Why a Coin That Landed Heads Five Times Is Still Fifty-Fifty Next Time

Probability measures how likely an event is, on a scale from $0$ to $1$:

- **$0$ — the event cannot happen
- $1$ — the event is certain
- Values in between** — the closer to $1$, the more likely

It cannot tell you what will happen on the next toss or draw, but it tells you what to expect over many repetitions. This chapter covers sample spaces, simple probabilities, complementary events and problems on cards, balls and tickets.

A random experiment has known possible outcomes but an unpredictable result, its sample space is the set of all those outcomes, and an event is any collection of outcomes from the sample space.

Sample spaces:

- Tossing a coin: $S=\{H,T\}$
- Throwing a die: $S=\{1,2,3,4,5,6\}$
- Tossing two coins: $S=\{HH,HT,TH,TT\}$
- Drawing a card from a pack of $52$: four suits of $13$ cards each — hearts and diamonds are red, spades and clubs are black; each suit has an ace, king, queen, jack and cards $2$ to $10$

Events. On a die, getting an even number is the event $E=\{2,4,6\}$.

Worked example. In a pack of $52$ cards there are $4\times 3=12$ face cards (king, queen, jack of each suit) and $2\times 13=26$ red cards.

An everyday example. The toss before a cricket match is a random experiment with sample space $\{H,T\}$.

The substance. **With two coins, $HT$ and $TH$ are different outcomes**, which is why the sample space has four outcomes, not three.

When all outcomes are equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes.

$$P(E)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$

Worked examples with one die.

$$P(\text{even})=\frac{3}{6}=\frac{1}{2}P(\text{prime})=\frac{3}{6}=\frac{1}{2}P(\text{more than }4)=\frac{2}{6}=\frac{1}{3}$$

The primes on a die are $2,3,5$; the numbers more than $4$ are $5,6$.

Worked example with two coins. At least one head means $HH,HT,TH$:

$$P(\text{at least one head})=\frac{3}{4}$$

An everyday example. **In a lucky draw at a school fete with $200$ tickets**, a family holding $5$ tickets has a probability of $\frac{5}{200}=\frac{1}{40}$ of winning.

The substance. The formula needs equally likely outcomes: passing or failing a test are two outcomes, but that does not make each one $\frac{1}{2}$.

**The probability that an event does not happen is $1$ minus the probability that it does; every probability lies from $0$ to $1$, with $0$ for an impossible event and $1$ for a sure event.**

$$P(\text{not }A)=1-P(A)0\le P(A)\le 1$$

Worked example 1. $P(\text{a king})=\frac{4}{52}=\frac{1}{13}$, so

$$P(\text{not a king})=1-\frac{1}{13}=\frac{12}{13}$$

Worked example 2. If the probability of rain tomorrow is $0.35$, the probability of no rain is $1-0.35=0.65$.

Impossible and sure events on a die.

- **Getting a $7$** — impossible, $P=0$
- **Getting a number less than $7$** — sure, $P=1$

An everyday example. **A weather forecast giving a $35\%$ chance of rain** means a $65\%$ chance of a dry day.

The substance. **A probability of $1.2$ or $-0.3$ is always a mistake**, and the probabilities of all outcomes of an experiment add up to $1$.

Count the total number of equally likely outcomes, count the favourable ones carefully without counting any outcome twice, and divide.

Cards from a pack of $52$:

$$P(\text{red})=\frac{26}{52}=\frac{1}{2}P(\text{face card})=\frac{12}{52}=\frac{3}{13}P(\text{black king})=\frac{2}{52}=\frac{1}{26}$$

Neither a heart nor a king: hearts $13$, plus kings that are not hearts $3$, gives $16$ cards to avoid.

$$P=\frac{52-16}{52}=\frac{36}{52}=\frac{9}{13}$$

Balls. A bag has $5$ red, $3$ blue and $2$ green balls.

$$P(\text{blue})=\frac{3}{10}P(\text{not green})=\frac{8}{10}=\frac{4}{5}$$

Missing number. A bag has $15$ white balls and some red balls, and $P(\text{red})=\frac{3}{8}$:

$$\frac{x}{x+15}=\frac{3}{8}8x=3x+45x=9$$

Tickets numbered $1$ to $30$: multiples of $3$ or $5$ are $10+6-2=14$ numbers, since $15$ and $30$ are counted twice.

$$P=\frac{14}{30}=\frac{7}{15}P(\text{perfect square})=\frac{5}{30}=\frac{1}{6}$$

An everyday example. Raffle tickets at a Diwali mela are drawn exactly like these numbered tickets.

The trap. Count numbers or cards that satisfy both conditions only once.