Why the Chance of Rain or Wind Is Not Just the Two Chances Added

Everyday probability relies on counting equally likely outcomes, but that breaks down when outcomes are not equally likely. The axiomatic approach defines probability through a few simple rules on sets, and every familiar formula — for 'not', 'and' and 'or' — follows from them.

This part covers the axiomatic definition, probabilities of combined events, and the addition theorem.

**In the axiomatic approach, probability is a rule P that assigns to each event E of a sample space S a number $P(E)$, with $P(E)\ge 0$, $P(S)=1$, and $P(E\cup F)=P(E)+P(F)$ whenever E and F are mutually exclusive.

The axioms:

- Non-negativity** — $P(E)\ge 0$ for every event E
- Certainty — $P(S)=1$
- Additivity — if $E\cap F=\varphi$, then $P(E\cup F)=P(E)+P(F)$

Consequences:

- $P(\varphi )=0$ and $0\le P(E)\le 1$
- The probability of an event is the sum of the probabilities of its sample points

Worked example. A die is loaded so that each of the faces 1 to 5 has probability 0.1 and the face 6 has probability 0.5. The assignment is valid, since every value is non-negative and $5\times 0.1+0.5=1$. For E = 'an even number', $P(E)=0.1+0.1+0.5=0.7$.

Equally likely case. When all n outcomes are equally likely, each has probability $\frac{1}{n}$ and $P(E)=\frac{n(E)}{n(S)}$ — the classical formula is a special case of the axioms.

An everyday example. A forecast giving chances of 0.5 for sun, 0.3 for cloud and 0.2 for rain in Pune tomorrow is a valid assignment, because the values are non-negative and add to 1.

The substance. Chances of 0.6, 0.3 and 0.2 for three exhaustive outcomes are invalid — they add to more than 1, whatever the story behind them.

**'Not A' is the complement $A^{\prime }$ with $P(A^{\prime })=1-P(A)$, 'A and B' is the intersection $A\cap B$, 'A or B' is the union $A\cup B$, and 'A but not B' is $A\cap B^{\prime }$ with $P(A\cap B^{\prime })=P(A)-P(A\cap B)$.

Translating words into sets:**

- 'not A' — $A^{\prime }$, with $P(A^{\prime })=1-P(A)$
- 'A and B' — $A\cap B$
- 'A or B', meaning at least one — $A\cup B$
- 'A but not B' — $A\cap B^{\prime }$
- 'neither A nor B' — $(A\cup B{)}^{\prime }$, with probability $1-P(A\cup B)$

Worked example. A card is drawn from a pack of 52. Let A = 'a heart' and B = 'a king'.

- $P(A)=\frac{13}{52}$, $P(B)=\frac{4}{52}$ and $P(A\cap B)=\frac{1}{52}$
- $P(\text{not a heart})=1-\frac{13}{52}=\frac{3}{4}$
- $P(\text{a heart but not a king})=\frac{13}{52}-\frac{1}{52}=\frac{12}{52}=\frac{3}{13}$

An everyday example. If the chance that a train to Chennai runs late is 0.15, the chance that it runs on time is $1-0.15=0.85$ — the complement rule in action.

The substance. 'Or' in probability includes both — 'a heart or a king' counts the king of hearts too, which is why the addition theorem subtracts the overlap.

**The addition theorem states that for any two events $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and for mutually exclusive events it reduces to $P(A\cup B)=P(A)+P(B)$.

Proof idea.** $A\cup B$ splits into the mutually exclusive parts A and $B\cap A^{\prime }$, and $P(B\cap A^{\prime })=P(B)-P(A\cap B)$; adding the two parts gives the theorem.

Three events:

$$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$$

Worked example. For the card drawn above,

$$P(\text{heart or king})=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\approx 0.308$$

Worked example 2. For a student chosen at random, the probability of studying Computer Science is 0.4, of studying Economics is 0.5, and of studying both is 0.2.

- At least one: $0.4+0.5-0.2=0.7$
- Neither: $1-0.7=0.3$
- Exactly one: $0.7-0.2=0.5$

An everyday example. If the chance of rain in Mumbai tomorrow is 0.6, of strong wind is 0.3, and of both is 0.2, the chance of rain or wind is $0.6+0.3-0.2=0.7$, not 0.9.

The substance. Given probabilities must be consistent — $P(A\cap B)$ can never exceed the smaller of $P(A)$ and $P(B)$, and the union can never exceed 1.