Answer :
Let's start by solving the question step by step.
### Finding the value of [tex]\( k \)[/tex]
We are given the probability mass function [tex]\( f(x) \)[/tex] for [tex]\( x = 0, 1, 2, 3, 4, 5, 6 \)[/tex] and zero otherwise. We need to find the value of [tex]\( k \)[/tex].
Since [tex]\( f(x) \)[/tex] represents a probability mass function, the sum of all probabilities must equal 1. Hence,
[tex]\[ \sum_{x=0}^{6} f(x) = 1 \][/tex]
Substituting the given probability values:
[tex]\[ k + k + k + k + k + k + k = 1 \][/tex]
This simplifies to:
[tex]\[ 7k = 1 \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{1}{7} \][/tex]
So, the value of [tex]\( k \)[/tex] is [tex]\( \frac{1}{7} \)[/tex].
### Finding [tex]\( P(C \times < 4 / x < 1) \)[/tex]
We interpret [tex]\( P(C \times < 4 / x < 1) \)[/tex] as [tex]\( P(x < 1) \)[/tex] because the given format suggests a probability less than a certain value.
Therefore, we need to find the probability that [tex]\( x < 1 \)[/tex].
From the probability mass function, we have:
[tex]\[ f(0) = k = \frac{1}{7} \][/tex]
Since [tex]\( x < 1 \)[/tex] includes only [tex]\( x = 0 \)[/tex], we have:
[tex]\[ P(x < 1) = P(x = 0) = f(0) = \frac{1}{7} \][/tex]
So, the detailed step-by-step procedure gives us the following:
1. Summing [tex]\( f(x) \)[/tex] over the possible values of [tex]\( x \)[/tex] to find [tex]\( k \)[/tex].
2. Setting up the equation [tex]\( 7k = 1 \)[/tex] and solving for [tex]\( k \)[/tex].
3. Finding [tex]\( P(x < 1) \)[/tex] by considering the probability mass function at [tex]\( x = 0 \)[/tex].
The value of [tex]\( k \)[/tex] is [tex]\( \frac{1}{7} \)[/tex], and the probability [tex]\( P(x < 1) \)[/tex] is [tex]\( \frac{1}{7} \)[/tex].
### Finding the value of [tex]\( k \)[/tex]
We are given the probability mass function [tex]\( f(x) \)[/tex] for [tex]\( x = 0, 1, 2, 3, 4, 5, 6 \)[/tex] and zero otherwise. We need to find the value of [tex]\( k \)[/tex].
Since [tex]\( f(x) \)[/tex] represents a probability mass function, the sum of all probabilities must equal 1. Hence,
[tex]\[ \sum_{x=0}^{6} f(x) = 1 \][/tex]
Substituting the given probability values:
[tex]\[ k + k + k + k + k + k + k = 1 \][/tex]
This simplifies to:
[tex]\[ 7k = 1 \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{1}{7} \][/tex]
So, the value of [tex]\( k \)[/tex] is [tex]\( \frac{1}{7} \)[/tex].
### Finding [tex]\( P(C \times < 4 / x < 1) \)[/tex]
We interpret [tex]\( P(C \times < 4 / x < 1) \)[/tex] as [tex]\( P(x < 1) \)[/tex] because the given format suggests a probability less than a certain value.
Therefore, we need to find the probability that [tex]\( x < 1 \)[/tex].
From the probability mass function, we have:
[tex]\[ f(0) = k = \frac{1}{7} \][/tex]
Since [tex]\( x < 1 \)[/tex] includes only [tex]\( x = 0 \)[/tex], we have:
[tex]\[ P(x < 1) = P(x = 0) = f(0) = \frac{1}{7} \][/tex]
So, the detailed step-by-step procedure gives us the following:
1. Summing [tex]\( f(x) \)[/tex] over the possible values of [tex]\( x \)[/tex] to find [tex]\( k \)[/tex].
2. Setting up the equation [tex]\( 7k = 1 \)[/tex] and solving for [tex]\( k \)[/tex].
3. Finding [tex]\( P(x < 1) \)[/tex] by considering the probability mass function at [tex]\( x = 0 \)[/tex].
The value of [tex]\( k \)[/tex] is [tex]\( \frac{1}{7} \)[/tex], and the probability [tex]\( P(x < 1) \)[/tex] is [tex]\( \frac{1}{7} \)[/tex].
