Probability · Answer guide

Lesson 1 · Count equally likely outcomes

Use favourable outcomes divided by all equally likely outcomes.

1. A fair spinner has 6 equal sectors, of which 1 are blue. What is P(blue)?

   Answer: 1/6. Each sector is equally likely. There are 1 favourable sectors out of 6, so P(blue) = 1/6 = 1/6.

2. A fair spinner has 8 equal sectors, of which 3 are blue. What is P(blue)?

   Answer: 3/8. Each sector is equally likely. There are 3 favourable sectors out of 8, so P(blue) = 3/8 = 3/8.

3. A fair spinner has 10 equal sectors, of which 4 are blue. What is P(blue)?

   Answer: 2/5. Each sector is equally likely. There are 4 favourable sectors out of 10, so P(blue) = 4/10 = 2/5.

Discuss: List an event on a fair die with probability 1/2, then create a different event with the same probability.

Teaching note: Two named events need not be equally likely. “Roll a 6” and “do not roll a 6” are not a 1/2–1/2 split for a fair die. Open responses need discussion; they are not automatically marked.

Lesson 2 · Use complements and bounds

Find “not” probabilities and distinguish impossible from certain.

1. A bag has 7 otherwise identical balls, 2 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   Answer: 5/7. P(not blue) = 1 − 2/7 = 5/7. There are 5 non-blue balls.

2. A bag has 8 otherwise identical balls, 3 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   Answer: 5/8. P(not blue) = 1 − 3/8 = 5/8. There are 5 non-blue balls.

3. A bag has 9 otherwise identical balls, 4 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   Answer: 5/9. P(not blue) = 1 − 4/9 = 5/9. There are 5 non-blue balls.

Discuss: Make a three-colour bag model. Explain why not red includes both of the other colours, then check complementary probabilities add to 1.

Teaching note: “Not blue” may include several colours if the bag contains more than two colours. The complement is everything outside the event. Open responses need discussion; they are not automatically marked.

Lesson 3 · Organise a sample space without losing outcomes

Use a table or tree for a small combined experiment.

1. A fair coin and an independent fair 2-sector spinner numbered 1 to 2 are used. What is P(heads and spinner number at most 1)?

   Answer: 1/4. There are 2×2 = 4 equally likely ordered outcomes. Exactly 1 have heads and a spinner number at most 1, so probability is 1/4.

2. A fair coin and an independent fair 3-sector spinner numbered 1 to 3 are used. What is P(heads and spinner number at most 1)?

   Answer: 1/6. There are 2×3 = 6 equally likely ordered outcomes. Exactly 1 have heads and a spinner number at most 1, so probability is 1/6.

3. A fair coin and an independent fair 4-sector spinner numbered 1 to 4 are used. What is P(heads and spinner number at most 1)?

   Answer: 1/8. There are 2×4 = 8 equally likely ordered outcomes. Exactly 1 have heads and a spinner number at most 1, so probability is 1/8.

Discuss: Make a 2-by-4 table for a coin and a four-sector spinner. Shade one “and” event and one “or” event, explaining any overlap.

Teaching note: “And” requires both conditions; “or” includes outcomes satisfying either condition. Overlap is counted once. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. A fair spinner has 9 equal sectors, of which 2 are blue. What is P(blue)?

   2/9. Each sector is equally likely. There are 2 favourable sectors out of 9, so P(blue) = 2/9 = 2/9.

2. A fair spinner has 15 equal sectors, of which 6 are blue. What is P(blue)?

   2/5. Each sector is equally likely. There are 6 favourable sectors out of 15, so P(blue) = 6/15 = 2/5.

3. A bag has 13 otherwise identical balls, 6 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   7/13. P(not blue) = 1 − 6/13 = 7/13. There are 7 non-blue balls.

4. A bag has 9 otherwise identical balls, 2 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   7/9. P(not blue) = 1 − 2/9 = 7/9. There are 7 non-blue balls.

5. A fair coin and an independent fair 6-sector spinner numbered 1 to 6 are used. What is P(heads and spinner number at most 1)?

   1/12. There are 2×6 = 12 equally likely ordered outcomes. Exactly 1 have heads and a spinner number at most 1, so probability is 1/12.

6. A fair coin and an independent fair 7-sector spinner numbered 1 to 7 are used. What is P(heads and spinner number at most 2)?

   1/7. There are 2×7 = 14 equally likely ordered outcomes. Exactly 2 have heads and a spinner number at most 2, so probability is 1/7.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. A fair spinner has 20 equal sectors, of which 7 are blue. What is P(blue)?

   7/20. Each sector is equally likely. There are 7 favourable sectors out of 20, so P(blue) = 7/20 = 7/20.

2. A fair spinner has 16 equal sectors, of which 5 are blue. What is P(blue)?

   5/16. Each sector is equally likely. There are 5 favourable sectors out of 16, so P(blue) = 5/16 = 5/16.

3. A bag has 15 otherwise identical balls, 7 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   8/15. P(not blue) = 1 − 7/15 = 8/15. There are 8 non-blue balls.

4. A bag has 11 otherwise identical balls, 3 blue and the rest red. One is selected uniformly at random. What is P(not blue)?

   8/11. P(not blue) = 1 − 3/11 = 8/11. There are 8 non-blue balls.

5. A fair coin and an independent fair 8-sector spinner numbered 1 to 8 are used. What is P(heads and spinner number at most 3)?

   3/16. There are 2×8 = 16 equally likely ordered outcomes. Exactly 3 have heads and a spinner number at most 3, so probability is 3/16.

6. A fair coin and an independent fair 9-sector spinner numbered 1 to 9 are used. What is P(heads and spinner number at most 2)?

   1/9. There are 2×9 = 18 equally likely ordered outcomes. Exactly 2 have heads and a spinner number at most 2, so probability is 1/9.