Probability

Equally likely outcomes

When every outcome of an experiment is equally likely, the probability of an event $A$ is
$$\mathrm{P}(A)=\frac{\text{number of outcomes in}A}{\text{total number of outcomes}}\mathrm{.}$$
The set of all possible outcomes is the sample space. For small experiments, list it in a table. The syllabus example is the total score when two fair dice are thrown: a $6\times 6$ grid has 36 equally likely cells, one for each ordered pair of scores. A total of 9 comes from $(3,6),(4,5),(5,4),(6,3)$, so $\mathrm{P}(\text{total}9)=\frac{4}{36}=\frac{1}{9}$.

Notice that $(4,5)$ and $(5,4)$ are different outcomes. Treating "a 4 and a 5" as one outcome makes the outcomes unequally likely, and the counting goes wrong.

For larger experiments, count with permutations or combinations instead of listing. If 3 bulbs are chosen at random from 12, of which 4 are faulty, every selection of 3 is equally likely. There are $\left(\genfrac{}{}{0ex}{}{12}{3}\right)=220$ selections, and $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ of them contain no faulty bulb, so $\mathrm{P}(\text{no faulty bulb})=\frac{56}{220}=\frac{14}{55}$. The numerator and the denominator must count the same kind of thing: both selections, or both arrangements.

Give probabilities as exact fractions or as decimals correct to 3 significant figures.

Adding and multiplying probabilities

Adding. Two events are mutually exclusive if they cannot both happen. For mutually exclusive events,
$$\mathrm{P}(A\text{or}B)=\mathrm{P}(A)+\mathrm{P}(B)\mathrm{.}$$
This is why you add the probabilities of separate paths on a tree diagram: two different paths can never both happen.

Multiplying. Along a path on a tree, you multiply:
$$\mathrm{P}(A\text{and}B)=\mathrm{P}(A)\times \mathrm{P}(B\text{given}A)\mathrm{.}$$
If $A$ happens a fraction $\mathrm{P}(A)$ of the time, and on those occasions $B$ then follows a fraction $\mathrm{P}(B\text{given}A)$ of the time, both happen a fraction equal to the product.

The complement. $\mathrm{P}(\text{not}A)=1-\mathrm{P}(A)$. For "at least one", it is usually quickest to find $1-\mathrm{P}(\text{none})$.

You do not need the general formula $\mathrm{P}(A\cup B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A\cap B)$ explicitly; questions can always be done by splitting into exclusive cases, or with a Venn diagram, a table or a tree. It is still correct to use it.

Tree diagrams

A tree diagram shows a sequence of events. Each set of branches from one point covers all the possibilities, so their probabilities add up to 1. To find the probability of a combined outcome, multiply along the path; to combine several paths, add them.

The second set of branches depends on what happened first when objects are taken without replacement. The syllabus example is drawing balls at random from a bag containing balls of different colours. If a bag has 3 red and 5 blue counters and two are taken without replacement, after a red counter the bag holds 2 red and 5 blue, so the second-stage probabilities are $\frac{2}{7}$ and $\frac{5}{7}$; after a blue counter they are $\frac{3}{7}$ and $\frac{4}{7}$. With replacement, the second stage would be the same as the first.

In longer problems you do not have to draw the whole tree. Write down only the paths you need, but say which outcomes they are, for example $\mathrm{P}(RB)+\mathrm{P}(BR)$.

Each pair of branches adds up to 1. The probability of red then blue is $\frac{3}{8}\times \frac{5}{7}=\frac{15}{56}$.

Exclusive and independent events

Events $A$ and $B$ are mutually exclusive (or exclusive) if they cannot happen together: $\mathrm{P}(A\cap B)=0$. Throwing a die once, "score 1" and "score 6" are exclusive.

Events $A$ and $B$ are independent if the occurrence of one does not change the probability of the other. The test the syllabus uses is
$A$ and $B$ are independent if and only if
$$\mathrm{P}(A\cap B)=\mathrm{P}(A)\times \mathrm{P}(B)\mathrm{.}$$
To determine independence, work out the two sides separately and compare them, and write a conclusion: "$\mathrm{P}(A\cap B)=0.15$ and $\mathrm{P}(A)\times \mathrm{P}(B)=0.12$. These are not equal, so $A$ and $B$ are not independent."

Do not confuse the two ideas. Exclusive events with non-zero probabilities are never independent: if $A$ happens, $B$ certainly cannot, so knowing $A$ changes the chance of $B$ to zero. Independence is about the numbers, not about whether the events "seem related". When a die is thrown, "even" and "a multiple of 3" are independent ($\frac{1}{6}=\frac{1}{2}\times \frac{1}{3}$), although both are about the same throw.

Conditional probability

The probability of $A$ given that $B$ has happened is written $\mathrm{P}(A|B)$. Knowing that $B$ happened shrinks the sample space to $B$ alone, and the part of $A$ that is still possible is $A\cap B$. So
$$\mathrm{P}(A|B)=\frac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}\mathrm{.}$$
With equally likely outcomes this is simply counting inside $B$: if 8 of the 20 chess players in a club also do drama, $\mathrm{P}(\text{drama}|\text{chess})=\frac{8}{20}$.

On a tree diagram the second-stage branches are already conditional probabilities. The formula is needed when the question asks the "reverse" way round: given what happened second, what is the probability of what happened first? Then

• the numerator is the probability of the one path where both happen;
• the denominator is the total probability of the given event, which is usually the sum of two or more paths.

If $A$ and $B$ are independent, $\mathrm{P}(A|B)=\mathrm{P}(A)$: knowing $B$ tells you nothing about $A$.