7. Table 17 gives the probability distribution of the random variable U. Table 17 k 0 1 3 4 Pr (U= k) 15 is is (a) Determine the probability that U = 4.
7. Table 17 gives the probability distribution of the random variable U. Table 17 k 0 1 3 4 Pr (U= k) 15 is is (a) Determine the probability that U = 4.
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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![The image presents a probability distribution table for the random variable \( U \).
### Probability Distribution Table (Table 17)
| \( k \) | \( \Pr(U = k) \) |
|---------|-------------------|
| 0 | \(\frac{4}{15}\) |
| 1 | \(\frac{2}{15}\) |
| 2 | \(\frac{4}{15}\) |
| 3 | \(\frac{3}{15}\) |
| 4 | ? |
### Problem Statement
(a) Determine the probability that \( U = 4 \).
### Explanation
The table provides probabilities associated with each possible value of the random variable \( U \). The sum of all probabilities should equal 1. To find the missing probability for \( U = 4 \), calculate:
\[ \Pr(U = 0) + \Pr(U = 1) + \Pr(U = 2) + \Pr(U = 3) + \Pr(U = 4) = 1 \]
Insert the known probabilities and solve for \( \Pr(U = 4) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5aa36f0-fa96-44c9-9490-28bf7ea812b5%2F41fdec78-ca22-47bd-9bf5-9ceb9b85d811%2F798gbw8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image presents a probability distribution table for the random variable \( U \).
### Probability Distribution Table (Table 17)
| \( k \) | \( \Pr(U = k) \) |
|---------|-------------------|
| 0 | \(\frac{4}{15}\) |
| 1 | \(\frac{2}{15}\) |
| 2 | \(\frac{4}{15}\) |
| 3 | \(\frac{3}{15}\) |
| 4 | ? |
### Problem Statement
(a) Determine the probability that \( U = 4 \).
### Explanation
The table provides probabilities associated with each possible value of the random variable \( U \). The sum of all probabilities should equal 1. To find the missing probability for \( U = 4 \), calculate:
\[ \Pr(U = 0) + \Pr(U = 1) + \Pr(U = 2) + \Pr(U = 3) + \Pr(U = 4) = 1 \]
Insert the known probabilities and solve for \( \Pr(U = 4) \).
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