Let X be a random variable with probability mass function

f(x) = c (0.59)x for  x = 0, 1, 2, .... 

Find P(X ≥ 6) (correct up to at least 3 decimal places)

Given that \( X \) is a discrete random variable with a probability mass function \( f(x) = c (0.59)^x \) for \( x = 0, 1, 2, \dots \), we need to find \( P(X \geq 6) \).

### Step 1: Determine the constant \( c \)
The sum of the probabilities over all possible values of \( X \) must equal 1, so:

\[
\sum_{x=0}^{\infty} f(x) = \sum_{x=0}^{\infty} c (0.59)^x = 1
\]

This is a geometric series with the first term \( a = c \) and common ratio \( r = 0.59 \). The sum of an infinite geometric series is given by:

\[
\sum_{x=0}^{\infty} ar^x = \frac{a}{1-r}
\]

Substituting the values, we get:

\[
\frac{c}{1 - 0.59} = 1
\]

\[
c = 1 - 0.59 = 0.41
\]

Thus, \( f(x) = 0.41 (0.59)^x \).

### Step 2: Calculate \( P(X \geq 6) \)
The probability \( P(X \geq 6) \) is the sum of the probabilities from \( X = 6 \) to infinity:

\[
P(X \geq 6) = \sum_{x=6}^{\infty} f(x) = \sum_{x=6}^{\infty} 0.41 (0.59)^x
\]

This is again a geometric series, but starting at \( x = 6 \). The sum of a geometric series starting at \( x = k \) is:

\[
\sum_{x=k}^{\infty} ar^x = \frac{ar^k}{1 - r}
\]

Here, \( a = 0.41 \) and \( r = 0.59 \). Plugging in \( k = 6 \), we get:

\[
P(X \geq 6) = \frac{0.41 \times (0.59)^6}{1 - 0.59}
\]

\[
P(X \geq 6) = \frac{0.41 \times (0.59)^6}{0.41} = (0.59)^6
\]

### Step 3: Compute the value
Now, we compute \( (0.59)^6 \):

\[
(0.59)^6 \approx 0.05305
\]

### Final Answer:
Thus, the probability \( P(X \geq 6) \) is approximately \( \boxed{0.053} \).

Would you like more details or have any questions? Here are 5 related questions you might find interesting:

1. How would the calculation change if \( r \) were a different value, such as 0.7?
2. What is the expected value \( E(X) \) of this random variable?
3. How would you compute \( P(X = 4) \) using the probability mass function?
4. What is the variance \( \text{Var}(X) \) for this distribution?
5. How does the geometric distribution differ from other discrete distributions like the Poisson distribution?

**Tip:** Understanding the properties of geometric series is crucial for solving problems involving infinite sums in probability theory.

Learn how to calculate the probability P(X ≥ 6) for a geometric series in probability theory. This problem involves finding the sum of probabilities for a discrete random variable with a given probability mass function. Suitable for advanced undergraduate students studying probability theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation