Probability

— AS & A Level Mathematics · Probability & Statistics 1 · Chapter 4 —

★ What you must be able to do: find probabilities by counting equally likely outcomes · use addition & multiplication laws · know mutually exclusive vs independent · test for independence · calculate & use conditional probability.

Warm-up (prerequisite check)

Check your skills recall

1) 6s in 180 rolls: 16 × 180 = 30

2) Total 4 on two dice: (1,3),(2,2),(3,1) → 336 = 112

3) A={1,2,5}, B′={2,4,5} ⇒ B={1,3}, A′={3,4} ⇒ ℰ={1,2,3,4,5}. A∪B′ = {1,2,4,5} → n = 4; A′∩B = {3} → n = 1

1. Experiments, Events & Outcomes

Probability = how likely an event is, on a scale from 0 (impossible) to 1 (certain). Written P(event) as a fraction, decimal or %. Bigger P ⇒ more likely.

Why it matters: insurance is based on risk (the probability of events); governments weigh the probable benefits of their spending.

the probability scale — every probability lives between 0 and 1

Experiment = something we do (roll a die). Outcome / elementary event = one result of it. Event = a combination of outcomes.

Example:

Roll a fair die → 6 outcomes: 1, 2, 3, 4, 5, 6.
Event “odd number” → 3 favourable outcomes: 1, 3, 5.

Random selection & equiprobable events

Selecting at random = every object has the same chance (called fair / unbiased) ⇒ selections are equally likely = equiprobable.

Key Point 4.1 : one object chosen at random from n objects
P(any particular object) = 1n

Key Point 4.2 : P(event) = number of favourable equally likely outcomestotal number of equally likely outcomes

Tip: “particular” = ONE object. A particular boy, girl or student each has 119 chance.

Example: pick 1 student from 19 (11 boys, 8 girls)

Event / outcome	Probability	Why
any particular boy / girl / student	119	all 19 outcomes equally likely
selecting a boy	1119	11 of the 19 favourable
selecting a girl	819	8 of the 19 favourable

Exhaustive events & complements

A set of events containing all possible outcomes is exhaustive. Special case: event A and its complement A′ (“not A”) — exactly one must happen, so their probabilities add to 1.

Key Point 4.3 : P(A) + P(not A) = 1 or P(A) + P(A′) = 1 ⇒ P(A′) = 1 − P(A)

Experiment	A	A′	Probabilities
Toss a fair coin	heads	tails	12 + 12 = 1
Roll a fair die	less than 2	2 or more	16 + 56 = 1
Play chess	win	not win	P(win) + P(not win) = 1

Trick: “at least one…” / “has to stop at any…” questions → find the complement (none) and do 1 − P(none). Much shorter!

Trials & expectation

Each repeat of an experiment = a trial. Relative frequency = proportion of trials in which the event occurred → an estimate of P(event). Knowing P, we predict how often it should happen = the expectation.

Key Point 4.4 : in n trials, event A is expected to occur n × P(A) times

Worked Example 4.1expectation

P(rain on any particular day) = 0.2. On how many days of a 365-day year is rain not expected?

n = 365 , P(no rain) = 1 − 0.2 = 0.8365 × 0.8= 292 days

Multiply the probability of the event by the number of days (trials).

Explore 4.1 (illustration): toss a coin 10, 20, 30… times — the proportion of heads wobbles at first, then settles near 0.5. More trials ⇒ closer to the expected value.

Note: Expectation is an average, not a promise. Getting 1 loss more than expected doesn't mean a team under-performed — results naturally vary around the expected value.

2. Mutually Exclusive Events & the Addition Law

To find P(A or B) we may just add the probabilities — but only if A and B are mutually exclusive.

Mutually exclusive = no common favourable outcomes ⇒ they cannot both happen ⇒ P(A and B) = 0 (intersection is the empty set ∅).

