Mathematics (9709)
Topic 7 of 9Cambridge A Levels

Statistics 1: Probability and Distributions

Probability rules, permutations and combinations, discrete random variables, expectation and variance, and the Binomial distribution.

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Statistics 1: Probability and Distributions — CIE A Level Mathematics 9709


1. Probability Fundamentals


Basic rules:

  • 0 ≤ P(A) ≤ 1 for any event A
  • P(A') = 1 − P(A) (complement)
  • P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (addition rule)
  • If A and B mutually exclusive: P(A ∪ B) = P(A) + P(B)

Conditional probability: P(A|B) = P(A ∩ B) / P(B)


Independent events: P(A ∩ B) = P(A) × P(B) ⟺ A and B are independent.


Tree diagrams: Multiply along branches (AND); add across branches (OR).


2. Permutations and Combinations


Permutation (order matters): Number of ways to arrange r from n: nPr = n!/(n−r)!


Combination (order doesn't matter): Number of ways to choose r from n: C(n,r) = n!/[r!(n−r)!]


Example: How many ways can 3 books be chosen from 8? C(8,3) = 56.

How many ways can they be arranged? P(8,3) = 8×7×6 = 336.


Arrangement with repetition: n! for n objects; n!/k! if k are identical; n!/(a!b!c!) if multiple groups of identical objects.


Example: How many arrangements of KARACHI? 7 letters, A appears twice.

7!/2! = 2520


3. Discrete Random Variables


A discrete random variable X takes specific values with given probabilities.


Probability distribution table: lists P(X = x) for all x.


Requirements:

  • 0 ≤ P(X = x) ≤ 1 for all x
  • ΣP(X = x) = 1

Expectation (mean): E(X) = Σ x·P(X = x)


Variance: Var(X) = E(X²) − [E(X)]² where E(X²) = Σ x²·P(X = x)


Standard deviation: σ = √Var(X)

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Linear transformations:

  • E(aX + b) = aE(X) + b
  • Var(aX + b) = a²Var(X)

Example:


| X | 1 | 2 | 3 | 4 |

|---|---|---|---|---|

| P | 0.1 | 0.3 | 0.4 | 0.2 |


E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7

E(X²) = 1(0.1) + 4(0.3) + 9(0.4) + 16(0.2) = 0.1 + 1.2 + 3.6 + 3.2 = 8.1

Var(X) = 8.1 − (2.7)² = 8.1 − 7.29 = 0.81


4. The Binomial Distribution


If X counts the number of successes in n independent trials each with probability p of success:


X ~ B(n, p)


$$P(X = r) = inom{n}{r} p^r (1-p)^{n-r}$$


Mean: E(X) = np

Variance: Var(X) = np(1−p)

Standard deviation: σ = √[np(1−p)]


When to use Binomial:

  1. Fixed number of trials (n)
  2. Each trial is independent
  3. Each trial has exactly two outcomes (success/failure)
  4. Probability of success (p) is constant

Example: A fair die is thrown 10 times. Find P(exactly 3 sixes).

X ~ B(10, 1/6). P(X = 3) = C(10,3)(1/6)³(5/6)⁷

= 120 × (1/216) × (78125/279936) = 0.155 (3 s.f.)


Cumulative binomial: Use tables or calculator. P(X ≤ k) = Σᵣ₌₀ᵏ P(X = r).


P(X ≥ k) = 1 − P(X ≤ k−1)


Example: X ~ B(12, 0.3). Find P(X ≥ 4).

P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − [P(0)+P(1)+P(2)+P(3)] ≈ 1 − 0.4925 = 0.5075

Key Points to Remember

  • 1P(A∪B) = P(A)+P(B)−P(A∩B); mutually exclusive: P(A∩B)=0; independent: P(A∩B)=P(A)P(B)
  • 2Conditional: P(A|B)=P(A∩B)/P(B); multiply along tree branches; add across
  • 3Permutations (order matters): nPr=n!/(n−r)!; Combinations (order doesn't): C(n,r)=n!/r!(n−r)!
  • 4E(X)=Σx·P(X=x); Var(X)=E(X²)−[E(X)]²; E(aX+b)=aE(X)+b; Var(aX+b)=a²Var(X)
  • 5Binomial X~B(n,p): P(X=r)=C(n,r)pʳ(1−p)ⁿ⁻ʳ; mean=np, variance=np(1−p)
  • 6Binomial conditions: fixed n trials, independent, two outcomes, constant p

Pakistan Example

Probability in Pakistan: Quality Control in Sialkot's Sports Industry

Sialkot manufactures 70% of the world's surgical instruments. A quality control inspector samples n=15 instruments per batch. Based on historical data, each instrument has probability p=0.05 of being defective. X~B(15, 0.05). P(X=0) = (0.95)¹⁵ ≈ 0.463 — nearly half the batches have zero defects. P(X≥2) = 1−P(X≤1) = 1−[P(0)+P(1)] = 1−[0.463+15(0.05)(0.95)¹⁴] ≈ 1−0.463−0.366 = 0.171. The company's acceptable quality limit (AQL) decision — accept batch if X≤1, reject if X≥2 — is a direct application of B(15,0.05). Expected defects per batch: np = 15×0.05 = 0.75. This quality control framework saves Pakistani exporters from costly returns from European clients.

Quick Revision Infographic

Mathematics — Quick Revision

Statistics 1: Probability and Distributions

Key Concepts

1P(A∪B) = P(A)+P(B)−P(A∩B); mutually exclusive: P(A∩B)=0; independent: P(A∩B)=P(A)P(B)
2Conditional: P(A|B)=P(A∩B)/P(B); multiply along tree branches; add across
3Permutations (order matters): nPr=n!/(n−r)!; Combinations (order doesn't): C(n,r)=n!/r!(n−r)!
4E(X)=Σx·P(X=x); Var(X)=E(X²)−[E(X)]²; E(aX+b)=aE(X)+b; Var(aX+b)=a²Var(X)
5Binomial X~B(n,p): P(X=r)=C(n,r)pʳ(1−p)ⁿ⁻ʳ; mean=np, variance=np(1−p)
6Binomial conditions: fixed n trials, independent, two outcomes, constant p

Formulas to Know

A∪B) = P(A)+P(B)−P(A∩B); mutually exclusive: P(A∩B)=0; independent: P(A∩B)=P(A)P(B)
P(A|B)=P(A∩B)/P(B); multiply along tree branches; add across
Pr=n!/(n−r)!; Combinations (order doesn't): C(n,r)=n!/r!(n−r)!
X)=Σx·P(X=x); Var(X)=E(X²)−[E(X)]²; E(aX+b)=aE(X)+b; Var(aX+b)=a²Var(X)
Pakistan Example

Probability in Pakistan: Quality Control in Sialkot's Sports Industry

Sialkot manufactures 70% of the world's surgical instruments. A quality control inspector samples n=15 instruments per batch. Based on historical data, each instrument has probability p=0.05 of being defective. X~B(15, 0.05). P(X=0) = (0.95)¹⁵ ≈ 0.463 — nearly half the batches have zero defects. P(X≥2) = 1−P(X≤1) = 1−[P(0)+P(1)] = 1−[0.463+15(0.05)(0.95)¹⁴] ≈ 1−0.463−0.366 = 0.171. The company's acceptable quality limit (AQL) decision — accept batch if X≤1, reject if X≥2 — is a direct application of B(15,0.05). Expected defects per batch: np = 15×0.05 = 0.75. This quality control framework saves Pakistani exporters from costly returns from European clients.

SeekhoAsaan.com — Free RevisionStatistics 1: Probability and Distributions Infographic

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