Two-way tables and multistage events

Reading probabilities from two-way tables, tree diagrams for multistage events, independent events and the product rule, and drawing with and without replacement.

Real questions rarely involve one coin. A survey sorts people by two things at once. A weekend has two days that might be wet. A pen comes out of a pencil case and doesn't go back in. Two-way tables and tree diagrams are the tools that keep these multistage situations organised, so you count everything once and nothing twice.

Two-way tables

A two-way table sorts a group by two categories. The totals at the ends of each row and column are your best friends: they tell you what to divide by.

A school surveyed 200 Year 12 students about whether they have a part-time job and whether they drive to school.

	Drives	Doesn't drive	Total
Has a job	46	64	110
No job	24	66	90
Total	70	130	200

Example 1●●

Using the survey, one student is chosen at random. Find the probability that the student (a) has a job and drives, (b) has a job or drives (or both), (c) drives, given that they have a job.

$P(\text{job and drives}) = \frac{46}{200} = 0.23$

'And' means both at once, which is the single cell where the 'Has a job' row meets the 'Drives' column. Divide by everyone surveyed.

$P(\text{job or drives}) = \frac{46 + 64 + 24}{200} = \frac{134}{200} = 0.67$

'Or' includes every cell in the job row or the drives column. Add the three cells so the 46 is counted only once.

$P(\text{drives, given job}) = \frac{46}{110} \approx 0.42$

'Given that they have a job' shrinks the group to the 110 students in that row. Only those students count, so 110 is the new total.

Tree diagrams

A tree diagram shows a multistage experiment one stage at a time. Each set of branches from one point must add to 1, because one of them has to happen. Each complete path from the start to the end is one outcome.

Reading a tree

$$P(\text{a path}) = \text{multiply along the branches} \qquad P(\text{an event}) = \text{add the paths in it}$$

Why multiply? Suppose Saturday has a $0.3$ chance of rain. Of all the wet Saturdays, Sunday is also wet $0.3$ of the time. So both days are wet in $0.3$ of $0.3$ of all weekends, and "of" means multiply: $0.3 \times 0.3 = 0.09$.

When the result of one stage doesn't change the chances at the next stage, the events are independent, and the branch probabilities are the same each time.

Product rule for independent events

$$P(A \text{ and } B) = P(A) \times P(B)$$

InteractiveGrow a rainy-weekend tree

Try this

Set the chance of rain each day and the number of days. Pick an event and the matching paths light up. The Check readout shows the shortcut calculation.

With a $30\%$ chance of rain, how likely is at least one wet day over 3 days? Now push the chance up to $50\%$. Notice how quickly "at least one" heads towards certain as you add days.

Example 2●●

The forecast gives a $30\%$ chance of rain on each of Saturday and Sunday, independently. Find the probability that (a) both days are wet, (b) exactly one day is wet, (c) at least one day is wet.

$P(W) = 0.3,\ P(D) = 0.7$

The branches for each day are W (0.3) and D (0.7), which add to 1.

$P(WW) = 0.3 \times 0.3 = 0.09$

Both wet is the single path W then W. Multiply along it.

$P(WD) + P(DW) = 0.3 \times 0.7 + 0.7 \times 0.3 = 0.21 + 0.21 = 0.42$

Exactly one wet day has two paths, WD and DW. Find each, then add.

$1 - P(DD) = 1 - 0.7 \times 0.7 = 1 - 0.49 = 0.51$

At least one wet day is the complement of DD (both dry), which is one path.

Check: the four paths are $0.09 + 0.21 + 0.21 + 0.49 = 1$. If your paths don't add to 1, something is wrong.

With and without replacement

If you draw an object, put it back and draw again, the second draw is exactly like the first: with replacement gives independent stages. If you keep the first object, there is one fewer object for the second draw and the probabilities change: without replacement gives dependent stages.

Example 3●●●

A pencil case holds 4 blue pens and 2 black pens. Two pens are taken out, one after the other, without replacement. Find the probability that (a) both pens are blue, (b) the pens are different colours.

B is blue and K is black. The second-stage fractions are out of 5, because one pen has already gone.

$P(B_1) = \frac{4}{6}$

First draw: 4 of the 6 pens are blue.

$P(B_2 \text{ after } B_1) = \frac{3}{5}$

If the first pen was blue, only 3 blue pens remain out of 5 pens.

$P(BB) = \frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5}$

Multiply along the BB path.

$P(BK) + P(KB) = \frac{8}{30} + \frac{8}{30} = \frac{16}{30} = \frac{8}{15}$

Different colours can happen two ways, blue then black or black then blue. Add both paths.

With replacement, the second draw would still be $\frac{4}{6}$ for blue, so $P(BB) = \frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$. That is a little higher, because putting the blue pen back keeps more blue pens in the case.

Your turn

A bag has 5 green and 3 white lollies. Two are eaten one after the other. Find the probability that both are green.

Show the answer

Eating is definitely without replacement. $P(GG) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \approx 0.36$.

Flashcards

Tree diagram rule

Multiply along a path. Add the paths that make up the event.

Independent events

$P(A \text{ and } B) = P(A) \times P(B)$, because one result doesn't change the other.

Without replacement

The second stage has one fewer object, so the denominator drops by 1.

Practice

Using the job and driving survey on this page, find the probability, as a decimal, that a randomly chosen student has a job.
Using the same survey, find the probability, as a decimal, that a randomly chosen student drives to school.
A health survey of 300 people found: vaccinated and got sick 12; vaccinated and stayed well 188; not vaccinated and got sick 30; not vaccinated and stayed well 70. Of the vaccinated people, what proportion got sick?
Events $A$ and $B$ are independent with $P(A) = 0.25$ and $P(B) = 0.4$. Find $P(A \text{ and } B)$.
On a tree diagram, the branches coming out of one point have probabilities $0.35$ and $x$. What is $x$?
The chance of rain on any day this week is $0.2$, independently. Find the probability of at least one wet day in the next 3 days.
A bag has 3 red and 5 blue marbles. Two are drawn without replacement. What is $P(\text{both red})$?
A basketballer makes $60\%$ of free throws, and each shot is independent. Find the probability, as a decimal, that she makes both of her next two free throws.
Exam style: $70\%$ of a Year 12 class study for a test. Of those who study, $90\%$ pass. Of those who don't study, $40\%$ pass. Use a tree diagram to find the probability, as a decimal, that a randomly chosen student passes.
Exam style: For the same class, find the probability, as a decimal, that a randomly chosen student did not study and failed.