Why a Room of 23 People Probably Holds a Shared Birthday

Probability measures how likely an event is, on a scale from $0$ (impossible) to $1$ (sure). Once events are written as subsets of a sample space, we need clear rules for attaching numbers to them.

Those rules are the axioms of probability. Everything else — including the familiar favourable-over-total formula — follows from them.

This part covers the axioms, equally likely outcomes, the addition rule, and the probability of "not A".

**A probability $P$ assigns a number to every event $E$ of a sample space $S$ so that $P(E)\ge 0$, $P(S)=1$, and $P(E\cup F)=P(E)+P(F)$ whenever $E$ and $F$ are mutually exclusive.**

For a finite $S=\{{\omega }_1,{\omega }_2,\dots ,{\omega }_n\}$, this means each $P({\omega }_i)$ lies between $0$ and $1$, all of them add to $1$, and **$P(A)$ is the sum of $P({\omega }_i)$ over the outcomes in $A$.** It also follows that $P(\varnothing )=0$.

Worked example 1 — valid or not? For $S=\{{\omega }_1,{\omega }_2,{\omega }_3\}$:

- $\frac{1}{2},\frac{1}{3},\frac{1}{6}$ — non-negative, sum $1$: valid
- $0.4,0.5,0.2$ — sum $1.1$: not valid
- $0.6,0.5,-0.1$ — sum $1$ but one value negative: not valid

Worked example 2 — a biased die. A $6$ is three times as likely as each other face. If each other face has probability $p$:

$$5p+3p=1\Rightarrow p=\frac{1}{8}P(6)=\frac{3}{8}$$

$$P(\text{even})=P(2)+P(4)+P(6)=\frac{1}{8}+\frac{1}{8}+\frac{3}{8}=\frac{5}{8}$$

An everyday example. A weather forecast giving chances of rain, clouds and clear skies must use non-negative values that add to $1$.

The substance. The axioms do not require outcomes to be equally likely, which is why they work for biased dice too.

**If all $n(S)$ outcomes are equally likely, each has probability $\frac{1}{n(S)}$, and the axioms give $P(E)=\frac{n(E)}{n(S)}$ — favourable outcomes divided by total outcomes.

Worked example 1 — two dice.** There are $36$ outcomes.

$$P(\text{sum }7)=\frac{6}{36}=\frac{1}{6}P(\text{sum }8)=\frac{5}{36}$$

Sum $8$ comes from $(2,6),(3,5),(4,4),(5,3),(6,2)$.

Worked example 2 — cards.

$$P(\text{king})=\frac{4}{52}=\frac{1}{13}P(\text{red face card})=\frac{6}{52}=\frac{3}{26}$$

Worked example 3 — letters. A letter is picked from ASSASSINATION, which has $13$ letters with vowels A, A, A, I, I, O:

$$P(\text{vowel})=\frac{6}{13}$$

Worked example 4 — using combinations. Two balls are drawn from a bag of $4$ red and $5$ blue balls.

$$P(\text{both red})=\frac{{}^4{C}_2}{{}^9{C}_2}=\frac{6}{36}=\frac{1}{6}$$

An everyday example. Drawing one name from a box of folded chits to pick a class monitor gives every student the same chance.

The substance. **"It will rain or it won't" does not make the chance of rain $\frac{1}{2}$** — the formula needs equally likely outcomes.

**$P(A\cup B)=P(A)+P(B)-P(A\cap B)$, because outcomes in both events would otherwise be counted twice; for mutually exclusive events the last term is $0$.

Worked example 1 — cards.** A king or a heart:

$$P=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$$

The king of hearts is subtracted once.

Worked example 2 — given values. $P(A)=0.42$, $P(B)=0.48$, $P(A\cap B)=0.16$.

$$P(A\cup B)=0.42+0.48-0.16=0.74$$

$$P(A\text{ but not }B)=P(A)-P(A\cap B)=0.26$$

Worked example 3 — mutually exclusive. One die: $P(1\text{ or }6)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$.

Worked example 4 — a survey. Suppose that in a class of $60$, $30$ play cricket, $20$ play football and $10$ play both. For a student picked at random:

$$P(\text{cricket or football})=\frac{30}{60}+\frac{20}{60}-\frac{10}{60}=\frac{40}{60}=\frac{2}{3}$$

An everyday example. Counting homes in a colony that take a Hindi paper or an English paper, homes taking both would be counted twice without the subtraction.

The substance. **If $P(A)+P(B)$ comes out above $1$**, the events cannot be mutually exclusive.

**Since $A$ and $A^{\prime }$ are mutually exclusive and exhaustive, $P(A)+P(A^{\prime })=1$, so $P(\text{not }A)=1-P(A)$.

Worked example 1.** $P(\text{no six on one die})=1-\frac{1}{6}=\frac{5}{6}$.

Worked example 2. Two dice: $P(\text{sum is not }7)=1-\frac{1}{6}=\frac{5}{6}$.

Worked example 3 — "at least one". Three coins are tossed. The only outcome with no head is $TTT$:

$$P(\text{at least one head})=1-\frac{1}{8}=\frac{7}{8}$$

Worked example 4. Two dice: $25$ of the $36$ outcomes contain no $6$, so

$$P(\text{at least one six})=1-\frac{25}{36}=\frac{11}{36}$$

**Worked example 5 — given $P(A\cup B)=0.74$.**

$$P(\text{neither }A\text{ nor }B)=1-0.74=0.26$$

An everyday example. **If the chance that a train runs late is $0.15$**, the chance that it is not late is $0.85$.

The substance. "At least one" is almost always easier through its complement, "none".