Chapter 3 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

M4.6.3 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

  • 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

  • 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.

A spinner drawn as a circle in four sectors: red 4, blue 3, green 2 and yellow 1, which are tenths of the circle.red4blue3green2yellow1
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. Three bars, each cut into 10 equal parts. Step 1: the bar named red 4/10 has 4 parts hatched. Step 2: the bar named blue 3/10 has the next 3 parts shaded with dots. Step 3: the bar named red or blue 7/10 has all 7 of those parts shaded.red 4/10

    1 Shade the red part of the spinner: 4 of the 10 tenths.

  2. Three bars, each cut into 10 equal parts. Step 1: the bar named red 4/10 has 4 parts hatched. Step 2: the bar named blue 3/10 has the next 3 parts shaded with dots. Step 3: the bar named red or blue 7/10 has all 7 of those parts shaded.red 4/10blue 3/10

    2 Shade the blue part next to it: 3 more tenths. They do not overlap.

  3. Three bars, each cut into 10 equal parts. Step 1: the bar named red 4/10 has 4 parts hatched. Step 2: the bar named blue 3/10 has the next 3 parts shaded with dots. Step 3: the bar named red or blue 7/10 has all 7 of those parts shaded.red 4/10blue 3/10red or blue 7/10

    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?

  1. One spin cannot land on red and blue together, so the events cannot both happen.
  2. P(red or blue) = P(red) + P(blue).
  3. 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.

Three rows of six dice faces. The row named even has the faces 2, 4 and 6 hatched. The row named more than 3 has the faces 4, 5 and 6 hatched. The row named even or more than 3 has the faces 2, 4, 5 and 6 hatched, which is 4 of the 6, not 3 + 3.123456even123456more than 3123456even or more than 3
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.

A table with the coin down the side, H and T, and the dice across the top, 1 to 6. Each cell holds its pair such as H1. One cell, H6, is hatched, so 1 cell of 12.dicecoin123456HTH1H2H3H4H5H6T1T2T3T4T5T6
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?

  1. The spinner does not change the coin, so the events are independent.
  2. P(red and head) = P(red) × P(head).
  3. 2/5 × 1/2 = 2/10.
  4. 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?

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

  1. Toss the coin and roll the dice together. Write down whether you got a head and a 6.
  2. Do this 60 times. Count the "head and 6" results.
  3. Divide that count by 60 for the relative frequency.
  4. 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 table with the head trials, both, estimate. Rows: 60, 4, 0.067.trialsbothestimate6040.067

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.

Check yourself

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

  1. One spin lands on only one colour. What is the probability of green or yellow?

    A spinner drawn as a circle in four sectors: red 4, blue 3, green 2 and yellow 1, which are tenths of the circle.red4blue3green2yellow1
    Tap your answer.

    1. Yes, that is right. The events cannot both happen, so add: 2/10 + 1/10 = 3/10.

    2. Not this one. You multiplied 2/10 × 1/10. "Or" for events that cannot both happen means add. Go to M4.6.3. Try another.

    3. Not this one. You took away 2/10 − 1/10. One spin cannot land on both, so you add the two chances. Go to M4.6.3. Try another.

    See the answer

    3/10

    The events cannot both happen, so add: 2/10 + 1/10 = 3/10.

    • If you chose B: You multiplied 2/10 × 1/10. "Or" for events that cannot both happen means add. Go to M4.6.3.
    • If you chose C: You took away 2/10 − 1/10. One spin cannot land on both, so you add the two chances. Go to M4.6.3.

  2. A dice is rolled. For which two events can you add the chances to find the chance of one or the other?

    Tap your answer.

    1. Not this one. A 4 and a 6 are in both events. Adding counts them twice. Go to M4.6.3. Try another.

    2. Not this one. A 3 and a 4 are in both events, so the events can both happen. Go to M4.6.3. Try another.

    3. Not this one. A 3 and a 5 are in both events. Prime scores are 2, 3 and 5, and odd scores are 1, 3 and 5. Go to M4.6.3. Try another.

