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Assume , PCA) = 0.52, and P (ĀnB) = 0.23. Find P(B) [10 3. P(ANB)= 0•3 %3D %3D 4. Assume, P (ANB) = 0.13 P (Ā O B) =0•28 and %3D P (AUB) = 0. 7. Find P(A). L10 5. Assume, P (AUB) = 0.73 P (A) = 0.5. P(B) = 0.42. Find P(ĀO B). 15 %3D

Assume , PCA) = 0.52, and P (ĀnB) = 0.23. Find P(B) [10 3. P(ANB)= 0•3 %3D %3D 4. Assume, P (ANB) = 0.13 P (Ā O B) =0•28 and %3D P (AUB) = 0. 7. Find P(A). L10 5. Assume, P (AUB) = 0.73 P (A) = 0.5. P(B) = 0.42. Find P(ĀO B). 15 %3D

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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*These questions are not for a graded assignment, but for extra practice and understanding.*

**Example Problems on Probability** 3. Assume \( P(A) = 0.52 \), \( P(A \cap B) = 0.3 \) and \( P(\overline{A} \cap \overline{B}) = 0.23 \). Find \( P(B) \). - Answer: 1.10 4. Assume \( P(A \cap B) = 0.13 \), \( P(\overline{A} \cap \overline{B}) = 0.28 \) and \( P(A \cup B) = 0.7 \). Find \( P(A) \). - Answer: 1.10 5. Assume \( P(A \cup B) = 0.73 \), \( P(A) = 0.51 \) and \( P(B) = 0.42 \). Find \( P(\overline{A} \cap B) \). - Answer: 1.5 **Explanation:** - \( P(A) \), \( P(B) \): Probability of events A and B occurring. - \( P(A \cap B) \): Probability of both events A and B occurring simultaneously. - \( P(\overline{A} \cap \overline{B}) \): Probability of neither event A nor event B occurring. - \( P(A \cup B) \): Probability of at least one of the events A or B occurring. - \( P(\overline{A} \cap B) \): Probability of event B occurring but not A. These examples involve applying formulas and transformations in probability to solve for missing probabilities using given information.
Transcribed Image Text:**Example Problems on Probability** 3. Assume \( P(A) = 0.52 \), \( P(A \cap B) = 0.3 \) and \( P(\overline{A} \cap \overline{B}) = 0.23 \). Find \( P(B) \). - Answer: 1.10 4. Assume \( P(A \cap B) = 0.13 \), \( P(\overline{A} \cap \overline{B}) = 0.28 \) and \( P(A \cup B) = 0.7 \). Find \( P(A) \). - Answer: 1.10 5. Assume \( P(A \cup B) = 0.73 \), \( P(A) = 0.51 \) and \( P(B) = 0.42 \). Find \( P(\overline{A} \cap B) \). - Answer: 1.5 **Explanation:** - \( P(A) \), \( P(B) \): Probability of events A and B occurring. - \( P(A \cap B) \): Probability of both events A and B occurring simultaneously. - \( P(\overline{A} \cap \overline{B}) \): Probability of neither event A nor event B occurring. - \( P(A \cup B) \): Probability of at least one of the events A or B occurring. - \( P(\overline{A} \cap B) \): Probability of event B occurring but not A. These examples involve applying formulas and transformations in probability to solve for missing probabilities using given information.
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