Consider the sample space Sequals=​{o 1 comma o 2 comma o 3 comma o 4o1, o2, o3, o4​}. Suppose that Pr left parenthesis o 1 right parenthesis plus Pr left parenthesis o 3 right parenthesis equals Pr left parenthesis o 2 right parenthesis plus Pr left parenthesis o 4 right parenthesisPro1+Pro3=Pro2+Pro4 and that Pr left parenthesis o 1 right parenthesis equals 0.07Pro1=0.07.
​(a)
Find the probability assignment for the probability space when o 2 and o 3o2 and o3 have the same probability.
​(b)
Find the probability assignment for the probability space when Pr left parenthesis o 2 right parenthesis equals Pro2=0.17

Given the problem, we have the sample space \( S = \{ o_1, o_2, o_3, o_4 \} \) and the following conditions:

1. \( \text{Pr}(o_1) + \text{Pr}(o_3) = \text{Pr}(o_2) + \text{Pr}(o_4) \)
2. \( \text{Pr}(o_1) = 0.07 \)

We are asked to solve for the probability assignments under two scenarios.

### Part (a): When \( \text{Pr}(o_2) = \text{Pr}(o_3) \)

Let \( \text{Pr}(o_2) = \text{Pr}(o_3) = p \) and \( \text{Pr}(o_4) = q \).

Using the first condition:

\[
\text{Pr}(o_1) + \text{Pr}(o_3) = \text{Pr}(o_2) + \text{Pr}(o_4)
\]

Substituting the given values:

\[
0.07 + p = p + q
\]

Simplifying this:

\[
0.07 = q
\]

Thus, \( \text{Pr}(o_4) = 0.07 \).

Now, the sum of all probabilities must equal 1:

\[
\text{Pr}(o_1) + \text{Pr}(o_2) + \text{Pr}(o_3) + \text{Pr}(o_4) = 1
\]

Substituting the known values:

\[
0.07 + p + p + 0.07 = 1
\]

\[
2p + 0.14 = 1
\]

\[
2p = 0.86 \quad \Rightarrow \quad p = 0.43
\]

Therefore, the probability assignments are:

\[
\text{Pr}(o_1) = 0.07, \quad \text{Pr}(o_2) = 0.43, \quad \text{Pr}(o_3) = 0.43, \quad \text{Pr}(o_4) = 0.07
\]

### Part (b): When \( \text{Pr}(o_2) = 0.17 \)

Let \( \text{Pr}(o_3) = p \) and \( \text{Pr}(o_4) = q \).

Using the first condition:

\[
\text{Pr}(o_1) + \text{Pr}(o_3) = \text{Pr}(o_2) + \text{Pr}(o_4)
\]

Substituting the given values:

\[
0.07 + p = 0.17 + q
\]

Simplifying:

\[
p - q = 0.10 \quad \text{(Equation 1)}
\]

Now, the sum of all probabilities must equal 1:

\[
\text{Pr}(o_1) + \text{Pr}(o_2) + \text{Pr}(o_3) + \text{Pr}(o_4) = 1
\]

Substituting the known values:

\[
0.07 + 0.17 + p + q = 1
\]

\[
p + q = 0.76 \quad \text{(Equation 2)}
\]

Now, solving the system of equations:

- From Equation 1: \( p = q + 0.10 \)
- Substituting into Equation 2: \( (q + 0.10) + q = 0.76 \)

\[
2q + 0.10 = 0.76
\]

\[
2q = 0.66 \quad \Rightarrow \quad q = 0.33
\]

Thus, \( p = q + 0.10 = 0.33 + 0.10 = 0.43 \).

Therefore, the probability assignments are:

\[
\text{Pr}(o_1) = 0.07, \quad \text{Pr}(o_2) = 0.17, \quad \text{Pr}(o_3) = 0.43, \quad \text{Pr}(o_4) = 0.33
\]

---

Would you like further details or have any questions?

Here are some related questions you might consider:

1. How would the probability distribution change if \( \text{Pr}(o_3) \) was given instead of \( \text{Pr}(o_2) \)?
2. What are the possible values of \( \text{Pr}(o_2) \) if \( \text{Pr}(o_3) = 0.5 \)?
3. Can the conditions be satisfied if \( \text{Pr}(o_1) \) was greater than 0.5?
4. What is the relationship between the probabilities if \( \text{Pr}(o_1) = \text{Pr}(o_2) \)?
5. How would the probability assignment change if the sample space had 5 elements instead of 4?

**Tip:** When solving probability problems, always ensure the sum of all probabilities equals 1 to verify your calculations.

Consider the sample space S = {o1, o2, o3, o4}. Suppose that Pr(o1) + Pr(o3) = Pr(o2) + Pr(o4) and that Pr(o1) = 0.07. (a) Find the probability assignment for the probability space when o2 and o3 have the same probability. (b) Find the probability assignment for the probability space when Pr(o2) = 0.17.

Learn how to solve a probability assignment problem with the conditions Pr(o1) + Pr(o3) = Pr(o2) + Pr(o4) and Pr(o1) = 0.07. This step-by-step guide explores two scenarios where o2 and o3 have the same probability and when Pr(o2) = 0.17. Suitable for students in Grades 10-12.

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