Find C, if P(X=n) = C^(2/3)^n, n=1,2,.......Solve this problem .

To solve for \( C \), we need to consider that the total probability must equal 1. This is because the probability distribution \( P(X=n) \) sums over all possible values of \( n \), and the sum of all probabilities must equal 1.

Given the probability mass function \( P(X=n) = C \left(\frac{2}{3}\right)^n \), where \( n = 1, 2, 3, \ldots \), we need to sum the series:

\[
\sum_{n=1}^{\infty} P(X=n) = \sum_{n=1}^{\infty} C \left(\frac{2}{3}\right)^n = 1
\]

This is a geometric series with the first term \( a = C \left(\frac{2}{3}\right) \) and the common ratio \( r = \frac{2}{3} \).

The sum of an infinite geometric series is given by:

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

Substituting \( a = C \left(\frac{2}{3}\right) \) and \( r = \frac{2}{3} \) into this formula:

\[
\sum_{n=1}^{\infty} C \left(\frac{2}{3}\right)^n = C \left(\frac{2}{3}\right) \frac{1}{1-\frac{2}{3}} = C \left(\frac{2}{3}\right) \frac{1}{\frac{1}{3}} = C \left(\frac{2}{3}\right) \times 3 = 2C
\]

Since the sum must equal 1, we have:

\[
2C = 1
\]

Solving for \( C \):

\[
C = \frac{1}{2}
\]

Thus, the value of \( C \) is \( \frac{1}{2} \).

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

1. What is the expected value \( E(X) \) for this probability distribution?
2. How would the solution change if the probability function had a different form?
3. What is the variance \( \text{Var}(X) \) for this distribution?
4. How does the sum of an infinite geometric series apply to probability theory in general?
5. How can we verify if a function is a valid probability mass function?

**Tip:** In probability, the sum of all probabilities in a discrete distribution must always equal 1. This principle is key to solving for constants in probability mass functions.

Learn how to find constant C in the probability distribution P(X=n) = C^(2/3)^n for n=1,2,... This problem involves solving a geometric series to ensure the total probability equals 1. Suitable for advanced high school students and beyond.

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