Probability. Attnom Academy

M4.6.1 Chapter 1

Listing all the outcomes

I roll two dice and add the scores. Which total comes up most often?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A All the totals come up equally often, because each total is one outcome.
You picked A. Read on. At the end, the chapter comes back to your guess.
B A total of 7, because more pairs of scores make it.
You picked B. Read on. At the end, the chapter comes back to your guess.
C A total of 12, because it is the biggest.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

List all the outcomes of two events in a table
Find a probability from a table of outcomes

An outcome is one possible result, such as a 4 on a dice. An event is the result you are interested in, such as "an even score". It can hold more than one outcome.

When every outcome is equally likely, the probability of an event is a count divided by a count. It is the number of outcomes in the event, divided by the number of outcomes in all. For an even score on a dice, that is 3/6, which is 1/2.

Three of the six equally likely scores are even.

Now two things happen together: a coin is tossed and a dice is rolled. To count the outcomes without missing any, list them in a table. Put the coin outcomes down the side and the dice scores across the top.

Draw it first

List every outcome of a coin and a dice

1 List the outcomes of each thing. A coin has two: H for head and T for tail. A dice has six: the scores 1 to 6.
2 Make a table. Every coin outcome meets every score in one cell. Hatch the cells that fit the event: a head and an even score.

12 outcomes. The 3 hatched cells are a head and an even score.

Worked example

A coin is tossed and a dice is rolled. What is the probability of a head and an even score?

The coin has 2 outcomes and the dice has 6. Each coin outcome goes with each score.
So there are 2 × 6 = 12 outcomes in all, and they are equally likely.
The cells with a head and an even score are H2, H4 and H6. That is 3 outcomes.
P(head and even) = 3/12 = 1/4.

Answer: 1/4

The same table works for two dice. The first dice goes across the top and the second dice goes down the side. Each cell shows the total of its two scores. There are 36 cells.

A total of 7 appears in 6 of the 36 cells.

The 36 pairs are equally likely, because each dice is fair and one dice cannot change the other. But the 11 totals from 2 to 12 are not equally likely. A total of 7 fills 6 cells. A total of 2 fills only one cell.

The number of ways to make each total. The most likely total is 7.

Worked example

Two dice are rolled and added. What is the probability that the total is 10 or more?

Use the table of 36 equally likely pairs.
A total of 10 appears 3 times, 11 appears 2 times and 12 appears once.
That is 3 + 2 + 1 = 6 cells.
P(10 or more) = 6/36 = 1/6.

Answer: 1/6

Three totals make 10 or more, and together they fill 6 cells.

Try it

Roll two dice 36 times

You need: two dice, or two sets of six pieces of paper numbered 1 to 6, each set in its own bag, a pencil and paper for a tally.

Roll both dice and add the scores. Make a mark in the tally against that total.
Do this 36 times.
Draw a bar for each total.
Compare with the bar chart of ways. Do not expect them to match exactly.

What you will see: The middle totals, 6, 7 and 8, come up most often. The totals 2 and 12 come up least. Your bars will not match the ways exactly, because the rolls are chance.

If you cannot try it, this is what happens.

A wrong idea many people have

There are 11 totals, so each total has a probability of 1/11.

Why it feels right: The totals 2 to 12 are 11 results, and it seems fair to give each one the same share.

The test that breaks it: List the pairs. A total of 12 needs a 6 and a 6, so it is 1 of 36. A total of 7 can be made six ways, so it is 6 of 36.

What is true: Count only outcomes that are equally likely. The 36 pairs are. The 11 totals are not.

Key words

outcome: one possible result, such as a 5 on a dice or a tail on a coin.
event: the result you are interested in. It is made of one or more outcomes.
equally likely: having the same chance of happening, like the six faces of a fair dice.
probability: a number from 0 to 1 that measures how likely an event is.

Back to your guess

The answer is B: A total of 7, because more pairs of scores make it.

There are 36 pairs of scores, and all 36 are equally likely. A total of 7 can be made 6 ways. A total of 2 can be made only 1 way, with a 1 and a 1. So 7 comes up most often.

Keep it

Copy this into your exercise book.

List every outcome in a table. Check that they are equally likely. P(event) = outcomes in the event ÷ outcomes in all. Two dice make 36 pairs, but the totals are not equally likely.

Draw: A 6 by 6 table of two dice with the cells that give a total of 7 hatched.

M4.6.2 Chapter 2

Relative frequency and expected results

A bottle top is dropped on a table. How can I find the chance that it lands top up?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A It can land two ways, so the chance is 1/2.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Drop it many times and see what fraction of the drops land top up.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Chance cannot be measured, so there is no way to find out.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Estimate a probability from the results of an experiment
Find how many times to expect an outcome in 200 tries

Some events have no list of equally likely outcomes. A bottle top lands top up or the other way up, but the two are not equally likely. The only way to learn the chance is to try it.