Roll a die	Common outcomes?	Mutually exclusive?
even {2,4,6} & factor of 5 {1,5}	none	YES ✓ P(even or factor of 5) = P(even) + P(factor of 5)
odd {1,3,5} & factor of 5 {1,5}	1 and 5	NO ✗ P(odd or factor of 5) ≠ P(odd) + P(factor of 5)

Key Point 4.5 : mutually exclusive ⇒ P(A or B) = P(A) + P(B)
extends: P(A or B or C or …) = P(A) + P(B) + P(C) + …

Tip: “A or B” = A occurs, or B occurs, or both.
A ∪ B means “A or B”; A ∩ B means “A and B”.

Venn diagrams & set notation

Venn diagrams show favourable outcomes, their counts or their probabilities. n(A) = number of outcomes favourable to A. A′ = outcomes not in A. The rectangle ℰ is the universal set = the complete possibility space.

shaded = A (unshaded = A′, “not A”)

shaded = A ∪ B (“A or B”)
unshaded = (A∪B)′ “neither A nor B”

shaded = A ∩ B (“A and B”)
unshaded = (A∩B)′ “not both A and B”

Key Point 4.6 : mutually exclusive ⇒ P(A ∪ B) = P(A) + P(B) | A, B mutually exclusive ⇔ P(A ∩ B) = 0 ⇔ A ∩ B = ∅

Events that are NOT mutually exclusive

Count the favourable outcomes by enumerating — never count the overlap twice. In general (this also works when the overlap is 0):

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Key Point 4.7 : P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

A + B counts the middle twice ⇒ subtract A ∩ B once

Worked Example 4.2 (a)which pairs are mutually exclusive?

Pick one digit from 1-9. A: multiple of 3 · B: factor of 8 · C: prime.

ℰ = {1,2,3,4,5,6,7,8,9}

A = {3,6,9} → P(A) = 39 B = {1,2,4,8} → P(B) = 49 C = {2,3,5,7} → P(C) = 49

A ∩ B = ∅

A ∩ C = {3}

B ∩ C = {2}

Worked Example 4.2 (a) — answeraddition law

A∩B = ∅ ⇒ A and B are mutually exclusive. A∩C ≠ ∅ and B∩C ≠ ∅ ⇒ the other two pairs are not.

P(A ∪ B) = P(A) + P(B) (addition law OK — mutually exclusive)= 39 + 49= 79

Worked Example 4.2 (b)not mutually exclusive → subtract the overlap

Use the lists / Venn diagrams above.

i) P(A ∪ C)n(A∪C) = n(A)+n(C)−n(A∩C)= 3 + 4 − 1 = 6P(A∪C) = 39+49−19= 69 = 23

ii) P(B ∪ C)n(B∪C) = n(B)+n(C)−n(B∩C)= 4 + 4 − 1 = 7P(B∪C) = 49+49−19= 79

We subtract the common element because it was counted in both n(A) and n(C).

Worked Example 4.3two-circle Venn with percentages

Survey: 50% own a desktop (D), 60% own a laptop (L), 15% own both. % owning neither?

outside the circles: whole = 1; D total = 0.5; L total = 0.6

p = 0.5 − 0.15 = 0.35 (desktop only)q = 0.6 − 0.15 = 0.45 (laptop only)x = 1 − (0.35 + 0.15 + 0.45)x = 0.05 = 5%

∴ 5% own neither a desktop nor a laptop. (∴ means “therefore”)

Method: Fill the overlap (both) first, then subtract it from each circle total to get “only” regions, finally outside = whole − everything inside.

Worked Example 4.4three-circle Venn

40 children chose fruits: apples (A), bananas (B), cherries (C). Find P(likes apples or bananas).

numbers = children in each region (total 40)

n(A) = 6+7+3+1 = 17n(B) = 3+1+2+2 = 8n(A∩B) = 3+1 = 4P(A∪B) = 1740+840−440= 2140n(A∪B) ≠ 17+8 because A∩B ≠ ∅. Alternatively add all regions in the A and B circles: 6+7+3+1+2+2 = 21.

