S1.3 - Probability

Probability measures how likely an event is on a scale from 0 to 1. An impossible event has probability 0, a certain event probability 1, and all probabilities for a complete set of mutually exclusive outcomes add to 1.

Define the event before counting. Check that every outcome belongs to the same sample space, that totals use the correct population or trial count, and that outcomes are genuinely equally likely before using a simple favourable-over-total ratio.

A table records 320 households, 200 with a driveway and 88 with exactly two cars. For a household chosen at random from this table,$$P(\text{driveway})=\frac{200}{320}=\frac58.$$Among the households with a driveway, suppose 40 have exactly one car. Then the joint event 'driveway and exactly one car' has probability $40/320=1/8$.

A sample space $\Omega$ contains every possible outcome. An event is a subset of that space; unions mean 'at least one', intersections mean 'both', and complements mean 'not'.

List outcomes or partition a Venn diagram into disjoint regions before adding. Every region of a Venn diagram must represent one exact combination of event membership, and all regions—including the outside region—must total 1.

Suppose $P(A)=0.55$, $P(B)=0.40$ and $P(A\cap B)=0.20$. Then$$P(A\cup B)=0.55+0.40-0.20=0.75,$$$$P((A\cup B)')=0.25,$$and, after restricting to $B$,$$P(A\mid B)=\frac{0.20}{0.40}=0.50.$$The denominator changes because the condition changes the relevant sample space.

Events $A$ and $B$ are independent when knowing that one occurred does not change the probability of the other.

To test independence, calculate both sides of one complete equality using the same data. Equality supports independence; a difference proves dependence. The conditional forms require the conditioning event to have positive probability.

For a fair six-sided die, let $A$ be 'even' and $B$ be 'a multiple of 3'. Then $P(A)=1/2$, $P(B)=1/3$ and $P(A\cap B)=P(6)=1/6$. Since $(1/2)(1/3)=1/6$, $A$ and $B$ are independent. If $C$ is 'greater than 3', then $P(A\cap C)=2/6=1/3$ but $P(A)P(C)=1/4$, so $A$ and $C$ are not independent.

Multi-stage probability uses the product law along one route and the sum law across distinct routes. A tree diagram records changing conditional probabilities; a Venn diagram partitions simultaneous event combinations.

At every node, outgoing branch probabilities total 1. Label branches with conditional probabilities for the situation at that node, not with unconditional probabilities copied from the start. In a Venn diagram, add only the disjoint regions that satisfy the event expression.

A bag contains 3 red and 2 blue counters. Two are selected without replacement. The probability of exactly one red is$$\frac35\frac24+\frac25\frac34=\frac35.$$The probability of at least one red is$$1-\frac25\frac14=\frac9{10}.$$With replacement, exactly one red instead has probability$$2\left(\frac35\right)\left(\frac25\right)=\frac{12}{25},$$because the second draw again starts from five counters.

When repeated routes have the same probability, count how many distinct orders produce the event before multiplying. When branch probabilities change without replacement, apparently similar orders may still need to be written separately so their numerators and denominators are correct.