Each go is a trial. The relative frequency of an event is the number of times it happened divided by the number of trials. It is an estimate of the probability.

Draw it first

From a tally to an estimate

1 Drop the bottle top 50 times. Count the drops that land top up. Here it is 18.
2 Divide the number of times it happened by the number of trials: 18 ÷ 50.
3 The result, 0.36, is the relative frequency. It is your estimate of the probability.

Worked example

A bottle top is dropped 200 times and lands top up 73 times. Estimate the probability that it lands top up.

Relative frequency = number of times it happened ÷ number of trials.
It happened 73 times in 200 trials.
73 ÷ 200 = 0.365.
The estimate of the probability is 0.365.

Answer: 0.365

One experiment gives one estimate. Here are the same bottle top's results after more and more drops. The estimates move around, then settle near 0.37.

The estimate is the last column. Early estimates swing. Later ones settle.

A fair coin shows the same thing, and here we know the true probability is 0.5. This record holds the first 40 of 100 tosses. H is a head and T is a tail.

The first 40 tosses, in order. Heads came up 19 times.

Now work out the relative frequency of heads after 10 tosses, 20 tosses, and so on to 100. Plot each one against the number of tosses.

With few tosses the estimate swings. With more tosses it stays close to 0.5.

More trials usually give a better estimate. But a relative frequency never proves the probability. After 100 tosses the coin shows 0.52, not 0.5, and a new set of 100 tosses would give a different number.

Now turn the idea round. If you know the probability, you can say what to expect. The expected number of times an event happens in n trials is the probability times n.

Worked example

A spinner lands on blue with probability 3/10. It is spun 200 times. How many times should you expect blue?

Expected number = probability × number of trials.
Here that is 3/10 × 200.
200 ÷ 10 = 20.
3 × 20 = 60.
You expect blue about 60 times.

Answer: 60

A fair coin tossed 200 times: 100 heads expected, 93 found. That is normal.

Worked example

A fair dice is rolled 200 times. How many sixes should you expect?

The probability of a six is 1/6.
Expected number = 1/6 × 200 = 200 ÷ 6. To 3 decimal places, 200 ÷ 6 = 33.333.
To 1 decimal place, 200 ÷ 6 = 33.3.
To the nearest whole number, you expect 33 sixes.
The expected number need not be a whole number, because it is an average over many sets of 200 rolls.

Answer: 33.3

Try it

Toss a coin and watch the estimate settle

You need: a coin, or a flat stone with a mark on one side, a pencil and paper.

Toss the coin 100 times. Write H or T each time.
After every 10 tosses, work out the relative frequency of heads so far.
Plot each value on a graph with tosses across and relative frequency up.
Compare your points at 10, 50 and 100 tosses. Which are closest to 0.5?

What you will see: The early points jump about. The later points move less and stay near 0.5. Nobody gets exactly 0.5 every time.

If you cannot try it, this is what happens.

A wrong idea many people have

The expected number is how many will happen.

Why it feels right: The word "expect" sounds like a promise, and 200 × 1/2 gives one clear number, 100.

The test that breaks it: Toss a fair coin 200 times, then do it again. The heads will be near 100 each time, but not exactly 100 each time.

What is true: The expected number is an average. The real number is usually close to it and often a little different.

Key words

trial: one go of an experiment, such as one toss of a coin.
relative frequency: the number of times an event happened divided by the number of trials. It estimates the probability.
expected number: the probability times the number of trials. It is an average, not a promise.

Back to your guess

The answer is B: Drop it many times and see what fraction of the drops land top up.

The two ways are not equally likely, because the top and the bottom are different shapes. Counting outcomes does not work. Instead, drop it many times. The fraction of drops that land top up is an estimate of the probability.

Keep it

Copy this into your exercise book.

Relative frequency = times it happened ÷ trials. It estimates a probability, and more trials usually improve it. Expected number = probability × trials, an average and not a promise.

Draw: A graph of relative frequency against trials, swinging at first and then settling near a dashed line.

M4.6.3 Chapter 3

The OR rule and the AND rule

A coin lands heads with chance 1/2 and a dice lands on 6 with chance 1/6. What is the chance of both?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Add them: 1/2 + 1/6 = 2/3.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Take away: 1/2 − 1/6 = 1/3.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Multiply them: 1/2 × 1/6 = 1/12.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Add probabilities for events that cannot both happen
Multiply probabilities for independent events

You often know the chance of each event. Then you need the chance of one event or another, or of both together. There are two rules, and each works only under a condition.

Take a spinner whose circle is split into tenths. Red takes 4 tenths, blue 3, green 2 and yellow 1. So P(red) = 4/10 and P(blue) = 3/10.

The spinner. One spin lands on one colour only.

What is the chance of red or blue? One spin cannot land on both. Events that cannot both happen are called mutually exclusive. For them, the chances simply add up.

