Math Problem Statement

Consider the sample space Sequals=​{o 1 comma o 2 comma o 3 comma o 4 comma o 5o1, o2, o3, o4, o5​}. Suppose that Pr left parenthesis o 1 right parenthesisPro1=0.19 and Pr left parenthesis o 2 right parenthesisPro2=0.27. ​(a) Find the probability assignment for the probability space when o 3o3​, o 4o4​, and o 5o5 all have the same probability. ​(b) Find the probability assignment for the probability space when Pr left parenthesis o 5 right parenthesisPro5=0.18 and o 3o3 has the same probability as o 4o4 and o 5o5 combined.

Solution

We are given the sample space S={o1,o2,o3,o4,o5}S = \{ o_1, o_2, o_3, o_4, o_5 \} and some information about the probabilities of certain events. Let's break down the problem step by step.

(a) Find the probability assignment when o3,o4,o5o_3, o_4, o_5 all have the same probability.

We are provided with:

  • Pr(o1)=0.19\text{Pr}(o_1) = 0.19
  • Pr(o2)=0.27\text{Pr}(o_2) = 0.27
  • The probabilities for o3,o4,o5o_3, o_4, o_5 are equal.

We know that the total probability of all events must sum to 1, i.e.,

Pr(o1)+Pr(o2)+Pr(o3)+Pr(o4)+Pr(o5)=1\text{Pr}(o_1) + \text{Pr}(o_2) + \text{Pr}(o_3) + \text{Pr}(o_4) + \text{Pr}(o_5) = 1

Let the probability of o3,o4,o5o_3, o_4, o_5 each be denoted by pp. Thus, we have the equation:

0.19+0.27+3p=10.19 + 0.27 + 3p = 1

Simplifying:

0.46+3p=10.46 + 3p = 1

3p=10.46=0.543p = 1 - 0.46 = 0.54

p=0.543=0.18p = \frac{0.54}{3} = 0.18

Thus, the probabilities are:

  • Pr(o3)=0.18\text{Pr}(o_3) = 0.18
  • Pr(o4)=0.18\text{Pr}(o_4) = 0.18
  • Pr(o5)=0.18\text{Pr}(o_5) = 0.18

(b) Find the probability assignment when Pr(o5)=0.18\text{Pr}(o_5) = 0.18 and o3o_3 has the same probability as o4o_4 and o5o_5 combined.

We are given:

  • Pr(o1)=0.19\text{Pr}(o_1) = 0.19
  • Pr(o2)=0.27\text{Pr}(o_2) = 0.27
  • Pr(o5)=0.18\text{Pr}(o_5) = 0.18
  • The probability of o3o_3 is equal to the sum of the probabilities of o4o_4 and o5o_5.

Let the probability of o4o_4 be pp. Then the probability of o3o_3 is Pr(o3)=p+0.18\text{Pr}(o_3) = p + 0.18.

Using the total probability constraint:

Pr(o1)+Pr(o2)+Pr(o3)+Pr(o4)+Pr(o5)=1\text{Pr}(o_1) + \text{Pr}(o_2) + \text{Pr}(o_3) + \text{Pr}(o_4) + \text{Pr}(o_5) = 1

Substituting the known values:

0.19+0.27+(p+0.18)+p+0.18=10.19 + 0.27 + (p + 0.18) + p + 0.18 = 1

Simplifying:

0.46+2p+0.36=10.46 + 2p + 0.36 = 1

0.82+2p=10.82 + 2p = 1

2p=10.82=0.182p = 1 - 0.82 = 0.18

p=0.182=0.09p = \frac{0.18}{2} = 0.09

Thus, the probabilities are:

  • Pr(o4)=0.09\text{Pr}(o_4) = 0.09
  • Pr(o3)=p+0.18=0.09+0.18=0.27\text{Pr}(o_3) = p + 0.18 = 0.09 + 0.18 = 0.27
  • Pr(o5)=0.18\text{Pr}(o_5) = 0.18

Final Answers:

(a) The probability assignment is:

  • Pr(o1)=0.19\text{Pr}(o_1) = 0.19
  • Pr(o2)=0.27\text{Pr}(o_2) = 0.27
  • Pr(o3)=0.18\text{Pr}(o_3) = 0.18
  • Pr(o4)=0.18\text{Pr}(o_4) = 0.18
  • Pr(o5)=0.18\text{Pr}(o_5) = 0.18

(b) The probability assignment is:

  • Pr(o1)=0.19\text{Pr}(o_1) = 0.19
  • Pr(o2)=0.27\text{Pr}(o_2) = 0.27
  • Pr(o3)=0.27\text{Pr}(o_3) = 0.27
  • Pr(o4)=0.09\text{Pr}(o_4) = 0.09
  • Pr(o5)=0.18\text{Pr}(o_5) = 0.18

Would you like further details or have any questions?

Here are 5 related questions to explore this topic further:

  1. How can we verify if a set of probabilities is valid for any probability space?
  2. What is the difference between a discrete and continuous probability space?
  3. Can events with unequal probabilities be considered equally likely if normalized in some way?
  4. How would the total probability change if one of the outcomes were removed?
  5. What are conditional probabilities, and how do they apply to sample spaces?

Tip: In any probability space, the sum of all individual probabilities must always equal 1.

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Math Problem Analysis

Mathematical Concepts

Probability
Sample Space
Probability Assignment

Formulas

Sum of probabilities of all events: Pr(o1) + Pr(o2) + Pr(o3) + Pr(o4) + Pr(o5) = 1

Theorems

Law of Total Probability

Suitable Grade Level

Grades 10-12