**Title: Probability Calculation of Events A and B****Introduction:**In probability theory, understanding the relationship between different events is essential for computing their combined probabilities. Below is a problem that demonstrates how to compute the probability of the union of two events, given their individual probabilities and the probability of their intersection.**Problem Statement:**An experiment can result in events A with a probability of P(A) = 0.6 and event B with a probability of P(B) = 0.7. Assuming that P(A ∩ Bᶜ) = 0.1, compute the following probabilities:**Question:**P(A ∪ B) is: - 0.30 - 0.70 - 0.20 - 0.90**Explanation:**To solve for P(A ∪ B), we need to understand the formula for the probability of the union of two events:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]We are given:\[ P(A) = 0.6 \]\[ P(B) = 0.7 \]\[ P(A ∩ Bᶜ) = 0.1 \]First, we need to find $ P(A ∩ B) $. From the complement rule, we know that:\[ P(A) = P(A ∩ B) + P(A ∩ Bᶜ) \]Using this, we find $ P(A ∩ B) $:\[ 0.6 = P(A ∩ B) + 0.1 \]\[ P(A ∩ B) = 0.6 - 0.1 = 0.5 \]Now, substituting the values into the union formula:\[ P(A \cup B) = 0.6 + 0.7 - 0.5 = 1.3 - 0.5 = 0.8 \]**Conclusion:**Given the choices provided, the correct answer is not listed. However, the correct calculation for $ P(A \cup B) $ is 0.8. Therefore, none of the provided options (0.30, 0.70, 0.20, 0.90) correctly represent the probability of the union of events A and B based on the given data.**Graphical

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An experiment can result in events A with probability, P(A) =0.6 and event B with probability, P(B) = 0.7. Assuming that P(A°nBC) = 0.1, compute the following probabilities: P(AUB) is: 0.30 0.70 0.20 0.90

An experiment can result in events A with probability, P(A) =0.6 and event B with probability, P(B) = 0.7. Assuming that P(A°nBC) = 0.1, compute the following probabilities: P(AUB) is: 0.30 0.70 0.20 0.90

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Author:Amos Gilat

Publisher:Amos Gilat

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Concept explainers

Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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Question

**Title: Probability Calculation of Events A and B**

**Introduction:**
In probability theory, understanding the relationship between different events is essential for computing their combined probabilities. Below is a problem that demonstrates how to compute the probability of the union of two events, given their individual probabilities and the probability of their intersection.

**Problem Statement:**
An experiment can result in events A with a probability of P(A) = 0.6 and event B with a probability of P(B) = 0.7. Assuming that P(A ∩ Bᶜ) = 0.1, compute the following probabilities:

**Question:**
P(A ∪ B) is:
- 0.30
- 0.70
- 0.20
- 0.90

**Explanation:**
To solve for P(A ∪ B), we need to understand the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

We are given:
\[ P(A) = 0.6 \]
\[ P(B) = 0.7 \]
\[ P(A ∩ Bᶜ) = 0.1 \]

First, we need to find $ P(A ∩ B) $.
From the complement rule, we know that:
\[ P(A) = P(A ∩ B) + P(A ∩ Bᶜ) \]

Using this, we find $ P(A ∩ B) $:
\[ 0.6 = P(A ∩ B) + 0.1 \]
\[ P(A ∩ B) = 0.6 - 0.1 = 0.5 \]

Now, substituting the values into the union formula:
\[ P(A \cup B) = 0.6 + 0.7 - 0.5 = 1.3 - 0.5 = 0.8 \]

**Conclusion:**
Given the choices provided, the correct answer is not listed. However, the correct calculation for $ P(A \cup B) $ is 0.8. Therefore, none of the provided options (0.30, 0.70, 0.20, 0.90) correctly represent the probability of the union of events A and B based on the given data.

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