Topic 8: Probability

Probability is the branch of mathematics that puts a number on uncertainty. Whether you are tossing a coin, drawing a coloured ball from a bag, or predicting how many faulty bulbs a factory makes, probability lets you reason precisely about chance. In IGCSE Mathematics you will move from simple single-event calculations to multi-stage problems handled with tree diagrams and Venn diagrams. The key skills are choosing the right rule (add or multiply), deciding whether events affect one another, and keeping your fractions accurate. Master a clear, organised method and probability becomes one of the most reliable sources of marks in the exam.

The probability scale and notation

Every probability is a number between 0 and 1 inclusive. A probability of 0 means the event is impossible, and a probability of 1 means it is certain. Values in between describe how likely an event is: $0.5$ means an even, fifty-fifty chance, while $0.1$ means unlikely and $0.9$ means very likely. We write the probability of an event $A$ as $P(A)$. Probabilities can be given as fractions, decimals or percentages, so $P(A)=\frac{1}{2}=0.5=50\%$ all mean the same thing. For equally likely outcomes the basic formula is $P(A)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$. For example, a fair six-sided die has six equally likely outcomes, so the probability of rolling a $4$ is $P(4)=\frac{1}{6}$, and the probability of rolling an even number is $P(\text{even})=\frac{3}{6}=\frac{1}{2}$. A crucial rule is that the probabilities of all possible outcomes must add up to $1$. This gives the complement rule: the probability that $A$ does not happen is $P(\text{not }A)=1-P(A)$. If the chance of rain is $0.3$, then the chance of no rain is $1-0.3=0.7$. The complement is often the fastest route to an answer, especially for phrases like 'at least one'.

Relative frequency and expected frequency

Theoretical probability assumes outcomes are equally likely, but real objects may be biased and some events have no obvious symmetry. In those cases we estimate probability from experiments using relative frequency: $\text{relative frequency}=\frac{\text{number of times the event happened}}{\text{total number of trials}}$. If a drawing pin is dropped $200$ times and lands point-up $130$ times, the estimated probability of point-up is $\frac{130}{200}=0.65$. The more trials you carry out, the closer the relative frequency tends to get to the true probability; this is sometimes called the law of large numbers. Expected frequency works in the opposite direction: once you know (or estimate) a probability, you can predict how many times an event will occur in a given number of trials using $\text{expected frequency}=P(\text{event})\times \text{number of trials}$. For instance, if the probability that a manufactured bulb is faulty is $0.02$ and the factory makes $4500$ bulbs, the expected number of faulty bulbs is $0.02\times 4500=90$. Expected frequency rarely matches the actual result exactly, but it gives the best single prediction. Always state expected frequencies as whole numbers when counting real objects, rounding sensibly if the calculation does not give a whole number.

Combined events and the addition rule

A combined event involves more than one outcome happening, joined by 'and' or 'or'. The addition rule deals with 'or'. Two events are mutually exclusive if they cannot both happen at the same time, such as rolling a $2$ or rolling a $5$ on one throw of a die. For mutually exclusive events, $P(A\text{ or }B)=P(A)+P(B)$. So the probability of rolling a $2$ or a $5$ is $\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$. When events are not mutually exclusive they can overlap, and adding the separate probabilities would count the overlap twice. The general addition rule corrects this: $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$. For example, drawing a card that is a king or a heart from a standard pack: $P(\text{king})=\frac{4}{52}$, $P(\text{heart})=\frac{13}{52}$, and the king of hearts is in both, so $P(\text{king or heart})=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$. Recognising whether events overlap is the key decision: if they cannot happen together, the subtracted term is zero and the rule simplifies.

