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Recently, Goran ran for president of the local school board. Of those who voted in the election, 80% had a high school diploma, and the other 20% did not. Goran got 70% of the vote of those with high school diplomas, while he got only 15% of the vote of those without high school diplomas. Let D denote the event that a randomly chosen voter (in the school board election) had a high school diploma and D denote the event that a randomly chosen voter did not have a high school diploma. Let V denote the event that a randomly chosen voter voted for Goran and V denote the event that a randomly chosen voter did not vote for Goran. Fill in the probabilities to complete the tree diagram below, and then answer the question that follows. Do not round any of your responses. (If necessary, consult a list of formulas.) (a) Fill in the missing probabilities. 4 P(VD) = 0.7 P(DnV) = [ P (D)=0.8 Submit Assign Continue

Recently, Goran ran for president of the local school board. Of those who voted in the election, 80% had a high school diploma, and the other 20% did not. Goran got 70% of the vote of those with high school diplomas, while he got only 15% of the vote of those without high school diplomas. Let D denote the event that a randomly chosen voter (in the school board election) had a high school diploma and D denote the event that a randomly chosen voter did not have a high school diploma. Let V denote the event that a randomly chosen voter voted for Goran and V denote the event that a randomly chosen voter did not vote for Goran. Fill in the probabilities to complete the tree diagram below, and then answer the question that follows. Do not round any of your responses. (If necessary, consult a list of formulas.) (a) Fill in the missing probabilities. 4 P(VD) = 0.7 P(DnV) = [ P (D)=0.8 Submit Assign Continue

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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**Probability of Voting Patterns in a Local School Board Election** Recently, Goran ran for president of the local school board. Of those who voted in the election, 80% had a high school diploma, and the other 20% did not. Goran got 70% of the vote of those with high school diplomas, while he got only 15% of the vote of those without high school diplomas. Let \( D \) denote the event that a randomly chosen voter (in the school board election) had a high school diploma and \( \overline{D} \) denote the event that a randomly chosen voter did not have a high school diploma. Let \( V \) denote the event that a randomly chosen voter voted for Goran and \( \overline{V} \) denote the event that a randomly chosen voter did not vote for Goran. Fill in the probabilities to complete the tree diagram below, and then answer the question that follows. Do not round any of your responses. (If necessary, consult a list of formulas.) **(a) Fill in the missing probabilities.** \[ P(V|D) = 0.7 \] \[ P(D) = 0.8 \] \[ P(\overline{V}|D) = \] \[P(D \cap \overline{V}) = \]
Transcribed Image Text:**Probability of Voting Patterns in a Local School Board Election** Recently, Goran ran for president of the local school board. Of those who voted in the election, 80% had a high school diploma, and the other 20% did not. Goran got 70% of the vote of those with high school diplomas, while he got only 15% of the vote of those without high school diplomas. Let \( D \) denote the event that a randomly chosen voter (in the school board election) had a high school diploma and \( \overline{D} \) denote the event that a randomly chosen voter did not have a high school diploma. Let \( V \) denote the event that a randomly chosen voter voted for Goran and \( \overline{V} \) denote the event that a randomly chosen voter did not vote for Goran. Fill in the probabilities to complete the tree diagram below, and then answer the question that follows. Do not round any of your responses. (If necessary, consult a list of formulas.) **(a) Fill in the missing probabilities.** \[ P(V|D) = 0.7 \] \[ P(D) = 0.8 \] \[ P(\overline{V}|D) = \] \[P(D \cap \overline{V}) = \]
### Probability Tree Diagram Below we have a probability tree diagram which outlines the probabilities related to certain events. Each branch in the tree represents a specific event with its associated probability. 1. **First Set of Branches:** - `P(D) = 0.8` - `P(Ď)` is not given and has to be calculated. The event `D` occurs with a probability of 0.8, and the event `Ď` is the complement of `D`. 2. **Second Set of Branches for Event `D`:** - `P(V|D) = 0.7` - `P(Ṽ|D)` is not given and has to be calculated. Given that event `D` has occurred, event `V` occurs with a probability of 0.7, and the event `Ṽ` is the complement of `V`. 3. **Second Set of Branches for Event `Ď`:** - `P(V|Ď) = 0.15` - `P(Ṽ|Ď)` is not given and has to be calculated. Given that event `Ď` has occurred, event `V` occurs with a probability of 0.15, and the event `Ṽ` is the complement of `V`. ### Unlabeled Probabilities in the Diagram The diagram contains several unlabeled probabilities, which need to be calculated: - `P(Ď) = 1 - P(D)` - `P(Ṽ|D) = 1 - P(V|D)` - `P(Ṽ|Ď) = 1 - P(V|Ď)` Additionally, the following joint probabilities are to be calculated: - `P(D ∧ V)` - `P(D ∧ Ṽ)` - `P(Ď ∧ V)` - `P(Ď ∧ Ṽ)` ### Calculation of Specific Probabilities 1. **Complementary Probabilities:** - `P(Ď) = 1 - P(D) = 1 - 0.8 = 0.2` - `P(Ṽ|D) = 1 - P(V|D) = 1 - 0.7 = 0.3` - `P(Ṽ|Ď) =
Transcribed Image Text:### Probability Tree Diagram Below we have a probability tree diagram which outlines the probabilities related to certain events. Each branch in the tree represents a specific event with its associated probability. 1. **First Set of Branches:** - `P(D) = 0.8` - `P(Ď)` is not given and has to be calculated. The event `D` occurs with a probability of 0.8, and the event `Ď` is the complement of `D`. 2. **Second Set of Branches for Event `D`:** - `P(V|D) = 0.7` - `P(Ṽ|D)` is not given and has to be calculated. Given that event `D` has occurred, event `V` occurs with a probability of 0.7, and the event `Ṽ` is the complement of `V`. 3. **Second Set of Branches for Event `Ď`:** - `P(V|Ď) = 0.15` - `P(Ṽ|Ď)` is not given and has to be calculated. Given that event `Ď` has occurred, event `V` occurs with a probability of 0.15, and the event `Ṽ` is the complement of `V`. ### Unlabeled Probabilities in the Diagram The diagram contains several unlabeled probabilities, which need to be calculated: - `P(Ď) = 1 - P(D)` - `P(Ṽ|D) = 1 - P(V|D)` - `P(Ṽ|Ď) = 1 - P(V|Ď)` Additionally, the following joint probabilities are to be calculated: - `P(D ∧ V)` - `P(D ∧ Ṽ)` - `P(Ď ∧ V)` - `P(Ď ∧ Ṽ)` ### Calculation of Specific Probabilities 1. **Complementary Probabilities:** - `P(Ď) = 1 - P(D) = 1 - 0.8 = 0.2` - `P(Ṽ|D) = 1 - P(V|D) = 1 - 0.7 = 0.3` - `P(Ṽ|Ď) =
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