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If a spinner has 12 equal sections, what is the probability of landing on a factor of 6 or on a number greater than 9

User Kronosjt
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Final answer:

The probability of landing on a factor of 6 or a number greater than 9 on a spinner with 12 equal sections is 1/2 or 50%, as there are 6 favorable outcomes out of 12 possible outcomes.

Step-by-step explanation:

If a spinner has 12 equal sections, the probability of landing on a factor of 6 or on a number greater than 9 can be calculated by identifying the favorable outcomes and dividing by the total number of outcomes, which is 12. The factors of 6 are 1, 2, 3, and 6. Since each section is equal, these factors provide four favorable outcomes. Additionally, the numbers greater than 9 are 10, 11, and 12, giving us three more favorable outcomes. However, note that the number 6 has already been counted as a factor of 6, so we have a total of 6 unique favorable outcomes. Therefore, the probability is calculated as 6/12, which simplifies to 1/2 or 50%.

Example 3.2 describes a simple scenario with a fair six-sided die, where the sample space S is {1, 2, 3, 4, 5, 6}. This example helps illustrate basic probability concepts relevant to our spinner question, but the scenario is different in details.

User Andriy Plokhotnyuk
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