✓ Quick Recap — Addition law

mutually exclusive: P(A∪B) = P(A) + P(B)
any two events: P(A∪B) = P(A) + P(B) − P(A∩B)
fill Venn diagrams from the middle outwards

3. Independent Events & the Multiplication Law

Two events are independent if either can occur without being affected by the other. Examples: selecting with replacement; separate actions like rolling two dice.

Key Point 4.8 : independent ⇒ P(A and B) = P(A ∩ B) = P(A) × P(B)
extends: P(A and B and C …) = P(A) × P(B) × P(C) × …

Tree diagram — selecting with replacement

Bag: 2 blue (B), 5 white (W). Pick one, replace, pick again. Both picks come from the same 7 balls ⇒ identical & independent: P(B) = 27, P(W) = 57.

1st pick → 2nd pick → combined outcome. Events BB, BW, WB, WW are exhaustive: 4+10+10+25 = 49 ⇒ sum to 1

Tree rules: MULTIPLY along the branches (left → right, “and”). ADD the ends of different branches (vertically, “or” — they are mutually exclusive).

P(different colours) = P(BW) + P(WB)= 1049 + 1049 = 2049P(same colour) = P(BB) + P(WW)= 449 + 2549 = 2949

Check: different + same = 2049 + 2949 = 1 ✓ (complementary)

Possibility diagram (outcome space)

7 × 7 = 49 equally likely cells; red BB (4), blue BW (10), green WB (10), plain WW (25)

Tip: A plain 2×2 diagram (B / W) would NOT work for counting — its 4 cells aren't equally likely. Make one cell per ball (equiprobable).

P(≥ 1 blue) = P(BB)+P(BW)+P(WB)
= 449+1049+1049 = 2449
or 1 − P(WW) = 1 − 2549 = 2449

To find combined-event probabilities, count the favourable cells out of 49. Probabilities = relative frequencies of the favourable outcomes.

Worked Example 4.5three dice

Sum of three rolls of a fair die is less than 5.

6×6×6 = 216 equally likely outcomes, each 1216. Lowest possible sum = 3.

P(sum < 5) = P(3) + P(4)sum 3: only (1,1,1) → P = 1216sum 4: (1,1,2),(1,2,1),(2,1,1) → P = 3216P(sum < 5) = 1216 + 3216 = 154

Worked Example 4.6repeated independent events

Abha passes 3 independent traffic lights; P(stop at any one) = 0.2. Let S = stops, X = doesn't: P(S) = 0.2, P(X) = 0.8.

a) first stop at the 2nd lightP(XS) = 0.8 × 0.2= 0.16

b) stops at exactly oneSXX, XSX, XXS equally likely3 × (0.2 × 0.8 × 0.8)= 0.384

c) P(has to stop) = 1 − P(does not have to stop)= 1 − P(XXX) = 1 − 0.8³= 0.488alternatively P(S) + P(XS) + P(XXS)

Using the multiplication law to TEST for independence

A and B independent ⇔ P(A ∩ B) = P(A) × P(B)

e.g. P(X) = 0.3, P(Y) = 0.4 ⇒ independent only if P(X∩Y) = 0.3 × 0.4 = 0.12. In words: enjoying sport is independent of gender if equal proportions of males and females enjoy it; if unequal proportions of employed/unemployed own cars, car ownership depends on employment status.

Worked Example 4.7J, K, L independent

P(J) = 0.5, P(K) = 0.6, P(J∩L) = 0.24. Find a) P(J∩K) b) P(L) c) P(K∩L).

a)P(J∩K) = P(J)×P(K)= 0.5 × 0.6 = 0.3

b)P(J) × P(L) = P(J∩L)0.5 × P(L) = 0.24P(L) = 0.240.5 = 0.48

c)P(K∩L) = P(K)×P(L)= 0.6 × 0.48 = 0.288

Worked Example 4.8show events are independent

60 students: 27 male (M), 20 study History (H). Show “a male is selected” and “a History student is selected” are independent.

