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.

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

### (a) Find the probability assignment when \( o_3, o_4, o_5 \) all have the same probability.

We are provided with:
- \( \text{Pr}(o_1) = 0.19 \)
- \( \text{Pr}(o_2) = 0.27 \)
- The probabilities for \( o_3, o_4, o_5 \) are equal.

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

\[
\text{Pr}(o_1) + \text{Pr}(o_2) + \text{Pr}(o_3) + \text{Pr}(o_4) + \text{Pr}(o_5) = 1
\]

Let the probability of \( o_3, o_4, o_5 \) each be denoted by \( p \). Thus, we have the equation:

\[
0.19 + 0.27 + 3p = 1
\]

Simplifying:

\[
0.46 + 3p = 1
\]

\[
3p = 1 - 0.46 = 0.54
\]

\[
p = \frac{0.54}{3} = 0.18
\]

Thus, the probabilities are:
- \( \text{Pr}(o_3) = 0.18 \)
- \( \text{Pr}(o_4) = 0.18 \)
- \( \text{Pr}(o_5) = 0.18 \)

### (b) Find the probability assignment when \( \text{Pr}(o_5) = 0.18 \) and \( o_3 \) has the same probability as \( o_4 \) and \( o_5 \) combined.

We are given:
- \( \text{Pr}(o_1) = 0.19 \)
- \( \text{Pr}(o_2) = 0.27 \)
- \( \text{Pr}(o_5) = 0.18 \)
- The probability of \( o_3 \) is equal to the sum of the probabilities of \( o_4 \) and \( o_5 \).

Let the probability of \( o_4 \) be \( p \). Then the probability of \( o_3 \) is \( \text{Pr}(o_3) = p + 0.18 \).

Using the total probability constraint:

\[
\text{Pr}(o_1) + \text{Pr}(o_2) + \text{Pr}(o_3) + \text{Pr}(o_4) + \text{Pr}(o_5) = 1
\]

Substituting the known values:

\[
0.19 + 0.27 + (p + 0.18) + p + 0.18 = 1
\]

Simplifying:

\[
0.46 + 2p + 0.36 = 1
\]

\[
0.82 + 2p = 1
\]

\[
2p = 1 - 0.82 = 0.18
\]

\[
p = \frac{0.18}{2} = 0.09
\]

Thus, the probabilities are:
- \( \text{Pr}(o_4) = 0.09 \)
- \( \text{Pr}(o_3) = p + 0.18 = 0.09 + 0.18 = 0.27 \)
- \( \text{Pr}(o_5) = 0.18 \)

### Final Answers:

#### (a) The probability assignment is:
- \( \text{Pr}(o_1) = 0.19 \)
- \( \text{Pr}(o_2) = 0.27 \)
- \( \text{Pr}(o_3) = 0.18 \)
- \( \text{Pr}(o_4) = 0.18 \)
- \( \text{Pr}(o_5) = 0.18 \)

#### (b) The probability assignment is:
- \( \text{Pr}(o_1) = 0.19 \)
- \( \text{Pr}(o_2) = 0.27 \)
- \( \text{Pr}(o_3) = 0.27 \)
- \( \text{Pr}(o_4) = 0.09 \)
- \( \text{Pr}(o_5) = 0.18 \)

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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.

Consider the sample space S = {o1, o2, o3, o4, o5}. Suppose that Pr(o1) = 0.19 and Pr(o2) = 0.27. (a) Find the probability assignment for the probability space when o3, o4, and o5 all have the same probability. (b) Find the probability assignment for the probability space when Pr(o5) = 0.18 and o3 has the same probability as o4 and o5 combined.

Learn how to assign probabilities to a sample space S = {o1, o2, o3, o4, o5} given specific conditions. This problem involves calculating the probability when o3, o4, and o5 have equal probabilities and when Pr(o3) equals the sum of Pr(o4) and Pr(o5).

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