14.1 Probability — A Theoretical Approach

Toss a coin: heads or tails? Throw a die: one of six faces. Close your eyes and pick a ball from a bag: red, blue or green? We cannot say in advance what will happen on one try, but we can say how likely each result is — and that is the whole of probability. A shopkeeper who knows that $1$ ticket in $500$ wins a prize, or a captain who knows a coin is fair, is using exactly this idea. The aim of this section is to turn the vague words "likely", "unlikely", "certain" and "impossible" into a number that you can calculate.

The method is wonderfully simple: list the possible outcomes, count the ones you want, and divide. It works whenever the outcomes are equally likely, so we spend time on that condition — it is where most mistakes are made. We then prove two rules that every probability obeys, and use them to solve problems about bags of balls, numbers, letters and tickets.

Experiments, outcomes and events

For a coin the sample space is $\{H, T\}$ (head, tail). For a die it is $\{1,2,3,4,5,6\}$. For a bag with $4$ red and $6$ blue balls, drawing one ball gives $10$ outcomes if we think of the balls as ten different objects — or two colours, red and blue. Which list is the right one? That depends on whether the outcomes are equally likely.

For example, when a die is thrown, the event "an even number" has the three favourable outcomes $2$, $4$, $6$; the event "a number greater than $4$" has two, $5$ and $6$. Notice that an event is described in words, but it is made of outcomes, and you must list them.

The probability of an event

So for one throw of a fair die, $P(\text{an even number})=\dfrac36=\dfrac12$: three favourable outcomes out of six. A probability may be written as a fraction (preferably in lowest terms), as a decimal or as a percentage: $\tfrac12=0.5=50\%$. The word "theoretical" tells you that the number is found by reasoning about the outcomes and not by actually performing the experiment. If you do toss a coin many times, the fraction of heads settles near $\tfrac12$ — you saw this in the chapter introduction — but you do not need to toss to know it.

Impossible events, sure events and complementary events

Return to Example 2. The probability of drawing a yellow ball was $0$: no outcome favours it. What would the probability be of drawing a ball that is red, white, green or blue? All $20$ balls are favourable, so $P=\dfrac{20}{20}=1$. These two extreme cases have names, and they are the ends of the scale on which every probability sits.

Proof. Let there be $n$ outcomes ($n\ge1$) and let $m$ of them be favourable to $E$. The number $m$ cannot be negative, and it cannot exceed the total, so $0\le m\le n$. Dividing every part by $n\gt0$ gives $0\le\dfrac mn\le1$. If $m=0$ the event has no favourable outcome, so it can never happen; conversely, an event that can never happen has $m=0$. If $m=n$ every outcome is favourable, so the event happens whatever the result; conversely, a sure event has $m=n$. $\blacksquare$

So "getting $7$ on a die" is impossible ($P=0$), and "getting a number less than $7$" is sure ($P=1$). Events in between are possible but not certain; the closer $P(E)$ is to $1$, the more likely $E$ is.

Proof. Let there be $n$ equally likely outcomes, of which $m$ are favourable to $E$. Every outcome is either favourable to $E$ or not, and never both, so exactly $n-m$ outcomes are favourable to $\overline{E}$. Therefore $$P(E)+P(\overline{E})=\frac mn+\frac{n-m}{n}=\frac{m+(n-m)}{n}=\frac nn=1.\ \blacksquare$$

The rule is very useful whenever "not $E$" is easier to count than $E$. If a lottery ticket wins with probability $0.04$, it loses with probability $1-0.04=0.96$. For a die, $P(\text{not a }6)=1-\tfrac16=\tfrac56$.

Exercises