AQA A level Mathematics [Stat 1] Probability Distributions Notes
Probability Distributions
— Maths · AS & A Level (9709) · P&S 1 · Chapter 6 —
★ By the end you must be able to: ① identify & use a
discrete random variable; ② build a probability distribution table for a situation; ③ calculate
E(X) and Var(X) (and SD). These are Paper 6 (Probability & Statistics 1)
bread-and-butter marks — every exam has a distribution-table question.
Why bother? (Tools of the trade)
A company planning a campaign builds a probability
distribution of change in sales (big fall · small fall · same · small rise · big rise). It shows the
worst case, best case and most likely case → informed decisions. e.g. needs ≥ $50 000 revenue to profit,
and the distribution says P(revenue < $50 000) = 40% ⇒ it knows its level of risk.
0. Quick Revision — what you need first
① Complement rule ()P(A) = 1 − P(A′)
② Independent vs dependent (Ch 4 & 5)Independent (e.g. with
replacement): P(A and B) = P(A) × P(B) Dependent (e.g. without replacement): the 2nd probability
changes after the 1st pick
Check skill 1complement rule
A game is won (W), lost (L) or drawn (D). P(W′) = 0.46, P(L′) = 0.65. Find P(D).
2 cubes picked from 3 red + 3 blue. Show P(both red) is bigger with replacement.
With replacementP(RR) = 36 × 36= 14 = 0.25
Without replacementP(RR) = 36 × 25= 15 = 0.2
0.25 > 0.2 ✓ — (without replacement a red is already gone, so fewer reds remain).
1. Discrete Random Variables (6.1)
Definition: a variable is discrete and random if it can take
only certain separate values that occur by chance.
Notation tip: variable = capital X; its values = small x.
“X can be 1, 2 or 3” ⇒ X ∈ {1, 2, 3}. The symbol ∈ means “is an element of”.
Where do they come from? — 3 typical sources
Source
Example
Variable & values
Counting in one carton
carton of 6 eggs, some broken
X = no. of broken eggs, X ∈ {0,1,2,3,4,5,6}
Independent trials
roll 4 dice, count the 6s
S ∈ {0,1,2,3,4}
Selection without replacement
pick 3 children from 4 boys + 2 girls
B (boys) ∈ {1,2,3} · G (girls) ∈ {0,1,2}
Why is B ∈ {1,2,3} and not 0? Only 2 girls exist, so picking 3
children must include ≥ 1 boy. Likewise G can’t be 3 (only 2 girls). Always ask: which values are actually
possible? Notice B + G = 3 — two linked variables.
2. Probability Distributions (6.2)
Definition: the probability distribution of a discrete random
variable = a display of all possible values and their probabilities.
Shown as: ① a table (usual) ② a vertical line graph ③ a bar chart.
KEY POINT 6.1Σp = 1 (all probabilities add to 1)
Tip: P(X = x) = the relative frequency of that value of X. Always
finish by checking Σp = 1.
Example — tossing 2 fair coins, X = number of heads
So in many tosses expect 25% with no head, 50% with one head, 25% with two.
vertical line graph — symmetrical
How to BUILD a table — 3 methods
Method map: (a) independent trials → multiply along branches
& add branches that give the same value · (b) equally-likely outcomes → draw a sample-space
grid, count · (c) selection without replacement → combinations ⁿCᵣ
Worked Example 6.1method (b) — grid
A fair 4-sided spinner (1,2,3,4) is spun twice; X = the total. Draw the distribution table.
2nd↑ / 1st→
1
2
3
4
4
5
6
7
8
3
4
5
6
7
2
3
4
5
6
1
2
3
4
5
16 equally likely outcomes, so X ∈ {2,…,8}. Count each total, ÷ 16.
x
2
3
4
5
6
7
8
P(X=x)
116
216
316
416
316
216
116
Check: (1+2+3+4+3+2+1)/16 = 16/16 = 1 ✓
Shape: symmetrical, peak at 5 (the most likely total).