Draw it first

Add the chances of red and blue

1 Shade the red part of the spinner: 4 of the 10 tenths.
2 Shade the blue part next to it: 3 more tenths. They do not overlap.
3 Red or blue is all the shaded tenths together: 4 + 3 = 7 out of 10.

This is the OR rule for events that cannot both happen: P(A or B) = P(A) + P(B).

Worked example

On the spinner, P(red) = 4/10 and P(blue) = 3/10. What is the probability of red or blue?

One spin cannot land on red and blue together, so the events cannot both happen.
P(red or blue) = P(red) + P(blue).
4/10 + 3/10 = 7/10.

Answer: 7/10

The condition matters. Roll a dice. P(even) = 3/6 and P(more than 3) = 3/6. Adding gives 1, a certainty, but the dice can show a 1 or a 3, which are neither. What went wrong?

A score of 4 is even and more than 3. So is a score of 6. Adding counted those two scores twice. The events can both happen, so you may not simply add.

Even or more than 3 is 4 faces out of 6, not 6 out of 6.

A wrong idea many people have

To find the chance of A or B, always add the two chances.

Why it feels right: The spinner example worked, and "or" sounds like "plus".

The test that breaks it: Roll a dice. Even and more than 3 each have chance 3/6. Their sum is 1, but a roll of 1 is neither. The faces 4 and 6 are in both, and adding counts them twice.

What is true: Add only for events that cannot both happen. If they can, take away the overlap. Chapter 6 shows how.

Now the other rule. A coin is tossed and a dice is rolled. The coin does not change the dice. Two events like this are independent: one happening does not change the chance of the other.

Half of all the cases have a head. Of those, 1 in 6 also has a six. So the chance of both is 1/6 of 1/2, which is 1/6 × 1/2 = 1/12. The table agrees.

One cell in 12 is a head and a 6. That is 1/2 × 1/6.

This is the AND rule for independent events: P(A and B) = P(A) × P(B).

Worked example

A spinner lands on red with probability 2/5. A coin lands heads with probability 1/2. Both are done once. What is the probability of red and a head?

The spinner does not change the coin, so the events are independent.
P(red and head) = P(red) × P(head).
2/5 × 1/2 = 2/10.
2/10 = 1/5.

Answer: 1/5

Worked example

Mei's bus is late with probability 1/10. Kofi's bus, in another town, is late with probability 1/5. What is the probability that both are late?

The two buses are in different towns, so one being late does not change the other.
The events are independent, so multiply.
Multiply the tops: 1 × 1 = 1.
Multiply the bottoms: 10 × 5 = 50.
So 1/10 × 1/5 = 1/50.

Answer: 1/50

Try it

Test the AND rule

You need: a coin, a dice, a pencil and paper.

Toss the coin and roll the dice together. Write down whether you got a head and a 6.
Do this 60 times. Count the "head and 6" results.
Divide that count by 60 for the relative frequency.
Compare it with 1/2 × 1/6, which is about 0.083.

What you will see: You should get about 5 results in 60. Yours may be 3 or 8. The rule gives the chance, and the experiment only estimates it.

If you cannot try it, this is what happens.

A wrong idea many people have

After a coin lands heads five times, tails is more likely next.

Why it feels right: It seems that the coin must "even things out".

The test that breaks it: Each toss is independent. The coin has no memory of the heads before. The chance of a head on the next toss is still 1/2.

What is true: Independent events do not change each other. Past tosses do not change the next toss.

Key words

mutually exclusive: events that cannot both happen at once, such as red and blue on one spin.
independent: events where one happening does not change the chance of the other.
OR rule: P(A or B) = P(A) + P(B), for events that cannot both happen.
AND rule: P(A and B) = P(A) × P(B), for independent events.

Back to your guess

The answer is C: Multiply them: 1/2 × 1/6 = 1/12.

For "this and that", with events that do not affect each other, multiply. The chance of both is 1/2 × 1/6 = 1/12. You can check it on the table from chapter 1: one cell, H6, out of 12.

Keep it

Copy this into your exercise book.

A or B: add, but only if they cannot both happen. A and B: multiply, but only if they are independent. A coin has no memory.

Draw: A bar of ten tenths with the red part and the blue part shaded and added together.

M4.6.4 Chapter 4

Tree diagrams

A bag has 3 red and 2 blue counters. I take one, look, put it back, and take another. What is the chance of one of each colour?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 6/25, from 3/5 × 2/5.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 12/25, because red then blue and blue then red both count.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 1/2, because there are two colours.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Draw a tree diagram for two events
Find probabilities by multiplying along the branches

A table lists the outcomes of two events well. When there are more events, or the chances are not equal, a tree diagram is better. It follows the events in order, one stage at a time.

Start with a coin tossed twice. The first toss is H or T, and each has chance 1/2. From each of those, the second toss is H or T again.