    4. Yes, that is right. One roll cannot be a one and a six at once, so the events cannot both happen. Then P(one or six) = 1/6 + 1/6.

    See the answer

    A score of one, or a score of six.

    One roll cannot be a one and a six at once, so the events cannot both happen. Then P(one or six) = 1/6 + 1/6.

    • If you chose A: A 4 and a 6 are in both events. Adding counts them twice. Go to M4.6.3.
    • If you chose B: A 3 and a 4 are in both events, so the events can both happen. Go to M4.6.3.
    • If you chose C: A 3 and a 5 are in both events. Prime scores are 2, 3 and 5, and odd scores are 1, 3 and 5. Go to M4.6.3.

  3. The chance that Priya's bus is late is 1/5. The chance that Tariq's bus is late is 1/4. The buses run on different roads. What is the probability that both are late?

    Tap your answer.

    1. Not this one. You added 1/5 + 1/4. For "and" with independent events, multiply. Go to M4.6.3. Try another.

    2. Yes, that is right. The events are independent, so multiply: 1/5 × 1/4 = 1/20.

    3. Not this one. Both must be late, so the chance must be smaller than either chance on its own. Go to M4.6.3. Try another.

    See the answer

    1/20

    The events are independent, so multiply: 1/5 × 1/4 = 1/20.

    • If you chose A: You added 1/5 + 1/4. For "and" with independent events, multiply. Go to M4.6.3.
    • If you chose C: Both must be late, so the chance must be smaller than either chance on its own. Go to M4.6.3.

  4. A fair coin has landed heads 5 times in a row. What is the chance that the next toss is a head?

    Tap your answer.

    1. Not this one. The coin has no memory. Past tosses do not change the next one. Go to M4.6.3. Try another.

    2. Not this one. A streak is chance. It does not change the next toss of a fair coin. Go to M4.6.3. Try another.

    3. Yes, that is right. Each toss is a new chance. The earlier tosses cannot change a fair coin, so the chance stays 1/2.

    See the answer

    1/2, because each toss is independent of the ones before.

    Each toss is a new chance. The earlier tosses cannot change a fair coin, so the chance stays 1/2.

    • If you chose A: The coin has no memory. Past tosses do not change the next one. Go to M4.6.3.
    • If you chose B: A streak is chance. It does not change the next toss of a fair coin. Go to M4.6.3.

  5. Which statement about independent events is true?

    Tap your answer.

    1. Yes, that is right. That is what independent means. Then you may multiply the chances to find the chance of both.

    2. Not this one. That describes events that cannot both happen, which are mutually exclusive. Independent events can happen together, like a head and a six. Go to M4.6.3. Try another.

    3. Not this one. A head has chance 1/2 and a six has chance 1/6, and they are independent. The chances need not match. Go to M4.6.3. Try another.

    See the answer

    One happening does not change the chance of the other.

    That is what independent means. Then you may multiply the chances to find the chance of both.

    • If you chose B: That describes events that cannot both happen, which are mutually exclusive. Independent events can happen together, like a head and a six. Go to M4.6.3.
    • If you chose C: A head has chance 1/2 and a six has chance 1/6, and they are independent. The chances need not match. Go to M4.6.3.

  6. From an earlier chapter

    A spinner lands on red with probability 2/5. It is spun 300 times. How many times should you expect red?

    See the answer

    120

    Expected number = probability × trials = 2/5 × 300 = 120.

  7. Two events are independent. P(A) = 2/3 and P(B) = 3/5. What is P(A and B)?

    See the answer

    2/5

    Independent, so multiply: 2/3 × 3/5 = 6/15, and 6/15 = 2/5.

  8. On a spinner, the probability of red is 3/10 and of blue is 2/10. One spin cannot land on both. What is the probability of red or blue?

    See the answer

    1/2

    The events cannot both happen, so add: 3/10 + 2/10 = 5/10, which is 1/2.

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.

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.