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) =
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter12: Probability
Section12.3: Conditional Probability; Independent Events; Bayes' Theorem
Problem 7YT
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F462ffa92-ffdf-4e48-9e22-7b7cea535b4f%2F271074a0-4bb5-40fd-a8c8-53797ee1182c%2Fj6qpdvr_processed.jpeg&w=3840&q=75)
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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