Math Problem Statement

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

Solution

Given the problem, we have the sample space S={o1,o2,o3,o4}S = \{ o_1, o_2, o_3, o_4 \} and the following conditions:

  1. Pr(o1)+Pr(o3)=Pr(o2)+Pr(o4)\text{Pr}(o_1) + \text{Pr}(o_3) = \text{Pr}(o_2) + \text{Pr}(o_4)
  2. Pr(o1)=0.07\text{Pr}(o_1) = 0.07

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

Part (a): When Pr(o2)=Pr(o3)\text{Pr}(o_2) = \text{Pr}(o_3)

Let Pr(o2)=Pr(o3)=p\text{Pr}(o_2) = \text{Pr}(o_3) = p and Pr(o4)=q\text{Pr}(o_4) = q.

Using the first condition:

Pr(o1)+Pr(o3)=Pr(o2)+Pr(o4)\text{Pr}(o_1) + \text{Pr}(o_3) = \text{Pr}(o_2) + \text{Pr}(o_4)

Substituting the given values:

0.07+p=p+q0.07 + p = p + q

Simplifying this:

0.07=q0.07 = q

Thus, Pr(o4)=0.07\text{Pr}(o_4) = 0.07.

Now, the sum of all probabilities must equal 1:

Pr(o1)+Pr(o2)+Pr(o3)+Pr(o4)=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=10.07 + p + p + 0.07 = 1

2p+0.14=12p + 0.14 = 1

2p=0.86p=0.432p = 0.86 \quad \Rightarrow \quad p = 0.43

Therefore, the probability assignments are:

Pr(o1)=0.07,Pr(o2)=0.43,Pr(o3)=0.43,Pr(o4)=0.07\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 Pr(o2)=0.17\text{Pr}(o_2) = 0.17

Let Pr(o3)=p\text{Pr}(o_3) = p and Pr(o4)=q\text{Pr}(o_4) = q.

Using the first condition:

Pr(o1)+Pr(o3)=Pr(o2)+Pr(o4)\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+q0.07 + p = 0.17 + q

Simplifying:

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

Now, the sum of all probabilities must equal 1:

Pr(o1)+Pr(o2)+Pr(o3)+Pr(o4)=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=10.07 + 0.17 + p + q = 1

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

Now, solving the system of equations:

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

2q+0.10=0.762q + 0.10 = 0.76

2q=0.66q=0.332q = 0.66 \quad \Rightarrow \quad q = 0.33

Thus, p=q+0.10=0.33+0.10=0.43p = q + 0.10 = 0.33 + 0.10 = 0.43.

Therefore, the probability assignments are:

Pr(o1)=0.07,Pr(o2)=0.17,Pr(o3)=0.43,Pr(o4)=0.33\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 Pr(o3)\text{Pr}(o_3) was given instead of Pr(o2)\text{Pr}(o_2)?
  2. What are the possible values of Pr(o2)\text{Pr}(o_2) if Pr(o3)=0.5\text{Pr}(o_3) = 0.5?
  3. Can the conditions be satisfied if Pr(o1)\text{Pr}(o_1) was greater than 0.5?
  4. What is the relationship between the probabilities if Pr(o1)=Pr(o2)\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.

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Math Problem Analysis

Mathematical Concepts

Probability
Equations
Sample Space

Formulas

Pr(o1) + Pr(o3) = Pr(o2) + Pr(o4)
Pr(o1) + Pr(o2) + Pr(o3) + Pr(o4) = 1

Theorems

Law of Total Probability

Suitable Grade Level

Grades 10-12