EY

Module 3: Statistics - Probability

Basic Terminology

• Experiment: A repeatable action whose results are observed.
– Examples: coin toss, die roll, drug study, survey.
• Trial: One performance of the experiment (e.g.
first die roll, second die roll).
• Outcome: Result of a trial (e.g. “Heads”, “4”).
• Sample Space (S): Set of all possible outcomes.
– Die roll ⇒ S = {1,2,3,4,5,6}
• Event (E): Any subset of the sample space.
– Even‐number event ⇒ E = {2,4,6}

Describing Sample Spaces: Example

• Toss a coin, then roll a die.
• First coordinate = coin face, second = die number.
• S = {H1,H2,\dots,H6,\;T1,T2,\dots,T6} (12 outcomes).

Fundamental Counting Principle (FCP)

• If one event can occur in m ways and a second in n ways, the ordered pair of events can occur in m\times n ways.
• Extends to any number of sequential events: multiply counts.
• Car-buying example: 3 makers × 2 sizes × 4 colours ⇒ 3\cdot2\cdot4 = 24 distinct cars.
• A tree diagram visually verifies counts.

Classical vs. Empirical Probability

• Classical (theoretical) – equally likely outcomes
P(E)=\dfrac{\text{Number of outcomes in }E}{\text{Number of outcomes in }S}
• Empirical (relative-frequency) – based on data
P(E)=\dfrac{f}{n} where f = frequency of event E, n = total trials.

Complements

• Complement of E: E' – all outcomes not in E.
• Key identities:
P(E')+P(E)=1, P(E)=1-P(E').
• Age survey example (1,000 respondents):
– Age 25–34 (event A) has 366 respondents.
– P(A)=\tfrac{366}{1000}=0.366; thus P(A')=1-0.366=0.634.

Conditional Probability

• Probability of B given A has occurred: P(B\mid A).
• Card example: second card is Queen given first card is King (no replacement).
– Remaining deck = 51 cards; 4 are Queens ⇒ P(\text{Queen}\mid\text{King})=\tfrac{4}{51}.

Multiplication Rule & Bayes’ Theorem

• General form:
P(A \text{ and } B)=P(A)\,P(B\mid A)=P(B)\,P(A\mid B)
• Rearranging gives Bayes:
P(B\mid A)=\dfrac{P(B)\,P(A\mid B)}{P(A)}

Independence

• A and B are independent if occurrence of one doesn’t affect the probability of the other:
P(B\mid A)=P(B) (or symmetrically).
• Then P(A \text{ and } B)=P(A)P(B).
• Example: drawing successive cards without replacement makes events dependent (King then Queen).

Using the Multiplication Rule – Example

• Draw 2 cards w/o replacement: King then Queen.
P(\text{King then Queen})=P(\text{King})\,P(\text{Queen}\mid\text{King})=\tfrac{4}{52}\cdot\tfrac{4}{51}.

Contingency (Two-Way) Tables

• Summarize joint frequency of two categorical variables.
• Example (Gene × IQ – total 102 children):
– P(\text{High IQ}\mid \text{Gene})=\tfrac{33}{72}.

Mutually Exclusive Events & Addition Rule

• Mutually exclusive: A\cap B=\varnothing (cannot occur together).
• General addition: P(A\cup B)=P(A)+P(B)-P(A\cap B).
• If exclusive: P(A\cup B)=P(A)+P(B).

Examples

  1. Deck card is 4 or Ace (exclusive):
    P=\tfrac{4}{52}+\tfrac{4}{52}=\tfrac{8}{52}=\tfrac{2}{13}.
  2. Die roll less than 3 or odd (not exclusive):
    – A={1,2},\;B={1,3,5},\;A\cap B={1}.
    – P=\tfrac{2}{6}+\tfrac{3}{6}-\tfrac{1}{6}=\tfrac{4}{6}=\tfrac{2}{3}.
  3. Blood-bank table, type O or A:
    P=\tfrac{184+164}{409}=\tfrac{348}{409}.
  4. Blood-bank, type B or Rh-negative:
    – B total = 45, Rh-neg = 65, overlap = 8.
    – P=\tfrac{45+65-8}{409}=\tfrac{102}{409}.

Permutations

• Permutation: ordered arrangement.
• All n objects: n! (by definition 0! = 1).
• n objects taken r at a time:
^nP_r=\dfrac{n!}{(n-r)!}.

Example – 3-digit codes, no repeat digits

• n=10, r=3\Rightarrow ^{10}P_3=10!\,/(7!)=10\cdot9\cdot8=720.

Distinguishable Permutations

• n objects with groups of identical items n1,n2,\dots,nk: \dfrac{n!}{n1!\,n2!\,\dots\,nk!} where \sum n_i=n.

Housing Example

• 6 one-story, 4 two-story, 2 split-level ⇒ n=12.
\dfrac{12!}{6!\,4!\,2!}=\text{13,860} distinguishable arrangements.

Combinations

• Combination: choose r objects from n without regard to order.
^nC_r=\dfrac{n!}{r!(n-r)!}.

Application Problems

Problem 1: Student Advisory Board

• 17 members, 3 distinct positions (ordered).
• Total possible assignments: ^{17}P_3=17\cdot16\cdot15=4,080.
• Favourable outcomes = 1 (selecting a specific trio).
• P=\tfrac{1}{4,080}.

Problem 2: Spelling “Mississippi”

• Letters: 1 M, 4 I’s, 4 S’s, 2 P’s (11 total).
• Total distinguishable permutations:
\dfrac{11!}{1!\,4!\,4!\,2!}=34,650.
• Exactly one spells “MISSISSIPPI”.
• P=\tfrac{1}{34,650}.

Problem 3: Corn‐Kernel Toxin

• 400 kernels; 3 toxic. Randomly pick 4.
• Favourable: choose exactly 1 toxic & 3 non-toxic.
– 3C1=3, 397C3=10,349,790.
– Ways = 3\times10,349,790=31,049,370.
• Total 4-kernel samples: 400C_4=1,050,739,900.
• Probability:
P=\dfrac{31,049,370}{1,050,739,900}\approx0.0296 (≈2.96 %).

Ethical & Practical Notes

• Proper randomisation ensures independence assumptions.
• In medical or genetic studies (e.g.
gene–IQ contingency table), ethical oversight is crucial to avoid stigma or misinterpretation of probabilities.
• Counting principles underpin cryptographic keyspace analysis and quality-control sampling.