Independent and dependent events, and the multiplication rule

The multiplication rule deals with 'and', meaning two events both happening, often one after another. Two events are independent if the outcome of one has no effect on the probability of the other, such as tossing a coin and then rolling a die. For independent events, $P(A\text{ and }B)=P(A)\times P(B)$. The probability of getting a head and then a $6$ is $\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}$. Events are dependent when the first outcome changes the probabilities for the second. The classic case is drawing objects without replacement: once you remove an item, both the favourable count and the total change. Suppose a bag holds $5$ red and $3$ blue counters. The probability of drawing two reds in a row without replacement is $P(\text{red then red})=\frac{5}{8}\times\frac{4}{7}=\frac{20}{56}=\frac{5}{14}$, because after one red is removed only $4$ reds remain out of $7$ counters. With replacement, the counter is returned, the bag resets, and the draws become independent: $\frac{5}{8}\times\frac{5}{8}=\frac{25}{64}$. Always check whether the problem says 'with replacement' or 'without replacement', as it completely changes the second fraction.

Tree diagrams

A tree diagram is the most reliable tool for multi-stage probability problems because it shows every possible path and keeps your working organised. Each set of branches represents one stage; the branches from a single point must have probabilities that add to $1$. To find the probability of a particular path, multiply along the branches. To find the probability of an event that can happen by several different paths, add the path probabilities together. Consider a bag of $5$ red and $3$ blue counters with two draws made without replacement. The four end paths are red-red, red-blue, blue-red and blue-blue. We already found $P(\text{RR})=\frac{5}{8}\times\frac{4}{7}=\frac{20}{56}$. Similarly $P(\text{RB})=\frac{5}{8}\times\frac{3}{7}=\frac{15}{56}$, $P(\text{BR})=\frac{3}{8}\times\frac{5}{7}=\frac{15}{56}$ and $P(\text{BB})=\frac{3}{8}\times\frac{2}{7}=\frac{6}{56}$. These four add to $\frac{56}{56}=1$, a useful check. The probability of one red and one blue in any order is $P(\text{RB})+P(\text{BR})=\frac{15}{56}+\frac{15}{56}=\frac{30}{56}=\frac{15}{28}$. For 'at least one' questions the complement is fastest: $P(\text{at least one red})=1-P(\text{BB})=1-\frac{6}{56}=\frac{50}{56}=\frac{25}{28}$.

Venn diagrams and set notation

Venn diagrams organise outcomes into overlapping sets and pair naturally with probability. The universal set, written $\xi$, contains all outcomes. The intersection $A\cap B$ is the region where both $A$ and $B$ occur, while the union $A\cup B$ is the region where $A$ or $B$ (or both) occur. The complement $A'$ is everything outside $A$. In set language the addition rule becomes $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Suppose in a class of $30$ students, $18$ study French, $12$ study Spanish, and $5$ study both. Place the $5$ in the overlap first, then $18-5=13$ in French only and $12-5=7$ in Spanish only, leaving $30-13-5-7=5$ who study neither. From this you can read off probabilities directly: a student chosen at random studies French only with probability $\frac{13}{30}$, and studies at least one language with probability $\frac{25}{30}=\frac{5}{6}$. Working from a completed Venn diagram is often easier than juggling formulae, because each region is a simple count. Always fill the intersection first and work outwards, checking that all regions sum to the total.

Conditional probability

Conditional probability is the probability of one event given that another has already happened, written $P(A\mid B)$ and read 'the probability of $A$ given $B$'. Knowing that $B$ has occurred restricts attention to only the outcomes inside $B$, so the formula is $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. With a Venn diagram this is simple: just use counts. Returning to the languages example, the probability that a student studies Spanish given that they study French is $P(\text{Spanish}\mid\text{French})=\frac{\text{both}}{\text{French total}}=\frac{5}{18}$, because among the $18$ French students, $5$ also take Spanish. Notice the denominator is the size of the condition, not the whole class. Dependent events on tree diagrams are really conditional probabilities in disguise: the second-stage branches show probabilities conditional on the first outcome. For example, in the counters problem $P(\text{second red}\mid\text{first red})=\frac{4}{7}$. The skill to practise is identifying the condition, narrowing your sample space to it, and then counting favourable outcomes within that reduced set.