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The image describes a probability problem where:

Given set \( S = \{1, 2, 3, \dots \} \), and the probability mass function (PMF) \( P(k) = \frac{c}{k^2} \) for \( k = 1, 2, 3, \dots \):

### (a) Find the value of \( c \) that makes this a valid probability distribution.

For a valid probability distribution, the sum of the probabilities must equal 1. Therefore, we need to solve:

\[
\sum_{k=1}^{\infty} P(k) = 1
\]

Substituting \( P(k) = \frac{c}{k^2} \):

\[
\sum_{k=1}^{\infty} \frac{c}{k^2} = 1
\]

The sum \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) is a well-known series, the Basel problem, which converges to \( \frac{\pi^2}{6} \). Thus:

\[
c \sum_{k=1}^{\infty} \frac{1}{k^2} = c \cdot \frac{\pi^2}{6} = 1
\]

Solving for \( c \):

\[
c = \frac{6}{\pi^2}
\]

### (b) Find \( P(\text{outcome is greater than 3}) \).

This is asking for the probability that \( k > 3 \). We can calculate it as:

\[
P(k > 3) = 1 - P(1) - P(2) - P(3)
\]

Substituting the values of \( P(k) = \frac{c}{k^2} \) and \( c = \frac{6}{\pi^2} \):

\[
P(1) = \frac{6}{\pi^2 \cdot 1^2}, \quad P(2) = \frac{6}{\pi^2 \cdot 2^2}, \quad P(3) = \frac{6}{\pi^2 \cdot 3^2}
\]

Now, we calculate:

\[
P(k > 3) = 1 - \left( \frac{6}{\pi^2} \cdot 1 + \frac{6}{\pi^2} \cdot \frac{1}{4} + \frac{6}{\pi^2} \cdot \frac{1}{9} \right)
\]

Let me compute this value.The probability that the outcome is greater than 3 is approximately \( 0.173 \).

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. How is the value of \( c \) affected if the probability formula changes to \( \frac{c}{k^n} \) with a different power \( n \)?
2. How would the distribution change if we limited \( k \) to a finite set, such as \( k \in \{1, 2, 3, \dots, 10\} \)?
3. What is the expected value \( E(k) \) for this probability distribution?
4. Can you extend this problem to continuous distributions with a similar form?
5. What would happen to the probabilities if we introduced a weighting factor on certain values of \( k \)?

**Tip:** The Basel problem's solution (\( \frac{\pi^2}{6} \)) is useful in various probability and series-summation problems in mathematics.

Given set S = {1, 2, 3, ...}, and the probability mass function (PMF) P(k) = c / k^2 for k = 1, 2, 3, ..., find: (a) The value of c that makes this a valid probability distribution. (b) The probability that the outcome is greater than 3.

Learn how to find the constant c that makes P(k) = c / k^2 a valid probability distribution, and calculate the probability that the outcome is greater than 3. This problem uses the solution to the Basel problem and key concepts from probability theory.

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