Draw it first

Draw a tree for two tosses of a coin

1 Draw the first toss: two branches, H and T, each marked 1/2. Then draw the second toss from the end of each.
2 At each end, write the outcome and its chance. Multiply the branches on the way: 1/2 × 1/2 = 1/4.

Two rules make a tree work. The branches that leave one point add up to 1, because one of them must happen. And along a path, you multiply, because each path is "this and then that".

Worked example

A fair coin is tossed twice. What is the probability of two heads?

The path for two heads is H then H.
Each branch has probability 1/2.
Multiply along the path: 1/2 × 1/2 = 1/4.

Answer: 1/4

See it three ways

With things

Toss a fair coin twice and write H or T for each toss. Do it a few times. Two heads comes up in 1 of every 4 goes, about.

As a picture

In words

List: the outcomes are HH, HT, TH and TT, and one of the 4 is HH. Table: one cell of 4 is HH. Tree: the path H then H has probability 1/2 times 1/2. All three ways give 1/4.

In symbols

P(HH) = 1/4 = 1/2 × 1/2.

Now the bag. It holds 3 red and 2 blue counters, so P(red) = 3/5 and P(blue) = 2/5. We put the counter back each time, so the second pick has the same chances as the first.

The bag holds 5 counters. Each counter is put back after it is taken.

The two hatched ends are one of each colour. Their chances are 6/25 each.

To find the chance of an event, find every end where it happens. Work out each end by multiplying. Then add those ends, because the ends cannot happen together.

Worked example

A counter is taken from the bag of 3 red and 2 blue, then put back, and a second is taken. What is the probability of one of each colour?

Red then blue: 3/5 × 2/5 = 6/25.
Blue then red: 2/5 × 3/5 = 6/25.
These are two different paths, so add them: 6/25 + 6/25 = 12/25.

Answer: 12/25

A good check: all the ends together must add to 1, because something must happen. Here 9/25 + 6/25 + 6/25 + 4/25 = 25/25 = 1.

A tree can have more stages. Toss a coin three times. Each stage doubles the number of ends, so there are 8 ends. Each end has chance 1/2 × 1/2 × 1/2 = 1/8.

Three tosses. The three hatched ends have exactly two heads.

Worked example

A fair coin is tossed three times. What is the probability of exactly two heads?

The paths with exactly two heads are HHT, HTH and THH. That is 3 paths.
Each path has probability 1/2 × 1/2 × 1/2 = 1/8.
Add the three paths: 3 × 1/8 = 3/8.

Answer: 3/8

Try it

Test the tree with a bag

You need: 3 white stones and 2 dark stones, a cloth bag or a cup, a pencil and paper.

Put the stones in the bag. Pick one without looking, write its colour, and put it back.
Pick again, write the colour, and put it back.
Do this 50 times. Count the times you got one white and one dark.
The tree says 12/25 of the pairs, so you expect about 24 of 50.

What you will see: Your count is near 24, but not exactly 24. The tree gives the chance. The experiment only estimates it.

If you cannot try it, this is what happens.

A wrong idea many people have

To find the chance of a path, add the branch chances.

Why it feels right: We add chances for "or", so it seems we should add them here too.

The test that breaks it: Try red then red with chances 3/5 and 3/5. Adding gives 6/5, which is more than 1. No chance can be more than 1.

What is true: A path is "this and then that". Multiply along it. Add only where different paths end in the same event.

Key words

tree diagram: a diagram of branches that shows the outcomes of events in order, with a chance on each branch.
branch: one line of a tree diagram, marked with a probability.

Back to your guess

The answer is B: 12/25, because red then blue and blue then red both count.

One of each can happen two ways: red then blue, and blue then red. Each way has chance 3/5 × 2/5 = 6/25. Add the two ways and you get 12/25. A tree diagram shows both ways at once.

Keep it

Copy this into your exercise book.

A tree shows events in order. Branches from one point add to 1. Multiply along a path. Add the ends that make the event. All ends add to 1.

Draw: A tree for two tosses of a coin, with 1/2 on each branch and 1/4 at each end.

M4.6.5 Chapter 5

Picking without putting back

A bag has 5 red and 3 blue counters. I take one, keep it, and take a second. What is the chance that both are red?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 25/64, because 5/8 × 5/8.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 5/14, because 5/8 × 4/7.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 5/16, because 5/8 × 4/8.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Draw a tree diagram where the first pick changes the second
Find the chance that two picks from a bag are both red

In chapter 4 each counter was put back, so the second pick had the same chances as the first. Now the counter is kept. This is picking without putting back.

The bag has changed, so the second pick has different chances. The two picks are dependent: the first changes the chance of the second.

Draw it first

How the bag changes

1 The bag at the start has 5 red and 3 blue counters, 8 in all.
2 A red counter is taken and kept. Now the bag has 4 red and 3 blue counters, 7 in all.

If a blue counter is taken instead, the bag has 5 red and 2 blue.

