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Suppose that A and B are independent events such that P(B) = 0.70 and P (B)=0.20. Find P(An B) and P (AUB). (a) P(A n B) = 11 (b) P(AUB) =

Suppose that A and B are independent events such that P(B) = 0.70 and P (B)=0.20. Find P(An B) and P (AUB). (a) P(A n B) = 11 (b) P(AUB) =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Title: Solving for Intersection and Union of Independent Events** Suppose that \( A \) and \( B \) are independent events such that \( P(\bar{B}) = 0.70 \) and \( P(B) = 0.20 \). **Objective:** Find \( P(A \cap B) \) and \( P(A \cup B) \). **Step-by-Step Solution:** First, let's note that: \[ P(\bar{B}) = 0.70 \] \[ P(B) = 0.20 \] Recall the properties of probability: - The complement rule: \( P(B) + P(\bar{B}) = 1 \) Since \( P(B) = 0.20 \), it is consistent that: \[ P(\bar{B}) = 1 - P(B) = 1 - 0.20 = 0.80 \] Now, for independent events \( A \) and \( B \), the probability of their intersection is given by the product of their probabilities: \[ P(A \cap B) = P(A) \times P(B) \] However, we are not directly given \( P(A) \). For now, we keep \( P(A) \) as is: a) **Finding \( P(A \cap B) \):** \[ P(A \cap B) = P(A) \times 0.20 \] b) **Finding \( P(A \cup B) \):** Use the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute \( P(A \cap B) \): \[ P(A \cup B) = P(A) + 0.20 - (P(A) \times 0.20) \] Thus, we derived the findings based on given probabilities and calculated the following: a) \( P(A \cap B) = P(A) \times 0.20 \) b) \( P(A \cup B) = P(A) + 0.20 - (P(A) \times 0.20) \) Below is the summarization of the calculations: \[ \boxed{ P(A \cap B) = P(A) \times 0.20 } \] \[ \boxed{ P(A \cup B) = P(A)
Transcribed Image Text:**Title: Solving for Intersection and Union of Independent Events** Suppose that \( A \) and \( B \) are independent events such that \( P(\bar{B}) = 0.70 \) and \( P(B) = 0.20 \). **Objective:** Find \( P(A \cap B) \) and \( P(A \cup B) \). **Step-by-Step Solution:** First, let's note that: \[ P(\bar{B}) = 0.70 \] \[ P(B) = 0.20 \] Recall the properties of probability: - The complement rule: \( P(B) + P(\bar{B}) = 1 \) Since \( P(B) = 0.20 \), it is consistent that: \[ P(\bar{B}) = 1 - P(B) = 1 - 0.20 = 0.80 \] Now, for independent events \( A \) and \( B \), the probability of their intersection is given by the product of their probabilities: \[ P(A \cap B) = P(A) \times P(B) \] However, we are not directly given \( P(A) \). For now, we keep \( P(A) \) as is: a) **Finding \( P(A \cap B) \):** \[ P(A \cap B) = P(A) \times 0.20 \] b) **Finding \( P(A \cup B) \):** Use the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute \( P(A \cap B) \): \[ P(A \cup B) = P(A) + 0.20 - (P(A) \times 0.20) \] Thus, we derived the findings based on given probabilities and calculated the following: a) \( P(A \cap B) = P(A) \times 0.20 \) b) \( P(A \cup B) = P(A) + 0.20 - (P(A) \times 0.20) \) Below is the summarization of the calculations: \[ \boxed{ P(A \cap B) = P(A) \times 0.20 } \] \[ \boxed{ P(A \cup B) = P(A)
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