Chapter 6 of 8

All 8 chapters
  1. 1Listing all the outcomes, readunderstood
  2. 2Relative frequency and expected results, readunderstood
  3. 3The OR rule and the AND rule, readunderstood
  4. 4Tree diagrams, readunderstood
  5. 5Picking without putting back, readunderstood
  6. 6Venn diagrams and sets, readunderstood
  7. 7Conditional probability, readunderstood
  8. 8Counting choices, readunderstood

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

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • 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. Step 1: a box labelled 30 pupils with two overlapping circles named football and running, and a ? in each of the four regions. Step 2: the overlap now holds 5 and the other three regions keep a ?. Step 3: the football circle only holds 12, the overlap 5, the running circle only 7, and 6 stand outside both.footballrunning30 pupils????

    1 Draw a box for all 30 pupils. Draw two overlapping circles, one for football and one for running.

  2. Step 1: a box labelled 30 pupils with two overlapping circles named football and running, and a ? in each of the four regions. Step 2: the overlap now holds 5 and the other three regions keep a ?. Step 3: the football circle only holds 12, the overlap 5, the running circle only 7, and 6 stand outside both.footballrunning30 pupils?5??

    2 Start in the overlap. 5 pupils like both, so write 5 there.

  3. Step 1: a box labelled 30 pupils with two overlapping circles named football and running, and a ? in each of the four regions. Step 2: the overlap now holds 5 and the other three regions keep a ?. Step 3: the football circle only holds 12, the overlap 5, the running circle only 7, and 6 stand outside both.footballrunning30 pupils12576

    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?

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

Answer: 6

A box labelled 30 pupils holding two overlapping circles named football and running. The football circle only holds 12, the overlap holds 5, the running circle only holds 7, and 6 stand outside both circles. Nothing is shaded.footballrunning30 pupils12576
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 box labelled 30 pupils holding two overlapping circles named football and running. The football circle only holds 12, the overlap holds 5, the running circle only holds 7, and 6 stand outside both circles. The overlap, the 5 pupils who like both, is hatched.footballrunning30 pupils12576
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 box labelled 30 pupils holding two overlapping circles named football and running. The football circle only holds 12, the overlap holds 5, the running circle only holds 7, and 6 stand outside both circles. Both circles are hatched, which is everyone who likes football or running or both.footballrunning30 pupils12576
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?

  1. A pupil likes football or running when the pupil is in either circle.
  2. Count the pupils in either circle: 12 + 5 + 7 = 24.
  3. P(A ∪ B) = 24/30.
  4. 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 box labelled 6 faces holding two overlapping circles named below 3 and above 4. The below 3 circle only holds 2, the overlap holds 0, the above 4 circle only holds 2, and 2 stand outside both circles. Both circles are hatched.below 3above 46 faces2022
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.

A box labelled 6 faces holding two overlapping circles named even and above 3. The even circle only holds 1, the overlap holds 2, the above 3 circle only holds 1, and 2 stand outside both circles. Both circles are hatched.evenabove 36 faces1212
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?

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

Answer: 13/30

A box labelled 30 pupils holding two overlapping circles named football and running. The football circle only holds 12, the overlap holds 5, the running circle only holds 7, and 6 stand outside both circles. Everything outside the football circle is hatched, which is A′, the 13 pupils who do not like football.footballrunning30 pupils12576
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.

  1. Ask each person two questions: do you have a dog at home, and do you have a cat?
  2. Draw a box and two overlapping circles, one for dogs and one for cats.
  3. Write each person in the right region. Someone with both goes in the overlap.
  4. 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 box labelled 10 people holding two overlapping circles named dog and cat. The dog circle only holds 3, the overlap holds 2, the cat circle only holds 4, and 1 stands outside both circles.dogcat10 people3241

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.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. One learner is chosen at random from the 40. What is the probability that the learner is in music or art or both?

    A box labelled 40 learners holding two overlapping circles named music and art. The music circle only holds 14, the overlap holds 8, the art circle only holds 7, and 11 stand outside both circles.musicart40 learners148711
    Tap your answer.

    1. Not this one. You added 22 + 15, the totals of the two circles. That counts the 8 in both twice. Add the three regions instead. Go to M4.6.6. Try another.

    2. Not this one. That is only the learners in music. Learners in art only are in the union too. Go to M4.6.6. Try another.

    3. Yes, that is right. Add the three regions that touch a circle: 14 + 8 + 7 = 29. Or 22 + 15 − 8 = 29. The chance is 29 out of the 40 in the box.

    4. Not this one. You divided by 22 + 15. The chance is out of all 40 learners in the box. Go to M4.6.6. Try another.

    See the answer

    29/40

    Add the three regions that touch a circle: 14 + 8 + 7 = 29. Or 22 + 15 − 8 = 29. The chance is 29 out of the 40 in the box.

