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 break down each part of the problem, one step at a time:

### Assumptions and Spinner Setup:
We are dealing with a circular spinner, and we assume that the spinner stops uniformly at random over the interval \( [0, 2\pi] \). Therefore, the probability that the spinner lands in a particular region is proportional to the length of that region.

### a) Probability of stopping in the region \( A = (\frac{\pi}{4}, \pi) \)

To find the probability of the spinner stopping in a region, we need to calculate the length of the region \( A \) and divide it by the total length of the spinner's circumference, which is \( 2\pi \).

- Length of \( A = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)
- Total length of the spinner's circumference = \( 2\pi \)

Thus, the probability of stopping in region \( A \) is:

\[
P(A) = \frac{\text{Length of } A}{2\pi} = \frac{\frac{3\pi}{4}}{2\pi} = \frac{3}{8}
\]

### b) Probability of stopping in the region \( A_1 \cup A_2 \), where \( A_1 = (0, \frac{\pi}{2}) \) and \( A_2 = (\frac{\pi}{4}, \pi) \)

First, we calculate the lengths of regions \( A_1 \) and \( A_2 \):

- Length of \( A_1 = \frac{\pi}{2} - 0 = \frac{\pi}{2} \)
- Length of \( A_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

Since the two regions overlap in the interval \( (\frac{\pi}{4}, \frac{\pi}{2}) \), we need to subtract the length of the overlap from the sum of the lengths of \( A_1 \) and \( A_2 \).

- Length of overlap \( = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \)

Now, the probability of stopping in \( A_1 \cup A_2 \) is:

\[
P(A_1 \cup A_2) = \frac{\frac{\pi}{2} + \frac{3\pi}{4} - \frac{\pi}{4}}{2\pi} = \frac{\frac{\pi}{2} + \frac{2\pi}{2}}{2\pi} = \frac{\frac{3\pi}{2}}{2\pi} = \frac{3}{4}
\]

### c) Probability of landing in region \( R = (0, \frac{\pi}{10}) \) on one of the first 10 spins

First, calculate the probability of landing in region \( R \) on a single spin. The length of \( R \) is:

- Length of \( R = \frac{\pi}{10} - 0 = \frac{\pi}{10} \)

So, the probability of stopping in \( R \) on one 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 \( 1 - P(R) = \frac{19}{20} \). Therefore, the probability of not landing in \( R \) on any of the first 10 spins is:

\[
P(\text{not landing in } R \text{ in 10 spins}) = \left(\frac{19}{20}\right)^{10}
\]

Thus, the probability of landing in \( R \) at least once in the first 10 spins is:

\[
P(\text{landing in } R \text{ in 10 spins}) = 1 - \left(\frac{19}{20}\right)^{10}
\]

### d) Expected number of spins to land in region \( R \)

The expected number of spins to land in a region is the reciprocal of the probability of landing in that region on a single spin. From part (c), we know that the probability of landing in \( R \) on a single spin is \( \frac{1}{20} \). Therefore, the expected number of spins to land in \( R \) is:

\[
E(\text{spins}) = \frac{1}{P(R)} = 20
\]

### e) Number of spins required to have a 95% chance of landing in region \( R \) at least once

We want the probability of landing in \( R \) at least once to be 95%. Let \( n \) be the number of spins. The probability of not landing in \( R \) in \( n \) spins is \( \left( \frac{19}{20} \right)^n \), and we want this to be 5% (since there's a 95% chance of landing in \( R \)):

\[
\left( \frac{19}{20} \right)^n = 0.05
\]

Taking the natural logarithm of both sides:

\[
n \ln \left( \frac{19}{20} \right) = \ln(0.05)
\]

Solving for \( n \):

\[
n = \frac{\ln(0.05)}{\ln \left( \frac{19}{20} \right)}
\]

### f) Expected number of times the spinner will stop in region \( R \) in 25 spins

The expected number of times the spinner will stop in a region is the product of the number of spins and the probability of stopping in that region on a single spin. From part (c), we know that the probability of stopping in \( R \) on a single spin is \( \frac{1}{20} \). Therefore, the expected number of times the spinner will stop in \( R \) in 25 spins is:

\[
E(\text{stops in } R) = 25 \times \frac{1}{20} = 1.25
\]

### g) Probability that the spinner stopped in \( A_1 = (0, \frac{\pi}{2}) \), given that it stopped somewhere in \( A_2 = (\frac{\pi}{4}, \pi) \)

This is a conditional probability question. We are asked to find \( P(A_1 | A_2) \), the probability that the spinner stopped in \( A_1 \) given that it stopped in \( A_2 \). Using the formula for conditional probability:

\[
P(A_1 | A_2) = \frac{P(A_1 \cap A_2)}{P(A_2)}
\]

We already know that:

- \( P(A_1 \cap A_2) \) is the probability of stopping in the overlap between \( A_1 \) and \( A_2 \), which is the region \( (\frac{\pi}{4}, \frac{\pi}{2}) \). The length of this region is \( \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \), so \( P(A_1 \cap A_2) = \frac{\frac{\pi}{4}}{2\pi} = \frac{1}{8} \).
- \( P(A_2) = \frac{3}{8} \) (from part a).

Thus:

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

---

### Summary:

- a) \( P(A) = \frac{3}{8} \)
- b) \( P(A_1 \cup A_2) = \frac{3}{4} \)
- c) \( P(\text{landing in } R \text{ in 10 spins}) = 1 - \left(\frac{19}{20}\right)^{10} \)
- d) Expected spins = 20
- e) \( n = \frac{\ln(0.05)}{\ln(19/20)} \)
- f) Expected stops in \( R \) in 25 spins = 1.25
- g) \( P(A_1 | A_2) = \frac{1}{3} \)

Let me know if you'd like any of the steps further clarified!

### Follow-up Questions:
1. How would the answers change if the spinner's total range was \( [0, \pi] \)?
2. What happens if the spinner has uneven probabilities for different regions?
3. Can we derive a general formula for expected spins to land in any region \( R \)?
4. What would happen if the spinner had multiple regions with different stopping probabilities?
5. How can we simulate the spinner problem using a computer program?

#### Tip: Conditional probability can be tricky—always ensure you're accounting for overlaps correctly when dealing with unions!

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)?

Explore detailed solutions for calculating probabilities and expected values when spinning a circular spinner. This problem covers probability theory, conditional probability, and expected spins to land in specific regions. Suitable for advanced high school students in Grades 11-12.

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