Page 445 #6b An experiment in which the three mutually exclusive events A, B, and C form a partition of the uniform sample space S is depicted in the diagram below. (Round your answers to three decimal places.

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Page 445 #6b

An experiment in which the three mutually exclusive events AB, and C form a partition of the uniform sample space S is depicted in the diagram below. (Round your answers to three decimal places.)

**Page 445 #6b**

An experiment in which the three mutually exclusive events \( A, B, \) and \( C \) form a partition of the uniform sample space \( S \) is depicted in the diagram below. (Round your answers to three decimal places.)

![Diagram](attachment_name)

The diagram is divided as follows:
- A rectangular space labeled \( S \) contains three vertical sections labeled \( A, B, \) and \( C \), each containing numerical values.
- Inside the rectangle, there is an oval labeled \( D \) with its own numerical values.

The numerical values are:
- In section \( A \) (outside the oval \( D \)): 25
- In section \( B \) (overlapping with \( D \)): 20 inside \( D \) and 20 outside \( D \)
- In section \( C \) (overlapping with \( D \)): 15 inside \( D \) and 20 outside \( D \) 

(b) **Find \( P(B \mid D^c) = \)**

**Explanation:**
- \( D^c \) refers to the complement of the event \( D \), which includes all outcomes not in \( D \). 
- The values relevant to \( D^c \) are those outside the oval, specifically: 25 from \( A \), 20 from \( B \), and 20 from \( C \).
- To calculate \( P(B \mid D^c) \), use the formula:
\[ 
P(B \mid D^c) = \frac{P(B \cap D^c)}{P(D^c)} 
\]
- Calculate \( P(B \cap D^c) \), which is the probability of event \( B \) occurring outside \( D \): 20.
- Calculate \( P(D^c) \), the total of all probabilities outside \( D \): 25 (from \( A \)) + 20 (from \( B \)) + 20 (from \( C \)) = 65.

Finally, substituting these into the formula:
\[ 
P(B \mid D^c) = \frac{20}{65} 
\]
- Simplify and round the calculation to three decimal places.
Transcribed Image Text:**Page 445 #6b** An experiment in which the three mutually exclusive events \( A, B, \) and \( C \) form a partition of the uniform sample space \( S \) is depicted in the diagram below. (Round your answers to three decimal places.) ![Diagram](attachment_name) The diagram is divided as follows: - A rectangular space labeled \( S \) contains three vertical sections labeled \( A, B, \) and \( C \), each containing numerical values. - Inside the rectangle, there is an oval labeled \( D \) with its own numerical values. The numerical values are: - In section \( A \) (outside the oval \( D \)): 25 - In section \( B \) (overlapping with \( D \)): 20 inside \( D \) and 20 outside \( D \) - In section \( C \) (overlapping with \( D \)): 15 inside \( D \) and 20 outside \( D \) (b) **Find \( P(B \mid D^c) = \)** **Explanation:** - \( D^c \) refers to the complement of the event \( D \), which includes all outcomes not in \( D \). - The values relevant to \( D^c \) are those outside the oval, specifically: 25 from \( A \), 20 from \( B \), and 20 from \( C \). - To calculate \( P(B \mid D^c) \), use the formula: \[ P(B \mid D^c) = \frac{P(B \cap D^c)}{P(D^c)} \] - Calculate \( P(B \cap D^c) \), which is the probability of event \( B \) occurring outside \( D \): 20. - Calculate \( P(D^c) \), the total of all probabilities outside \( D \): 25 (from \( A \)) + 20 (from \( B \)) + 20 (from \( C \)) = 65. Finally, substituting these into the formula: \[ P(B \mid D^c) = \frac{20}{65} \] - Simplify and round the calculation to three decimal places.
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