    • If you chose A: You added 22 + 15, the totals of the two circles. That counts the 8 in both twice. Add the three regions instead. Go to M4.6.6.
    • If you chose B: That is only the learners in music. Learners in art only are in the union too. Go to M4.6.6.
    • If you chose D: You divided by 22 + 15. The chance is out of all 40 learners in the box. Go to M4.6.6.

  2. One learner is chosen at random from the 40. What is the probability that the learner is in both music and art?

    A box labelled 40 learners holding two overlapping circles named music and art. The music circle only holds 14, the overlap holds 8, the art circle only holds 7, and 11 stand outside both circles.musicart40 learners148711
    Tap your answer.

    1. Yes, that is right. 8 of the 40 learners are in both, so the probability is 8/40 = 1/5.

    2. Not this one. You divided by the learners in music. The chance is out of all 40 learners in the box. Go to M4.6.6. Try another.

    3. Not this one. You divided by the learners in music or art. The whole group is the 40 in the box. Go to M4.6.6. Try another.

    See the answer

    1/5

    8 of the 40 learners are in both, so the probability is 8/40 = 1/5.

    • If you chose B: You divided by the learners in music. The chance is out of all 40 learners in the box. Go to M4.6.6.
    • If you chose C: You divided by the learners in music or art. The whole group is the 40 in the box. Go to M4.6.6.

  3. In set notation, what does A ∩ B mean?

    Tap your answer.

    1. Not this one. That is the union, A ∪ B. The intersection is smaller: only what is in both. Go to M4.6.6. Try another.

    2. Yes, that is right. A ∩ B, the intersection, is the overlap of the two circles. Its chance is the chance of A and B.

    3. Not this one. That is A′, written with a small mark after the letter. The sign ∩ joins two sets. Go to M4.6.6. Try another.

    See the answer

    Everything in both A and B.

    A ∩ B, the intersection, is the overlap of the two circles. Its chance is the chance of A and B.

    • If you chose A: That is the union, A ∪ B. The intersection is smaller: only what is in both. Go to M4.6.6.
    • If you chose C: That is A′, written with a small mark after the letter. The sign ∩ joins two sets. Go to M4.6.6.

  4. 14 learners play chess, and 6 of them also play draughts. In a Venn diagram, what number goes in the region for chess only?

    Tap your answer.

    1. Not this one. That is everyone who plays chess. It includes the 6 who play both, and they sit in the overlap, not in the chess only region. Go to M4.6.6. Try another.

    2. Not this one. You added the 6 to the 14. The 14 already includes those 6, so take them off. Go to M4.6.6. Try another.

    3. Not this one. That is the overlap, the learners who play both. Chess only is the rest of the chess circle. Go to M4.6.6. Try another.

    4. Yes, that is right. The 14 who play chess include the 6 who play both. The chess only region holds 14 − 6 = 8.

    See the answer

    8

    The 14 who play chess include the 6 who play both. The chess only region holds 14 − 6 = 8.

    • If you chose A: That is everyone who plays chess. It includes the 6 who play both, and they sit in the overlap, not in the chess only region. Go to M4.6.6.
    • If you chose B: You added the 6 to the 14. The 14 already includes those 6, so take them off. Go to M4.6.6.
    • If you chose C: That is the overlap, the learners who play both. Chess only is the rest of the chess circle. Go to M4.6.6.

  5. P(A) = 7/10. What is the probability of A′, which means not A?

    Tap your answer.

    1. Not this one. That is P(A) again. A′ means not A, which is what is left. Go to M4.6.6. Try another.

    2. Not this one. That compares 3 with 7. A chance is out of the whole, which is 10. Go to M4.6.6. Try another.

    3. Yes, that is right. P(A′) = 1 − P(A) = 10/10 − 7/10 = 3/10.

    See the answer

    3/10

    P(A′) = 1 − P(A) = 10/10 − 7/10 = 3/10.

    • If you chose A: That is P(A) again. A′ means not A, which is what is left. Go to M4.6.6.
    • If you chose B: That compares 3 with 7. A chance is out of the whole, which is 10. Go to M4.6.6.

  6. From an earlier chapter

    A bag has 3 red and 2 blue counters. Two are taken without putting back. What is the probability that both are red?

    See the answer

    3/10

    P(both red) = 3/5 × 2/4 = 6/20 = 3/10.

  7. In a group of 40 learners, 25 play football, 18 play netball and 9 play both. How many play neither?

    See the answer

    6

    Football or netball: 25 + 18 − 9 = 34. Neither: 40 − 34 = 6.

  8. P(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.1. What is P(A ∪ B)?

    See the answer

    0.8

    P(A ∪ B) = 0.6 + 0.3 − 0.1 = 0.8.

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.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (868 KB). It opens on any phone.