M only 18 · both 9 · H only 11 · neither 22

P(M) = 2760 , P(H) = 2060 , P(M∩H) = 960Does P(M)×P(H) = P(M∩H)?P(M)×P(H) = 2760×2060 = 960 = P(M∩H)

The multiplication law holds for M and H, therefore they are independent. Q.E.D.

Tip — always end with a sentence! State your conclusion in words after the maths (“…therefore they are independent”). Writing Q.E.D. is optional. Q.E.D. = quod erat demonstrandum (Latin) = “which is what had to be shown”; modern alternative W⁵ = “Which Was What Was Wanted”.

Explore 4.2: for the same 60 students test the pairs (M, H′), (M′, H), (M′, H′) too. If M and H are independent then all these pairs are independent as well.

4. Conditional Probability

Conditional = a probability that depends on extra information about an outcome/event. Written with | meaning “given that”.

P(A | B) = “probability that A occurs, given that B occurs”

Example:

Pick a letter from the word ACE: P(E) = 13.
If we are told it's a vowel (A or E): P(E | vowel) = 12 — the extra info shrinks the possibility space.

Idea: “Given B” ⇒ the new total is only the outcomes in B. Then count how many of those are also A.

Worked Example 4.9given information

Pick a child from 11 boys and 9 girls (one girl is Rose). P(Rose | a girl is selected)?

P(Rose | girl) = 19

“Given a girl” cuts the possibilities from 20 to 9, and Rose is one of those nine.

Worked Example 4.10table → Venn → conditional

	B	B′	Total
C	9	8	17
C′	7	1	8
Total	16	9	25

B = Biology, C = Chemistry; 25 students

a) P(C | B)16 study Biology, 9 of them Chemistry= 916b) P(B′ | C′)8 don't study Chemistry, 1 of them doesn't study Biology= 18

Worked Example 4.11without replacement

Two children chosen from 5 boys, 7 girls. P(2nd is a boy | 1st is …)

a) 1st = boyleft: 4 boys + 7 girlsP = 411

b) 1st = girlleft: 5 boys + 6 girlsP = 511

Independence ⇔ conditional probabilities are equal

Events X, Y are independent if each is unaffected by the other, i.e. X has the same probability whether Y occurs or not:

X and Y independent ⇔ P(X | Y) = P(X | Y′) ( = P(X) )

Illustrationfair die: X = square {1,4}, Y = odd {1,3,5}

Not mutually exclusive (1 is odd and square) — but independent?

Y occurs: die shows 1,3,5 → P(X | Y) = 13Y doesn't: die shows 2,4,6 → P(X | Y′) = 13P(X) = 26 = 13 whether Y occurs or not

Equal ⇒ P(X) unaffected by Y ⇒ X, Y are NOT mutually exclusive, but they ARE independent.

Common mistake: “not mutually exclusive” does not mean “dependent”. Mutually exclusive and independent are different ideas — always test with the numbers.

Explore 4.3: rolling dice / 12-sided die / integers 1–20 — for each pair of events decide (i) mutually exclusive? (ii) independent? Use P(A∩B) = P(A)P(B) to check.

5. Dependent Events & Conditional Probability

Events are dependent when neither can occur without being affected by the other — e.g. selecting without replacement (the 2nd selection depends on the 1st). The independent law is a special case of the general multiplication law:

Key Point 4.9 :
P(A and B) = P(A ∩ B) = P(A) × P(B | A)
P(B and A) = P(B ∩ A) = P(B) × P(A | B)
P(A) × P(B | A) ≡ P(B) × P(A | B)

Tip: ≡ means “is identical to”. P(A∩B) ≡ P(B∩A).