Worked Example 6.2unknown constant c
Distribution of V: v = 2,3,4,5,6 with P = 0.05, c², c+0.1, 2c+0.05, 0.16. Find c and P(V > 4).
Always check validity: after solving a quadratic for an unknown,
substitute back — reject any root that makes a probability < 0 or > 1. (Examiners award a mark for
explaining why one root is rejected.)
Worked Example 6.3method (c) — combinations
3 spaces on a bus; 8 youths, 1 man, 1 woman want to board. Driver picks 3 at random. Y = number of youths
selected. Draw the table.
No replacement ⇒ use combinations. Total ways = 10C3 = 120. Only 2
non-youths (Y′) exist ⇒ at least 1 youth, so Y ∈ {1,2,3}.
Pattern: P = (ways to pick the wanted type × ways to pick the rest) ÷ (ways to pick all). The
top r’s add up to the number picked; the bottom n’s add up to the group size.
Explore 6.1 — how does the shape change if the coins are unfair? (P(heads) = p)
Two coins, X = no. of heads. P(X=0) = (1−p)² · P(X=1) = 2p(1−p) · P(X=2) = p².
Pattern: p and 1−p give mirror images (0.4 ↔ 0.6, 0.3 ↔ 0.7 …).
Only p = 0.5 is symmetric.
p = 0.4 → skewed towards 0 heads; p = 0.6 → skewed towards 2 heads; swapping p ↔ 1−p reverses
the bar chart.
P(X=1) = 2p(1−p) is biggest (= 0.5) at p = 0.5, and → 0 when p → 0 or p → 1.
as p goes 0→1: P(0) falls, P(2) rises, P(1) peaks (= 0.5) at p = 0.5
Did you know? (Fermat & Pascal)
Pierre de Fermat (1607–1665) and Blaise Pascal exchanged
letters that developed the basic probability ideas of equally likely outcomes and expected values.
Fermat’s Last Theorem (xⁿ + yⁿ = zⁿ has no positive-integer solutions for n > 2) resisted proof for ~350 years
until Andrew Wiles proved it in 1995 using 20th-century mathematics.
3. Expectation of a Discrete Random Variable (6.3)
The mean of a discrete random variable X is called its expectation, written E(X). High-probability values occur more often; after many trials
the frequency distribution has a mean = the “expected” value.
Remember: E(X) = the long-term average value of X over a large
number of trials. It need not be a possible value of X (e.g. 1.7).
Build-up with a biased spinner (scores 0, 1, 2, 3)
x
0
1
2
3
P(X=x)
0.1
0.3
0.4
0.2
Expected frequencies in 1600 spins = P × 1600:
x
0
1
2
3
freq f
160
480
640
320
red ▲ = balance point = mean
Mean = ΣxfΣf = (0×160)+(1×480)+(2×640)+(3×320)1600=
27201600 = 1.7
Using relative frequencies (probabilities) instead gives the same answer, and the denominator Σp =
1 disappears:
Trick: since the denominator Σp = 1, just multiply each value by
its probability and add. If all probabilities share a denominator (e.g. /12), add the numerators first,
divide once.
Explore 6.2Adam vs Priya — does choosing cards
change the distribution?
Adam and Priya each have bags of cards 1–5, pick one at random simultaneously; X = numerical difference ∈
{0,1,2,3,4}. Repeat 200 times. Adam: “if we choose instead of random the distribution will be very
different.” Priya: “very similar, maybe exactly the same.”
Random case (25 equally likely pairs): P(0)=5/25, P(1)=8/25, P(2)=6/25, P(3)=4/25, P(4)=2/25 (sum
25/25 ✓).
x
0
1
2
3
4
P(X=x)
0.20
0.32
0.24
0.16
0.08
Whether they agree: it depends on how they choose. If both deliberately pick the same card (e.g.
always 5) then X = 0 every time → very different (Adam right). If their choices are effectively unpredictable
and every card is equally likely to be picked, the distribution is the same as random (Priya right). →
“Choosing” only matters if the choices are biased.
