change the probability of the other.

A, B independent ⟺ P (A \cap B) = P (A) \times P (B)

To test for independence, work out both sides and compare.
Two events with non-zero probabilities that are mutually exclusive can never be independent: if one happens, the other cannot.

Worked example Probability · Venn diagrams and independence

Testing for independence with a two-way table

Question

Of 200 commuters, 80 cycle to work, 50 are under 30, and 20 are under-30 cyclists. Are the events

C

"cycles" and

U

"under 30" independent?

1
Convert to probabilities: $P (C) = \frac{80}{200} = 0.4$ P(C) = 80/200 = 0.4, $P (U) = \frac{50}{200} = 0.25$ P(U) = 50/200 = 0.25
2
Joint probability: $P (C \cap U) = \frac{20}{200} = 0.1$ P(C ∩ U) = 20/200 = 0.1
3
Multiply: $P (C) \times P (U) = 0.4 \times 0.25 = 0.1$ P(C) × P(U) = 0.4 × 0.25 = 0.1
4
Compare: $P (C \cap U) = P (C) \times P (U)$ P(C ∩ U) = P(C) × P(U)

Answer

yes,

C

and

U

are independent.

⚠️ Watch out

always state the comparison and conclude in words. "0.1 = 0.1" alone is not a conclusion.

Quick check

?

A

and

B

are independent, with

P (A) = 0.3

and

P (B) = 0.5

. Find the probability that neither

A

nor

B

happens.

Probability · Venn diagrams and independence

Sample spaces, trees and three events

A sample space diagram lists every equally likely outcome; count the outcomes in an event and divide by the total.
On a tree diagram, multiply along the branches for "this then that", and add the probabilities of separate paths.
For independent events the branch probabilities stay the same on every stage.
With three events, list every path that fits (for "exactly two", there are three) before adding.

Calculate

?On a drive, Kai passes three sets of traffic lights. They show green independently with probabilities 0.6, 0.5 and 0.8. Find the probability that exactly two of the three show green. Give an exact decimal.

Sort it

?Classify each pair of events.

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

Mission check

?Is a dice total of 7 linked to what the first die shows?

Mission 2

A 95% accurate test says you have a rare disease. How likely is it?

Make a guess: A 95% accurate test says you have a rare disease. How likely is it?

Probability · conditional probability

Conditional probability (A-level only)

$P (A ∣ B)$ P(A | B) means "the probability of $A$ A given that $B$ B has happened". You restrict attention to $B$ B only.

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (A \cap B) = P (A) \times P (B ∣ A)

From a two-way table: divide by the total of the row or column you are given, not the grand total.
$P (A ∣ B)$ P(A | B) and $P (B ∣ A)$ P(B | A) are usually different: check which event is "given".
Addition law: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Independence means $P (A ∣ B) = P (A)$ P(A | B) = P(A).

Worked example Probability · conditional probability

Two-way table: which way round?

Question

A sixth form has 70 Year 12 students (7 drive) and 50 Year 13 students (18 drive). One student is chosen at random. Find

P (drives ∣ Year 13)

and

P (Year 13 ∣ drives)

.

1
Given Year 13, the total is 50: $P (drives ∣ Y13) = \frac{18}{50} = 0.36$ P(drives | Y13) = 18/50 = 0.36
2
Given a driver, the total is $7 + 18 = 25$ 7 + 18 = 25: $P (Y13 ∣ drives) = \frac{18}{25} = 0.72$ P(Y13 | drives) = 18/25 = 0.72
3
Check with the formula: $\frac{18/120}{25/120} = \frac{18}{25}$ (18/120)/(25/120) = 18/25

Answer

P (drives ∣ Y13) = 0.36

and

P (Y13 ∣ drives) = 0.72

.

⚠️ Watch out

\frac{18}{120} = 0.15

is

P (drives \cap Y13)

, not a conditional probability.

Quick check

?

P (A) = 0.45

,

P (B) = 0.3

and

P (A \cap B) = 0.12

. Find

P (B ∣ A)

.

Calculate

?Events

A

and

B

have

P (A) = 0.6

,

P (B) = 0.4

and

P (A \cup B) = 0.75

. Find

P (A ∣ B)

. Give an exact decimal.

Probability · conditional probability

Conditional probability on tree diagrams

On a tree, the second-stage branches are conditional: they are $P (B ∣ A)$ P(B | A) and $P (B ∣ A^{'})$ P(B | A').
Multiply along a path: $P (A \cap B) = P (A) \times P (B ∣ A)$ P(A ∩ B) = P(A) × P(B | A).
To find $P (B)$ P(B), add every path that ends in $B$ B.
For "given $B$ B, find the chance of $A$ A": the path through $A$ A and $B$ B, divided by the total of all paths ending in $B$ B.

Worked example Probability · conditional probability

Working backwards on a tree

Question

On any school day

P (rain) = 0.3

. Sam is late with probability 0.2 if it rains and 0.05 if it is dry. Given that Sam is late, find the probability that it rained.

1
Path rain then late: $0.3 \times 0.2 = 0.06$ 0.3 × 0.2 = 0.06
2
Path dry then late: $0.7 \times 0.05 = 0.035$ 0.7 × 0.05 = 0.035
3
Add the late paths: $P (late) = 0.06 + 0.035 = 0.095$ P(late) = 0.06 + 0.035 = 0.095
4
Restrict to "late": $P (rain ∣ late) = \frac{0.06}{0.095} = \frac{12}{19}$ P(rain | late) = 0.06/0.095 = 12/19

Answer

P (rain ∣ late) = \frac{12}{19} \approx 0.632

.

⚠️ Watch out

0.2 is

P (late ∣ rain)

, the other way round. The question gives "late" as the condition.

Calculate

?A factory's machine A makes 60% of its parts and machine B makes 40%. 3% of A's parts and 8% of B's parts are faulty. A part is chosen at random and is faulty. Find the probability that it was made by machine B. Give an exact decimal.

Probability · conditional probability

Modelling with probability

A probability model rests on assumptions: outcomes equally likely (a fair coin or die), trials independent, probabilities constant, or values estimated from past data.
Critique each assumption in context: is it realistic here?
Say how a more realistic assumption would change the answer. If wet days tend to follow wet days, the chance of two wet days in a row is higher than the independent model gives.
Probabilities from past data are only estimates; bigger samples give better estimates.

Sort it

?Is each modelling assumption reasonable, or questionable?

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

Mission check

?A 95% accurate test says you have a rare disease. How likely is it?

1% have the disease; the test is positive for 95% of people with it and negative for 95% of people without it.

Recap

Big ideas

Venn diagrams: fill the overlap first; $P (A^{'}) = 1 - P (A)$ P(A') = 1 − P(A).
Mutually exclusive: $P (A \cap B) = 0$ P(A ∩ B) = 0. Independent: $P (A \cap B) = P (A) \times P (B)$ P(A ∩ B) = P(A) × P(B). Test by computing both sides.
Trees and sample spaces: multiply along paths, add separate paths, list every path that fits.
$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ P(A | B) = (P(A ∩ B))/(P(B)): restrict to the "given" event and divide by its total.
On a tree, second branches are conditional; to reverse a condition, divide one path by the sum of all paths ending in the given event.
Every probability model rests on assumptions (fair, independent, constant): critique them in context and say how the answer would change.

🏆

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You've worked through Probability for AQA A-level Mathematics (7357). 🎉

Mission 1Is a dice total of 7 linked to what the first die shows?
Mission 2A 95% accurate test says you have a rare disease. How likely is it?

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