Worked Example 4.12without replacement

Two children chosen from 11 boys (B), 14 girls (G) (total 25). Find P(a) two boys, (b) a boy and a girl in any order.

a) P(2 boys) = P(B₁) × P(B₂ | B₁)= 1125 × 1024= 1160b) P(B and G) + P(G and B)= 1125×1424 + 1425×1124= 77150

1st chosen from 25, 2nd from the remaining 24 ⇒ dependent ⇒ conditional probs. Two orders (BG, GB) but P(B and G) = P(G and B). Suffixes 1, 2 = 1st, 2nd pick.

Worked Example 4.13tree with conditional branches

Each Saturday a man invites his sister to the Theatre (70%) or Cinema. She accepts 90% of theatre invitations and rejects 40% of cinema ones. P(invitation accepted)?

second-stage branches are conditional on where she was invited (they sum to 1 at each node)

P(accepts) = P(T and acc) + P(C and acc)= 0.63 + 0.18= 0.81

Her acceptance depends on where she's invited ⇒ use conditional probabilities. She can accept the theatre or the cinema ⇒ add.

Finding a conditional probability

Rearrange P(A∩B) = P(A) × P(B | A):

Key Point 4.10 :
P(B | A) = P(A ∩ B)P(A) and P(A | B) = P(B ∩ A)P(B) where P(A∩B) ≡ P(B∩A)

Remember: Divide by the probability of the “given” event (the one after the |). Its probability is the new “total”.

Worked Example 4.14using the formula

P(A∩B) = 0.36, P(B) = 0.9. Find P(A | B).

P(A | B) = P(B∩A)P(B)= 0.360.9= 0.4

Worked Example 4.15fair die

Number is prime, given that it is odd.

odd = {1,3,5} → P(odd) = 36odd and prime = {3,5} → P = 26P(prime | odd) = 2/63/6 = 23

Shortcut: 2 of the 3 odd numbers are prime ⇒ 23 (count inside the “given” set).

Worked Example 4.16reverse conditional (tree + formula)

A boy walks (W) 60% of the time, cycles (C) 40%. Late (L) on 5% of walks, 2% of cycles. Given he is late, P(he cycled) = P(C | L)?

multiply along branches to get each “and” probability

P(L) = P(W and L) + P(C and L)= 0.030 + 0.008 = 0.038 (late by walking or cycling)P(C | L) = P(C and L)P(L)= 0.0080.038= 419 ≈ 0.211

Method for “given the end result, what happened first?”: ① draw the tree ② add the branch-products that give the end result (denominator) ③ divide the product for the path you want by it.

Chapter Cheat-Sheet

which law? — pick by the key word “or” / “and” and ask the question in the middle box

✓ Full Chapter 4 Checklist

Probabilities run from 0 (impossible) to 1 (certain).
One of n chosen at random: P(particular object) = 1/n.
P(event) = favourable equally likely outcomes ÷ total equally likely outcomes.
P(A) + P(A′) = 1 ⇒ P(A′) = 1 − P(A).
In n trials, A is expected n × P(A) times.
A ∪ B means “A or B”; A ∩ B means “A and B”.
Mutually exclusive: no common outcome, P(A∩B) = 0 ⇒ P(A∪B) = P(A) + P(B).
Not mutually exclusive: P(A∪B) = P(A) + P(B) − P(A∩B).
Independent: unaffected by each other. P(A∩B) = P(A) × P(B), i.e. P(A | B) = P(A | B′).
Any two events: P(A∩B) = P(A) × P(B | A) and P(B | A) = P(A∩B) ÷ P(A).

Common mistakes: ① adding P(A)+P(B) without checking for overlap · ② confusing mutually exclusive with independent · ③ forgetting that “without replacement” changes the numerator and denominator · ④ dividing by the wrong event in P(B | A) · ⑤ no concluding sentence when showing independence.