4. Variance & Standard Deviation
Variance and SD measure the spread of values about the mean E(X). Take the old data
formula and replace f by p and x̄ by E(X):
KEY POINT 6.3Var(X) = Σx²p − {E(X)}² SD(X) = √Var(X)
Rewind: variance = “mean of the squares minus square of the mean”
(Ch 3, Section 3.3).
Common mistakes: ① forgetting to subtract {E(X)}² — remember it
every time! ② Σx²p means square the x first, then × p — not (Σxp)². ③ A variance can never be
negative — if you get one, recheck.
Worked Example 6.4E(X), Var(X), SD
x = 0, 5, 15, 20 with P = 1/12, 3/12, 5/12, 3/12. Find E(X), Var(X), SD(X).
Common denominator 12 ⇒ add numerators, divide once.
Tip: same-denominator fractions make the working much simpler. Remember
to subtract the square of E(X).
Extra shortcut (used in Exercise 6B, Q10–11): when you pick n items
at random without replacement from a group, the expected number of a given type = n × (proportion of
that type). e.g. 2 students from 12 girls + 18 boys: E(girls) = 2 × 12/30 = 0.8, E(boys) = 2 × 18/30
= 1.2 (ratio 2 : 3 = 12 : 18, the same as the class ratio!). It saves building the full table when only E
is asked — but you still need the table for Var.
5. Explore 6.3 — Falling-ball devices (a preview of Ch 7)
A ball drops through rows of nails (red dots). At every nail it goes Left (L) or Right (R), each
with probability ½. Each trial has only two outcomes — exactly the situation of Chapter 7 (binomial).
3 cups: LL→A, LR/RL→B, RR→C4 cups: 1, 3, 3, 1 paths out of 8
Why (½+½)? Each row multiplies the paths by (½ + ½) = 1 (L or R).
Expanding (½+½)² and (½+½)³ gives the coefficients 1,2,1 and 1,3,3,1 — the numbers of paths to each cup. (Printed
as a power of a bracket in the book — scan is hard to read, check your copy.)
Check each time: probabilities add to 1 — ¼+½+¼ = 1 ✓, (1+3+3+1)/8 = 1 ✓, (1+4+6+4+1)/16 = 1 ✓. The
pattern is Pascal’s triangle.
Fast forward: Chapter 7 (7.1, 7.2) studies the expectation of
these two special random variables (and the variance of one of them) built from independent two-outcome trials —
success/failure.
6. Method & Worked Exam-style Patterns
the whole chapter in one flow
Checklist for any question: state possible values →
probabilities (grid / multiply / ⁿCᵣ) → Σp = 1 check → E(X) = Σxp → Var = Σx²p − {E(X)}² → SD = √Var. Give
3 s.f. unless exact. Show every line — method marks!
Ex 6A Q1unknown from ratios
V ∈ {1,2,3}; P(V=1) = P(V=2) = 2 × P(V=3). Draw the table.
Let P(V=3) = p, then the other two are 2p each.
2p + 2p + p = 15p = 1p =
0.2
v
1
2
3
P(V=v)
0.4
0.4
0.2
Ex 6A Q4independent trials
P(basket) = 7/9, two shots, S = successful shots. Show P(S=0) = 4/81 and draw the table.
multiply along branches; add branches giving the same x
Equally likely outcomes (two dice, spinners)
draw a grid, count outcomes ÷ total
Selection without replacement
ⁿCᵣ ways ÷ total ways (or conditional probabilities)
Unknown constant (k, c, p …)
Σp = 1 ⇒ solve; reject roots giving p < 0 or p > 1
Missing value from a given E(X)
write E(X) = Σxp = given value, solve
★ Don’t lose marks: ① list only possible values ② Σp = 1
check ③ validity of solutions ④ subtract {E(X)}² ⑤ units / 3 s.f. ⑥ E(X) need not be a possible value — it’s the
long-run average.