2.	You spin a spinner (as discussed in class).
a)	What is the probability of it stopping in the region: A = (π/4, π)?
b)	What is the probability that it stops in the region: A1 ∪ A2, 
	where: A1 = (0, π/2), A2 = (π/4, π)? 
c)	 What is the probability that you will land in the region R = (0, π/10) on one of the first 10 spins?
d)	How many spins would you expect it to take to land in this region R?
e)	How many spins would it take to have a 95% chance to land in the region R at least once?
f)	 How many times would you expect the spinner to stop in the region R in 25 spins?
g) What is the probability that the spinner stopped in A1 = (0, π/2), given that it stopped somewhere in A2 = (π/4, π)?

Let's work through each part of the question methodically, starting with an interpretation of the spinner's range and probability principles.

### Assumptions:
The spinner rotates uniformly over the interval \([0, 2\pi]\), meaning the probability of landing in any region is proportional to the size of that region.

### a) Probability of stopping in region \(A = (\pi/4, \pi)\)
The probability of landing in a region on the spinner is the ratio of the length of that region to the total length of the spinner's range.

- The length of region \(A = (\pi/4, \pi)\) is:
  \[
  \text{Length of } A = \pi - \frac{\pi}{4} = \frac{3\pi}{4}
  \]
- The total length of the spinner's range is \(2\pi\).

So, the probability is:
\[
P(A) = \frac{\frac{3\pi}{4}}{2\pi} = \frac{3}{8}
\]

### b) Probability of stopping in region \(A_1 \cup A_2\), where \(A_1 = (0, \pi/2)\) and \(A_2 = (\pi/4, \pi)\)
First, we calculate the lengths of \(A_1\) and \(A_2\):
- The length of \(A_1 = (0, \pi/2)\) is:
  \[
  \text{Length of } A_1 = \frac{\pi}{2}
  \]
- The length of \(A_2 = (\pi/4, \pi)\) is the same as in part (a):
  \[
  \text{Length of } A_2 = \frac{3\pi}{4}
  \]

Now, we need to find the union of \(A_1\) and \(A_2\). The intersection between \(A_1\) and \(A_2\) is \((\pi/4, \pi/2)\), which has length:
\[
\text{Length of } A_1 \cap A_2 = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
\]

Therefore, the total length of the union \(A_1 \cup A_2\) is:
\[
\text{Length of } A_1 \cup A_2 = \frac{\pi}{2} + \frac{3\pi}{4} - \frac{\pi}{4} = \frac{3\pi}{4}
\]

Thus, the probability is:
\[
P(A_1 \cup A_2) = \frac{\frac{3\pi}{4}}{2\pi} = \frac{3}{8}
\]

### c) Probability of landing in region \(R = (0, \pi/10)\) on one of the first 10 spins
The probability of landing in \(R\) on a single spin is:
\[
P(R) = \frac{\frac{\pi}{10}}{2\pi} = \frac{1}{20}
\]

The probability of not landing in \(R\) on a single spin is:
\[
P(\text{not } R) = 1 - \frac{1}{20} = \frac{19}{20}
\]

The probability of not landing in \(R\) after 10 spins is:
\[
P(\text{not } R \text{ in 10 spins}) = \left(\frac{19}{20}\right)^{10}
\]

Thus, the probability of landing in \(R\) at least once in 10 spins is:
\[
P(\text{R in 10 spins}) = 1 - \left(\frac{19}{20}\right)^{10}
\]

### d) Expected number of spins to land in region \(R = (0, \pi/10)\)
The expected number of spins to land in \(R\) is the reciprocal of the probability of landing in \(R\) on a single spin:
\[
E(\text{spins to land in } R) = \frac{1}{P(R)} = \frac{1}{\frac{1}{20}} = 20
\]

### e) Number of spins for a 95% chance to land in region \(R\) at least once
We want to find the number of spins, \(n\), such that the probability of landing in \(R\) at least once is 95%. That is:
\[
P(\text{not } R \text{ in } n \text{ spins}) = \left(\frac{19}{20}\right)^n = 0.05
\]

Taking the natural logarithm on both sides:
\[
n \ln\left(\frac{19}{20}\right) = \ln(0.05)
\]
\[
n = \frac{\ln(0.05)}{\ln\left(\frac{19}{20}\right)}
\]

We can calculate this to find \(n\).

### f) Expected number of times the spinner stops in \(R\) in 25 spins
The expected number of times the spinner stops in \(R\) in 25 spins is:
\[
E(\text{stops in } R) = 25 \times P(R) = 25 \times \frac{1}{20} = 1.25
\]

### g) Probability that the spinner stopped in \(A_1 = (0, \pi/2)\), given that it stopped somewhere in \(A_2 = (\pi/4, \pi)\)
This is a conditional probability problem:
\[
P(A_1 | A_2) = \frac{P(A_1 \cap A_2)}{P(A_2)}
\]
We have already calculated:
- \(P(A_1 \cap A_2) = \frac{\frac{\pi}{4}}{2\pi} = \frac{1}{8}\)
- \(P(A_2) = \frac{3}{8}\)

Thus:
\[
P(A_1 | A_2) = \frac{\frac{1}{8}}{\frac{3}{8}} = \frac{1}{3}
\]

---

Do you want details or have any questions?

Here are five related questions:
1. How would probabilities change if the spinner had a different range, say \([0, \pi]\)?
2. What happens to probabilities when we add more regions or overlaps?
3. Can we calculate expected spins for smaller regions like \(R = (0, \pi/100)\)?
4. How would probabilities adjust with weighted spinners (non-uniform distribution)?
5. How can conditional probabilities help in decision-making with overlapping events?

**Tip:** Always consider the total range when calculating probabilities in a continuous setting, as it helps ensure correct ratios for different regions.

You spin a spinner (as discussed in class).
a) What is the probability of it stopping in the region: A = (π/4, π)?
b) What is the probability that it stops in the region: A1 ∪ A2, where: A1 = (0, π/2), A2 = (π/4, π)?
c) What is the probability that you will land in the region R = (0, π/10) on one of the first 10 spins?
d) How many spins would you expect it to take to land in this region R?
e) How many spins would it take to have a 95% chance to land in the region R at least once?
f) How many times would you expect the spinner to stop in the region R in 25 spins?
g) What is the probability that the spinner stopped in A1 = (0, π/2), given that it stopped somewhere in A2 = (π/4, π)?

This detailed solution covers various probability questions related to spinning a spinner with a uniform probability distribution. Learn how to calculate the probability of landing in specific regions, the expected number of spins, and the chances of success in multiple trials. Suitable for Grades 10-12 students studying probability.

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