Read the chances for the second pick from the bag as it now is. After a red is kept: red is 4/7 and blue is 3/7. After a blue is kept: red is 5/7 and blue is 2/7. In each case the two add to 1.

Draw it first

Draw the tree for picking without putting back

1 Draw the first pick: red 5/8 and blue 3/8.
2 Draw the second pick from each end. Use the bag as it is now: after red, 4/7 and 3/7. After blue, 5/7 and 2/7.
3 Multiply along each path to write the chance at each end.

Worked example

A bag has 5 red and 3 blue counters. Two counters are taken without putting back. What is the probability that both are red?

The first counter is red with probability 5/8.
Then 4 red and 3 blue are left, so the second is red with probability 4/7.
P(both red) = 5/8 × 4/7 = 20/56.
20/56 = 5/14.

Answer: 5/14

The tree for picking without putting back. The hatched end is both red.

Worked example

The same bag, two counters taken and kept. What is the probability of one counter of each colour?

Red then blue: 5/8 × 3/7 = 15/56.
Blue then red: 3/8 × 5/7 = 15/56.
The two paths are different, so add: 15/56 + 15/56 = 30/56.
30/56 = 15/28.

Answer: 15/28

Check the tree: the four ends must add to 1. Over 56, they are 20 + 15 + 15 + 6 = 56, so 56/56 = 1. This is a good way to catch a slip.

Compare with putting the counter back. The chance of both red falls from 25/64 to 5/14, because taking a red leaves fewer reds in the bag. The chance of one of each rises from 15/32 to 15/28.

The same bag, put back or kept. Keeping the first counter makes "same colour" less likely.

Try it

Pick two stones and keep them

You need: 5 white stones and 3 dark stones, a cloth bag or a cup, a pencil and paper.

Put the 8 stones in the bag. Pick two without looking, and write the colours.
Put both back, shake the bag, and do it again.
Do this 40 times. Count the times both were white.
The tree says 5/14 of the pairs, so you expect about 14 of 40.

What you will see: Your count is near 14. If you put the first stone back each time, you would see more pairs of white stones, about 16 in 40.

If you cannot try it, this is what happens.

A wrong idea many people have

The second pick has the same chances as the first.

Why it feels right: The bag looks the same, and the same colours are in it.

The test that breaks it: Take a red counter and keep it. The bag now has 7 counters, not 8, and only 4 are red. The chance of red is 4/7, not 5/8.

What is true: When a counter is kept, change both numbers for the second pick. There is one fewer in all, and one fewer of the colour taken.

Key words

without putting back: taking a counter and keeping it out, so the next pick is from a smaller bag.
dependent: events where one happening changes the chance of the other.

Back to your guess

The answer is B: 5/14, because 5/8 × 4/7.

After a red counter is taken and kept, 7 counters are left, and only 4 of them are red. So the second pick has chance 4/7, not 5/8. The chance of both is 5/8 × 4/7 = 20/56 = 5/14.

Keep it

Copy this into your exercise book.

Without putting back, the bag shrinks. Second pick: one fewer counter in all, and one fewer of the colour taken. Multiply along paths. All ends add to 1.

Draw: A tree for two picks from a bag of 5 red and 3 blue, with 5/8, 3/8 and the changed second branches.

M4.6.6 Chapter 6

Venn diagrams and sets

In a class of 30, 17 like football, 12 like running, and 5 like both. What is the chance that a pupil chosen at random likes football or running?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 29/30, from 17/30 + 12/30.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 4/5, because 5 pupils are counted twice and must be taken off once.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 1/6, the pupils who like both.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Draw a Venn diagram for two sets
Find P(A or B) and P(A and B) from a Venn diagram
Use set notation like A ∪ B and A ∩ B

A set is a collection of things, such as the pupils who like football. A Venn diagram shows sets as circles inside a box. The box holds everything we are talking about, and each circle holds one set.

Let A be the set of pupils who like football and B the set who like running. Some pupils like both, so the circles overlap.

Draw it first

Fill in a Venn diagram

1 Draw a box for all 30 pupils. Draw two overlapping circles, one for football and one for running.
2 Start in the overlap. 5 pupils like both, so write 5 there.
3 Football has 17 in all, so 17 − 5 = 12 are in its circle only. Running has 12 in all, so 12 − 5 = 7 are in its circle only. The rest, 30 − 24 = 6, are outside both.

Always begin with the overlap. The totals for each set, 17 and 12, include the overlap, so each circle needs the overlap taken off.

Worked example

In a class of 30, 17 like football, 12 like running, and 5 like both. How many pupils like neither?

Both: 5 pupils. Football only: 17 − 5 = 12 pupils.
Running only: 12 − 5 = 7 pupils.
Football or running or both: 12 + 5 + 7 = 24 pupils.
Neither: 30 − 24 = 6 pupils.

Answer: 6

The four regions add up to 30: 12 + 5 + 7 + 6.

Now read chances from the diagram. Choose one pupil at random. Each of the 30 pupils is equally likely, so a chance is a count of pupils divided by 30.

The intersection of A and B is the pupils in both circles. It is written A ∩ B. Its probability is P(A ∩ B) = 5/30 = 1/6. This is the chance of A and B.

A ∩ B: the overlap, in both circles. 5 of 30.

The union of A and B is the pupils in A or B or both. It is written A ∪ B. All three regions that touch a circle count, so P(A ∪ B) = 24/30 = 4/5. This is the chance of A or B.

A ∪ B: everything in either circle. 24 of 30.

Worked example

A pupil is chosen at random from the class of 30. What is the probability that the pupil likes football or running?

A pupil likes football or running when the pupil is in either circle.
Count the pupils in either circle: 12 + 5 + 7 = 24.
P(A ∪ B) = 24/30.
24/30 = 4/5.

Answer: 4/5

You can also reach this from the totals. P(A) = 17/30 and P(B) = 12/30. Adding counts the overlap twice, so subtract it once.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Here that is 17/30 + 12/30 − 5/30 = 24/30. This matches the diagram.

In chapter 3 you added chances for events that cannot both happen. Then the overlap is empty, P(A ∩ B) = 0, and nothing is subtracted. The rule above covers that case too.

A dice score below 3 and a score above 4 cannot both happen. The overlap holds 0, so you add: 2/6 + 2/6.

Now the dice from chapter 3. Even scores are 2, 4 and 6. Scores above 3 are 4, 5 and 6. The scores 4 and 6 are in both.

Even or above 3 is 1 + 2 + 1 = 4 faces out of 6, which is 2/3.

One more piece of notation. A′ means "not A": everything in the box outside circle A. Its chance is P(A′) = 1 − P(A).

Worked example

A pupil is chosen at random from the class of 30. What is the probability that the pupil does not like football?

A′ is every pupil outside the football circle.
P(A) = 17/30, so A′ is the other 30 − 17 = 13 pupils.
P(A′) = 13/30.
Check with 1 − P(A) = 30/30 − 17/30 = 13/30.

Answer: 13/30

A′: everything outside the football circle. 13 of 30.

Try it

A Venn diagram of ten people

You need: a piece of paper, a pencil, ten people to ask, such as your family and friends.

Ask each person two questions: do you have a dog at home, and do you have a cat?
Draw a box and two overlapping circles, one for dogs and one for cats.
Write each person in the right region. Someone with both goes in the overlap.
Count each region and check that the counts add to 10.

What you will see: Every person goes in exactly one region. That is why the four counts must add up to the number of people.

If you cannot try it, this is what happens.

A wrong idea many people have

Write 17 in the football circle and 12 in the running circle.

Why it feels right: The question gives 17 for football and 12 for running, and a circle is where football pupils go.

The test that breaks it: The regions would then hold 17 + 5 + 12 + 6 = 40 pupils, but the class has only 30. The 17 already includes the 5 in the overlap.

What is true: Start with the overlap. In the rest of each circle, write the total for that set minus the overlap.

Key words

set: a collection of things, such as the pupils who like football.
Venn diagram: a diagram of circles in a box that shows how sets overlap.
union: everything in A or B or both. It is written A ∪ B.
intersection: everything in both A and B. It is written A ∩ B.

Back to your guess

The answer is B: 4/5, because 5 pupils are counted twice and must be taken off once.

The 5 pupils who like both are in the 17 and in the 12, so adding counts them twice. Take 5 off once: 17 + 12 − 5 = 24 pupils like football or running. 24/30 = 4/5.

Keep it

Copy this into your exercise book.

Start with the overlap. Regions add to the whole. A ∩ B is both. A ∪ B is either or both. P(A ∪ B) = P(A) + P(B) − P(A ∩ B). P(A′) = 1 − P(A).

Draw: A box with two overlapping circles, the overlap filled in first, and the four counts adding to the whole.

M4.6.7 Chapter 7

Conditional probability

In a class of 60, some pupils walk to school and some take the bus. If I pick a girl, what is the chance that she walks?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 32/60, because 32 of the 60 pupils walk.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 18/60, because 18 girls walk.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 18/30, because I only look at the 30 girls.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find a probability given that something has already happened
Use a two way table to find a conditional probability

Sometimes you already know something. You are told that a pupil is a girl, or that the first counter was red. This changes the chance of what comes next.

The chance of B when you know that A has already happened is called a conditional probability. We write it P(B given A).

A two way table is the best way to see it. Here are the 60 pupils, sorted by girl or boy and by walk or bus. The totals are worked out for you.

The class of 60. 30 girls, 30 boys, 32 who walk and 28 who take the bus.

Draw it first

Find P(walks given girl) from the table

1 Start with the table of all 60 pupils.
2 You know the pupil is a girl, so look only at the Girls row. It holds 30 pupils.
3 In that row, count the girls who walk: 18. The chance is 18 out of 30.

Worked example

A girl is chosen at random from the class of 60. What is the probability that she walks?

We know she is a girl, so look at the Girls row only. It has 18 + 12 = 30 pupils.
In that row, 18 pupils walk.
P(walks given girl) = 18/30.
18/30 = 3/5.

Answer: 3/5

Notice what changed. The chance that a pupil in the whole class walks is 32/60, which is 8/15. For a girl it is 3/5. Knowing the pupil is a girl changed the chance.

Why does the table method work? The chance that a pupil is a girl and walks is P(A and B) = 18/60. The chance of a girl is P(A) = 30/60. Divide the first by the second and the 60 cancels.

This gives the rule P(B given A) = P(A and B) ÷ P(A). Here, 18/60 ÷ 30/60 = 18/30. Dividing by P(A) shrinks the world to the pupils in A.

Worked example

A pupil who walks is chosen at random. What is the probability that the pupil is a girl?

We know the pupil walks, so look at the Walk column only. It has 18 + 14 = 32 pupils.
In that column, 18 are girls.
P(girl given walks) = 18/32.
18/32 = 9/16.

Answer: 9/16

P(girl given walks): look only at the Walk column. 18 of its 32.

Compare the two answers. P(walks given girl) = 3/5. P(girl given walks) = 9/16. They are different, because they look at different groups: the girls, and the walkers.

A wrong idea many people have

P(A given B) is the same as P(B given A).

Why it feels right: Both mention the same two events, and A and B seem to be sharing the same overlap.

The test that breaks it: The overlap is the same, 18 pupils, but it is divided by different totals. 18 out of 30 girls walk, while 18 out of 32 walkers are girls.

What is true: The event after "given" is the one you know. It sets the group you divide by. Always check which one it is.

You have met conditional chances before. In chapter 5, a bag held 5 red and 3 blue counters, and the first counter was kept. The branches for the second pick are conditional chances.

After a red first, the second branches are 4/7 and 3/7. They are P(second given first is red).

Worked example

Use the rule P(B given A) = P(A and B) ÷ P(A). A is "the first counter is red" and B is "the second is red". What is P(B given A)?

P(A and B) = P(red, red) = 5/8 × 4/7 = 5/14.
P(A) = P(first is red) = 5/8.
P(B given A) = 5/14 ÷ 5/8 = 5/14 × 8/5 = 40/70.
40/70 = 4/7. This is the branch on the tree.

Answer: 4/7

There is a link with independence. Events are independent when knowing one does not change the chance of the other: P(B given A) = P(B). In the pupil table, P(walks) = 8/15 but P(walks given girl) = 3/5. These differ, so being a girl and walking are not independent in this class.

Try it

Make a two way table of twenty people

You need: a pencil and paper, twenty people to ask.

Ask each person two questions: do you drink tea, and do you drink milk?
Draw a table with Tea and No tea down the side, and Milk and No milk across the top.
Make a mark in the right cell for each person, then count and add the totals.
Find P(milk given tea): look only at the tea row.

What you will see: The row for tea has its own total. That total, and not 20, is what you divide by.

If you cannot try it, this is what happens.

Key words

conditional probability: the chance of one event when you already know that another has happened.
given: the word that names what you already know. P(B given A) is the chance of B when A has happened.

Back to your guess

The answer is C: 18/30, because I only look at the 30 girls.

You already know the pupil is a girl, so only the 30 girls matter. Of those 30, 18 walk. The chance is 18/30 = 3/5.

Keep it

Copy this into your exercise book.

P(B given A): look only at the row or column of A. P(B given A) = P(A and B) ÷ P(A). P(B given A) and P(A given B) differ. Independent: P(B given A) = P(B).

Draw: A two way table with the row of the known event hatched and the matching cell marked.

M4.6.8 Chapter 8

Counting choices

Mei has 3 scarves and 2 hats. How many different outfits can she make with one scarf and one hat?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 3, one for each scarf.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 5, because 3 + 2.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 6, because each of the 3 scarves goes with each of the 2 hats.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Count outcomes with the product rule
Find how many ways there are to arrange 4 different letters

To find a probability by counting, you often need to know how many outcomes there are in all. Listing them works for small cases. For bigger cases, you need a faster way to count.

Start with Mei's clothes. A tree with no chances on its branches lists the outfits. Each scarf has two branches, one for each hat.

Draw it first

List the outfits as a tree

1 Draw the first choice: three branches for the three scarves.
2 From each scarf, draw two branches for the two hats. Number the ends: there are 6 outfits.

Notice that the tree has 3 groups of 2 ends. The number of ends is 3 × 2. This is the product rule. Suppose one choice can be made in a ways, and another in b ways. Then both can be made in a × b ways.

One box for each choice. Multiply the numbers in the boxes.

