What is the probability that both events B and C will occur? Now, find the probability of event C. 1 4 3H P(C) A B = 35 25 35 25 5 C D C D Enter
What is the probability that both events B and C will occur? Now, find the probability of event C. 1 4 3H P(C) A B = 35 25 35 25 5 C D C D Enter
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![### What is the probability that both events B and C will occur?
Let's dive into how we determine this probability.
Now, find the probability of event C.
#### Diagram Explanation
The diagram is a tree diagram that represents different paths and their associated probabilities. Here's a detailed breakdown:
1. **Starting Point**:
- **Probability Path 1**: From the starting point to A to C
- The probability of going from the starting point to A is \( \frac{1}{4} \).
- The probability of going from A to C is \( \frac{3}{5} \).
- **Probability Path 2**: From the starting point to A to D
- The probability of going from the starting point to A is \( \frac{1}{4} \).
- The probability of going from A to D is \( \frac{2}{5} \).
- **Probability Path 3**: From the starting point to B to C
- The probability of going from the starting point to B is \( \frac{3}{4} \).
- The probability of going from B to C is \( \frac{3}{5} \).
- **Probability Path 4**: From the starting point to B to D
- The probability of going from the starting point to B is \( \frac{3}{4} \).
- The probability of going from B to D is \( \frac{2}{5} \).
The diagram is organized into two main branches marked as A and B, and these branches split further into C and D with their respective probabilities.
#### Calculating P(C)
To find the overall probability of event C occurring, we need to consider all the paths that lead to C:
- Path 1 (A -> C): \( \frac{1}{4} \times \frac{3}{5} = \frac{3}{20} \)
- Path 2 (B -> C): \( \frac{3}{4} \times \frac{3}{5} = \frac{9}{20} \)
Adding the probabilities of these independent paths gives us the total probability of C:
\[ P(C) = \frac{3}{20} + \frac{9}{20} = \frac{12}{20} = \frac{3}{5} \]
Hence, the probability of event C occurring](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9711575-caba-439c-803f-f18a214b35cf%2Fd870e841-1bee-442e-bfca-218ef63994a9%2Fu16t4ns_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### What is the probability that both events B and C will occur?
Let's dive into how we determine this probability.
Now, find the probability of event C.
#### Diagram Explanation
The diagram is a tree diagram that represents different paths and their associated probabilities. Here's a detailed breakdown:
1. **Starting Point**:
- **Probability Path 1**: From the starting point to A to C
- The probability of going from the starting point to A is \( \frac{1}{4} \).
- The probability of going from A to C is \( \frac{3}{5} \).
- **Probability Path 2**: From the starting point to A to D
- The probability of going from the starting point to A is \( \frac{1}{4} \).
- The probability of going from A to D is \( \frac{2}{5} \).
- **Probability Path 3**: From the starting point to B to C
- The probability of going from the starting point to B is \( \frac{3}{4} \).
- The probability of going from B to C is \( \frac{3}{5} \).
- **Probability Path 4**: From the starting point to B to D
- The probability of going from the starting point to B is \( \frac{3}{4} \).
- The probability of going from B to D is \( \frac{2}{5} \).
The diagram is organized into two main branches marked as A and B, and these branches split further into C and D with their respective probabilities.
#### Calculating P(C)
To find the overall probability of event C occurring, we need to consider all the paths that lead to C:
- Path 1 (A -> C): \( \frac{1}{4} \times \frac{3}{5} = \frac{3}{20} \)
- Path 2 (B -> C): \( \frac{3}{4} \times \frac{3}{5} = \frac{9}{20} \)
Adding the probabilities of these independent paths gives us the total probability of C:
\[ P(C) = \frac{3}{20} + \frac{9}{20} = \frac{12}{20} = \frac{3}{5} \]
Hence, the probability of event C occurring
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