Two dice are rolled. Find the probabilities of the following, as reduced fractions, and justify. Use the formula sheet if needed for help. a. Both dice are ≥4, given that the sum is ≥9.
Two dice are rolled. Find the probabilities of the following, as reduced fractions, and justify. Use the formula sheet if needed for help. a. Both dice are ≥4, given that the sum is ≥9.
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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2. Two dice are rolled. Find the probabilities of the following, as reduced fractions, and justify. Use the formula sheet if needed for help.
a. Both dice are ≥4, given that the sum is ≥9.
![**Basic Probability Principle**
\[ P(E) = \frac{n(E)}{n(S)} \]
**Conditional Probability**
\[ P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{P(E \cap F)}{P(F)} \]
**Union Rule**
- For sets: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
- For probability: \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)
**Number of Subsets**
- If a set has \( n \) elements, then it has \( 2^n \) subsets.
**Product Rule**
\[ P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F) \]
**Odds**
\[ \text{Odds in favor of } E = \frac{P(E)}{P(E')} \]
If odds in favor of \( E \) are \( m : n \), then
\[ P(E) = \frac{m}{m+n} \]
**Complement Rule (Probability)**
\[ P(E') = 1 - P(E) \]
\[ P(E) = 1 - P(E') \]
**Complement Rule (Number of Outcomes)**
\[ n(E) = n(S) - n(E') \]
\[ n(E') = n(S) - n(E) \]
**Bayes' Theorem**
\[ P(F|E) = \frac{P(E|F)P(F)}{P(E)} \]
Special Case:
\[ P(F|E) = \frac{P(F)P(E|F)}{P(E|F)P(F) + P(E|F')P(F')} \]
**Independent Events**
\[ P(E|F) = P(E) \]
\[ P(F|E) = P(F) \]
\[ P(E \cap F) = P(E) \cdot P(F) \]
**Permutations**
\[ P(n, k) = \frac{n!}{(n-k)!} \]
**Combinations**
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
**Distinguishable Permutations**
\[ \frac{n!}{n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F985fbbf0-f8fa-4c87-a1a5-f926559935c9%2F381f71a0-a06b-4127-9795-d25731eafbf2%2Ff7sl9i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Basic Probability Principle**
\[ P(E) = \frac{n(E)}{n(S)} \]
**Conditional Probability**
\[ P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{P(E \cap F)}{P(F)} \]
**Union Rule**
- For sets: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
- For probability: \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)
**Number of Subsets**
- If a set has \( n \) elements, then it has \( 2^n \) subsets.
**Product Rule**
\[ P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F) \]
**Odds**
\[ \text{Odds in favor of } E = \frac{P(E)}{P(E')} \]
If odds in favor of \( E \) are \( m : n \), then
\[ P(E) = \frac{m}{m+n} \]
**Complement Rule (Probability)**
\[ P(E') = 1 - P(E) \]
\[ P(E) = 1 - P(E') \]
**Complement Rule (Number of Outcomes)**
\[ n(E) = n(S) - n(E') \]
\[ n(E') = n(S) - n(E) \]
**Bayes' Theorem**
\[ P(F|E) = \frac{P(E|F)P(F)}{P(E)} \]
Special Case:
\[ P(F|E) = \frac{P(F)P(E|F)}{P(E|F)P(F) + P(E|F')P(F')} \]
**Independent Events**
\[ P(E|F) = P(E) \]
\[ P(F|E) = P(F) \]
\[ P(E \cap F) = P(E) \cdot P(F) \]
**Permutations**
\[ P(n, k) = \frac{n!}{(n-k)!} \]
**Combinations**
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
**Distinguishable Permutations**
\[ \frac{n!}{n
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