Worked example

Mei has 3 scarves and 2 hats. How many outfits can she make with one scarf and one hat?

There are 3 ways to choose the scarf.
For each scarf there are 2 ways to choose the hat.
Multiply: 3 × 2 = 6 outfits.

Answer: 6

The rule works for any number of choices. Multiply the number of ways for each one. A lock has 3 wheels, and each wheel shows a digit from 0 to 9. Each wheel has 10 ways.

A lock with three wheels of ten digits. 10 × 10 × 10 codes.

Worked example

A lock has 3 wheels. Each wheel shows a digit from 0 to 9. How many different codes are there?

Each wheel has 10 ways, and a wheel can repeat a digit shown by another wheel.
The first two wheels together have 10 × 10 = 100 ways.
With the third wheel, 100 × 10 = 1 000 codes.

Answer: 1 000

Worked example

Joseph guesses a code for the lock at random. What is the probability that he opens it on his first try?

All 1 000 codes are equally likely.
Only one code opens the lock.
P(opens it) = 1/1000.
As a decimal, 1 ÷ 1 000 = 0.001.

Answer: 1/1000

Now a different kind of choice: putting things in order. Each way of doing it is an arrangement. How many arrangements of the letters A, B and C in a row are there, each letter used once? A tree lists them. After you choose the first letter, only two letters are left, and after the second, only one.

Three letters in a row: 6 arrangements.

The number of choices falls each time, because a letter that is used cannot be used again. So the number of ways is 3 × 2 × 1 = 6, which agrees with the tree.

Three letters: 3 choices, then 2, then 1.

Four letters A, B, C and D work the same way. There are 4 ways to choose the first letter. Then 3 are left for the second, 2 for the third and 1 for the last.

Four different letters can be arranged in 4 × 3 × 2 × 1 = 24 ways.

Worked example

In how many ways can 4 different letters be arranged in a row?

The first place can be filled in 4 ways.
Then 3 letters are left. The first two places can be filled in 4 × 3 = 12 ways.
Then 2 letters are left, so the first three places can be filled in 12 × 2 = 24 ways.
The last place has 1 way, so 4 × 3 × 2 × 1 = 24.

Answer: 24

Counting gives probabilities. If the 4 letters are put in a random order, all 24 arrangements are equally likely. The chance that they come out as ABCD is 1 out of 24.

Try it

Find all 24 arrangements

You need: four small pieces of paper, a pencil.

Write A, B, C and D on the four pieces.
Put A first. List every way to arrange B, C and D after it. There should be 6.
Do the same with B first, then C first, then D first.
Add up. Each first letter gives 6, so the total is 4 × 6 = 24.

What you will see: Each first letter gives the same count, 3 × 2 × 1 = 6. That is the product rule at work.

If you cannot try it, this is what happens.

A wrong idea many people have

To count the outfits, add the numbers of choices: 3 + 2.

Why it feels right: We add when we put groups together, and here there are two groups, scarves and hats.

The test that breaks it: Add and you get 5, but the tree shows 6 outfits. Each scarf is worn with each hat, so every scarf is counted twice, once for each hat.

What is true: When you make one choice and then another, multiply. Add only when you choose from one group or from the other, but not both.

A wrong idea many people have

Each place in a row of 4 letters has 4 choices, so there are 4 × 4 × 4 × 4 ways.

Why it feels right: The first place does have 4 choices, and it seems the others should too.

The test that breaks it: A letter used in the first place cannot be used again. The second place has only 3 letters left, then 2, then 1.

What is true: When nothing can be used twice, each choice has one fewer option than the last: 4 × 3 × 2 × 1.

Key words

product rule: if one choice has a ways and another has b ways, both together have a × b ways.
arrangement: a way of putting things in order, such as the letters ABCD in a row.

Back to your guess

The answer is C: 6, because each of the 3 scarves goes with each of the 2 hats.

Each scarf can be worn with either hat. That makes 2 outfits for the first scarf, 2 for the second and 2 for the third. 2 + 2 + 2 = 6, which is 3 × 2.

Keep it

Copy this into your exercise book.

Product rule: a ways then b ways make a × b ways. Nothing used twice: the choices fall by one each time. 4 letters: 4 × 3 × 2 × 1 = 24.

Draw: A row of four boxes holding 4, 3, 2 and 1, with times signs between and 24 at the end.

List every outcome of a coin and a dice

Roll two dice 36 times

From a tally to an estimate

Toss a coin and watch the estimate settle

Add the chances of red and blue

Test the AND rule

Draw a tree for two tosses of a coin

Test the tree with a bag

How the bag changes

Draw the tree for picking without putting back

Pick two stones and keep them

Fill in a Venn diagram

A Venn diagram of ten people

Find P(walks given girl) from the table

Make a two way table of twenty people

List the outfits as a tree

Find all 24 arrangements

Key words, A to Z