Question 1: A spinner divided into 4 equal sections is spun once. Calculate the probability that it lands on the red section.

Answer: 1/4

Question 2: A spinner has 6 sections: 2 are blue, 2 are green, and 2 are yellow. What is the probability that the spinner lands on a green or yellow section?

Answer: 2/6 + 2/6 = 4/6 = 2/3

Question 3: Construct a spinner with 8 equal sections: 3 red, 2 blue, and 3 green. What is the probability of landing on a red or green section?

Answer: 3/8 + 3/8 = 6/8 = 3/4

Question 4: A student claims that the probability of landing on a red section is always 1/3 for any spinner with three equal sections. Is this correct? Explain your reasoning.

Answer: No, because the probability depends on the number of sections. If the spinner has 4 sections, the probability of red is 1/4, not 1/3.

Question 5: A spinner is divided into 5 equal parts: 2 are labeled 'A', 2 are 'B', and 1 is 'C'. What is the probability it lands on 'A'?

Answer: 2/5

Question 6: A spinner with 10 sections has 3 red, 4 blue, and 3 green. If the spinner is spun twice, what is the probability that both spins land on blue?

Answer: (4/10) * (4/10) = 16/100 = 4/25

Question 7: Identify the common mistake in the following reasoning: 'Since there are 5 sections with equal probability, the chance of landing on any one is 1/5. If I spin twice, the probability of landing on red both times is (1/5) + (1/5) = 2/5.'

Answer: The mistake is adding probabilities instead of multiplying. For independent spins, probabilities should multiply.

Question 8: A spinner has 3 sections: one each labeled 'X', 'Y', and 'Z'. A student guesses the probability of landing on 'Y' is 1/2. Is this possible? Why or why not?

Answer: No, because with 3 equal sections, the probability of each is 1/3, not 1/2.

Question 9: Plot the graph of the probability function p = 1/n, where n is the number of equal sections on the spinner, for n = 1 to 6.

Answer: Students should plot points at (1,1), (2,1/2), (3,1/3), (4,1/4), (5,1/5), (6,1/6).

Question 10: A spinner is divided into 12 equal sections. If a student claims the probability of landing on a specific section is 1/12, but they only label 10 sections, what is the mistake?

Answer: The mistake is assuming all sections are equally likely when only 10 are labeled; unlabeled sections are still part of the spinner.

Question 11: Challenge: Design a spinner with 7 sections such that the probability of landing on an even number section is greater than 1/2. Describe its setup.

Answer: Have at least 4 even-numbered sections and 3 odd. For example, 4 even (2,4,6,8) and 3 odd (1,3,5) in total.