a. the spinner is spun twice. find each probability.\na) p(square both times)\nb) p(12, then circle)\nc)…

Explanation:

Step1: Determine total number of outcomes

The spinner has 12 sections, so when spun once, there are 12 possible outcomes. When spun twice, by the fundamental counting principle, the total number of outcomes for two - spins is $12\times12 = 144$.

Step2: Calculate $P(\text{square both times})$

There are 6 squares on the spinner. The probability of getting a square on the first spin is $\frac{6}{12}=\frac{1}{2}$, and the probability of getting a square on the second spin is also $\frac{6}{12}=\frac{1}{2}$. Since the spins are independent events, the probability of getting a square both times is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.

Step3: Calculate $P(12,\text{ then circle})$

The probability of getting 12 on the first spin is $\frac{1}{12}$. There are 6 circles on the spinner, so the probability of getting a circle on the second spin is $\frac{6}{12}=\frac{1}{2}$. Since the spins are independent, the probability is $\frac{1}{12}\times\frac{1}{2}=\frac{1}{24}$.

Step4: Calculate $P(\text{square, then even number})$

The probability of getting a square on the first spin is $\frac{6}{12}=\frac{1}{2}$. There are 6 even - numbered sections on the spinner, so the probability of getting an even number on the second spin is $\frac{6}{12}=\frac{1}{2}$. Since the spins are independent, the probability is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.

Answer:

a) $\frac{1}{4}$ b) $\frac{1}{24}$ c) $\frac{1}{4}$