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To determine if the given table represents a valid probability distribution for the random variable \( X \), two conditions must be met:

1. **Each probability must be between 0 and 1**: This ensures that the values are valid probabilities.
2. **The sum of all probabilities must equal 1**: This ensures that the total probability across all outcomes is correctly normalized.

Given the table:

\[
\begin{array}{|c|c|}
\hline
x & P(x) \\
\hline
1 & 0.2869 \\
2 & 0.0705 \\
3 & 0.2241 \\
4 & 0.192 \\
5 & 0.2265 \\
\hline
\end{array}
\]

### Step 1: Check if each probability is valid.
All probabilities \( P(x) \) are between 0 and 1, so this condition is satisfied.

### Step 2: Check if the sum of probabilities equals 1.
Let's sum the probabilities:

\[
0.2869 + 0.0705 + 0.2241 + 0.192 + 0.2265 = 0.9999999999999999
\]

This is extremely close to 1 (due to rounding error), so this condition is also satisfied.

### Conclusion:
The table represents a valid probability distribution. The answer to the question is **"yes"**.

Would you like any further details or explanations on this?

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Here are five related questions for further exploration:
1. What is the expected value of a discrete random variable?
2. How do you calculate the variance of a probability distribution?
3. What is the difference between a probability distribution and a cumulative distribution function (CDF)?
4. Can a probability distribution have negative values for probabilities?
5. How would the distribution change if another value of \( x \) was added?

**Tip:** When dealing with probabilities, always ensure the sum of all possible outcomes equals 1!

For families with 5 children, let X be the number of children with Genetic Condition B. Can the following table be a probability distribution for the random variable X?

Determine if the given table of probabilities for families with 5 children and Genetic Condition B represents a valid probability distribution. Explore the conditions for probability distributions and key probability rules.

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