Probability of Combined Events

Two events at once: listing the sample space

A combined event is a question where two things happen and you are asked about them together: a dice rolled and a coin tossed, a spinner spun twice, two counters taken from a bag. The first job is always the same - write down everything that can happen. That list is the sample space, and every outcome in it is equally likely when the question says fair, unbiased or at random.

A sample space diagram is that list written as a table, which stops you missing an outcome. The table below shows one spin of a three-section spinner numbered 1, 2, 3 together with one toss of a coin: three columns and two rows give six outcomes, and $P(\text{head and a 2})=\frac{1}{6}$ because exactly one of the six cells is H2.

Most one-mark questions in this topic just want the count. Two dice give $6\times 6=36$ outcomes; a four-section spinner spun twice gives $4\times 4=16$; two counters taken from a bag of 10 without replacement give $10\times 9=90$, because the second choice is made from nine counters. Multiply the number of choices at each step.

• Two dice: 36 equally likely outcomes
• A dice and a coin: 12 outcomes
• A fair spinner with 8 sections, spun twice: 64 outcomes
• Two counters from a bag of 10, without replacement: 90 outcomes

And means multiply: independent events

Two events are independent when the outcome of one does not change the probability of the other. A dice and a coin are independent; so are two spins of the same spinner, and two counters taken from a bag when the first is put back before the second is taken. The test is whether knowing the first result would change your prediction for the second.

For independent events, $P(A\text{ and }B)=P(A)\times P(B)$. A fair dice showing a 5 and a fair coin showing a head: $\frac{1}{6}\times \frac{1}{2}=\frac{1}{12}$. Two sixes with two dice: $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$. To multiply fractions, multiply the numerators and multiply the denominators, then cancel down.

The rule stretches to three events. Three heads in a row is $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$. If the same spinner is spun twice and red comes up with probability $\frac{3}{5}$, then two reds have probability $\frac{3}{5}\times \frac{3}{5}=\frac{9}{25}$.

Watch the wording, because 'both' is an and. 'Both counters are red' and 'the first is red and the second is red' describe the same event and use the same rule. A question that says 'or' needs a different rule, which is the next block.

• Two sixes: $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$
• Red twice on a spinner where $P(\text{red})=\frac{3}{5}$: $\frac{3}{5}\times \frac{3}{5}=\frac{9}{25}$
• Three heads in a row: $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$
• An 'and' answer is always smaller than either separate probability, so a value bigger than one of them is a warning sign

Or means add: mutually exclusive events

Mutually exclusive events cannot both happen in the same trial. A counter cannot be red and blue at once, and a dice cannot show a 3 and a 5 on the same roll. When two events are mutually exclusive, $P(A\text{ or }B)=P(A)+P(B)$.

A bag holds 8 red, 5 blue and 7 green counters, so 20 counters altogether. One counter is taken at random. The probability that it is red or blue is $\frac{8}{20}+\frac{5}{20}=\frac{13}{20}$, which you can check by counting: 13 of the 20 counters are red or blue.

If the two events can happen together, adding counts the overlap twice, so you take it off once. That is the general rule for two events: $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$. Rolling an even number and rolling a multiple of 3 on a fair dice gives $\frac{3}{6}+\frac{2}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$, because 6 is both even and a multiple of 3 and has been counted twice.

The two rules work as a pair. The 'and' rule produces $P(A\text{ and }B)$, which the 'or' rule then needs. When the events are mutually exclusive there is nothing in the overlap, so $P(A\text{ and }B)=0$ and the general rule collapses back to the simple one.

Tree diagrams: every branch labelled

A tree diagram draws one level of branches for each event. Cambridge's tree diagrams put the probability beside each branch and the outcome at the end of each branch, so the picture carries the numbers for you.

Reading a tree takes two moves. Multiply along a path to get the probability of the outcome written at the end of it. Add the paths that together make up the event the question asks about.

A bag holds 3 red and 2 blue counters. A counter is taken, its colour noted, and it is put back; then a second counter is taken. Both draws use the same pair of branch probabilities, $\frac{3}{5}$ for red and $\frac{2}{5}$ for blue, because the bag is back to five counters before the second draw.

The four paths give RR $\frac{3}{5}\times \frac{3}{5}=\frac{9}{25}$, RB $\frac{3}{5}\times \frac{2}{5}=\frac{6}{25}$, BR $\frac{2}{5}\times \frac{3}{5}=\frac{6}{25}$ and BB $\frac{2}{5}\times \frac{2}{5}=\frac{4}{25}$. The counters are different colours on the RB and BR paths, so $P(\text{different colours})=\frac{6}{25}+\frac{6}{25}=\frac{12}{25}$.

Two checks are worth doing every time. The two branches at any fork add to 1, and all the end-of-path probabilities add to 1: $\frac{9}{25}+\frac{6}{25}+\frac{6}{25}+\frac{4}{25}=1$.

• Branch probability multiplied by the next branch probability gives the path probability
• The paths that answer the question are added together
• At any fork the two branches sum to 1, and so do all the final outcomes
• RR and BB mean the same colour; RB and BR mean different colours

Tree diagrams without replacement (Extended)

If the first counter is not put back, the second draw is made from a smaller bag that is missing one counter of one colour. The second set of branch probabilities is therefore different from the first, and the two draws are not independent.

Take the same bag of 3 red and 2 blue counters, but this time do not replace the first counter. The first pair of branches is still $\frac{3}{5}$ and $\frac{2}{5}$. If the first counter was red, four counters remain and only 2 are red, so the second pair is $\frac{2}{4}$ and $\frac{2}{4}$. If the first counter was blue, four counters remain with 3 red, so that pair is $\frac{3}{4}$ and $\frac{1}{4}$.

So $P(\text{both red})=\frac{3}{5}\times \frac{2}{4}=\frac{6}{20}=\frac{3}{10}$, and $P(\text{one of each colour})=\frac{3}{5}\times \frac{2}{4}+\frac{2}{5}\times \frac{3}{4}=\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$. With $P(\text{both blue})=\frac{2}{5}\times \frac{1}{4}=\frac{1}{10}$, the four paths total $\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{1}{10}=1$.

The words to look for are with replacement (also written as 'put back' or 'replaced') against without replacement ('not replaced', or simply taking two counters at once). C8.3 says the Core combined-events questions are with replacement only; without replacement is Extended content (E8.3), and it is the version that makes the second set of branches unequal.

Venn diagrams: A or B, A and B, and not A

Two overlapping circles inside a rectangle draw the whole situation: the part of A outside B, the overlap, the part of B outside A, and the space outside both circles. Probability questions read the numbers in those four regions, so label the regions rather than the circles.

For two events, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The subtraction is there because the outcomes in the overlap sit inside both circles and have been counted twice. The complement $A^{\prime }$ is everything outside A, and $P(A^{\prime })=1-P(A)$.

A class of 25 students: 14 study History, 11 study Geography and 5 study both. Then 5 sits in the overlap, $14-5=9$ study History only, $11-5=6$ study Geography only, and $25-9-5-6=5$ study neither. A student chosen at random studies History or Geography with probability $\frac{9+5+6}{25}=\frac{20}{25}=\frac{4}{5}$, which is the same as $\frac{14}{25}+\frac{11}{25}-\frac{5}{25}$.

Venn diagrams are also the quickest way to see the complement. If 20 of the 25 study at least one of the two subjects, then $P(\text{neither})=1-\frac{4}{5}=\frac{1}{5}$, which is the 5 students outside both circles.

Reading a two-way table

A two-way table sorts one group by two features at once, such as boys and girls against walking and taking the bus. Each cell is a count, the row and column totals are given, and the bottom-right cell is the grand total.

The probability of an event is the number of people it covers divided by the group the question names. For 'a student chosen at random from the year group' the denominator is the grand total, 60. For a combined event such as 'a boy who walks', the numerator is the single cell, 18, and the denominator is still 60.

So $P(\text{a boy})=\frac{30}{60}=\frac{1}{2}$, $P(\text{walks})=\frac{24}{60}=\frac{2}{5}$, and $P(\text{a boy who walks})=\frac{18}{60}=\frac{3}{10}$. A combined event that uses 'and' picks out one cell; an event that uses 'or' picks out two cells or a whole row or column.

Use the totals printed in the table rather than adding cells up yourself. They are the numbers the question was built from, and they give you a free check that you are reading the right row.