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To determine if the given functions can define a probability distribution, we need to verify the following conditions:

1. **Non-negativity**: \( f(x) \geq 0 \) for all possible values of \( x \).
2. **Normalization**: The sum of all probabilities must equal 1, i.e., \( \sum f(x) = 1 \).

Let's break down and check each function in the image you uploaded.

### (a) \( f(x) = \frac{x}{14} \) for \( x = 0, 1, 2, 3, 4 \)

We need to check:
- **Non-negativity**: \( f(x) = \frac{x}{14} \geq 0 \) for all values of \( x \), which holds true.
- **Normalization**: We sum the values of \( f(x) \) over all possible values of \( x \):

\[
\sum_{x=0}^{4} \frac{x}{14} = \frac{0}{14} + \frac{1}{14} + \frac{2}{14} + \frac{3}{14} + \frac{4}{14} = \frac{0 + 1 + 2 + 3 + 4}{14} = \frac{10}{14} = \frac{5}{7}
\]

Since the sum is not equal to 1, this function **does not** define a valid probability distribution.

### (b) \( f(x) = \frac{3 - x^2}{4} \) for \( x = 0, 1, 2 \)

We need to check:
- **Non-negativity**: For each value of \( x \), we check if \( f(x) \geq 0 \):

For \( x = 0 \), \( f(0) = \frac{3 - 0^2}{4} = \frac{3}{4} \).
For \( x = 1 \), \( f(1) = \frac{3 - 1^2}{4} = \frac{2}{4} = \frac{1}{2} \).
For \( x = 2 \), \( f(2) = \frac{3 - 2^2}{4} = \frac{-1}{4} \).

Since \( f(2) \) is negative, this function **does not** define a valid probability distribution because it violates the non-negativity condition.

### (c) \( f(x) = \frac{2x + 1}{50} \) for \( x = 1, 2, 3, 4, 5 \)

We need to check:
- **Non-negativity**: For each value of \( x \), check if \( f(x) \geq 0 \):

For \( x = 1 \), \( f(1) = \frac{2(1) + 1}{50} = \frac{3}{50} \).
For \( x = 2 \), \( f(2) = \frac{2(2) + 1}{50} = \frac{5}{50} \).
For \( x = 3 \), \( f(3) = \frac{2(3) + 1}{50} = \frac{7}{50} \).
For \( x = 4 \), \( f(4) = \frac{2(4) + 1}{50} = \frac{9}{50} \).
For \( x = 5 \), \( f(5) = \frac{2(5) + 1}{50} = \frac{11}{50} \).

Since all values are positive, the non-negativity condition holds.

- **Normalization**: We check if the sum of these probabilities equals 1:

\[
\sum_{x=1}^{5} \frac{2x + 1}{50} = \frac{3}{50} + \frac{5}{50} + \frac{7}{50} + \frac{9}{50} + \frac{11}{50} = \frac{3 + 5 + 7 + 9 + 11}{50} = \frac{35}{50} = 0.7
\]

Since the sum is 0.7 and not 1, this function **does not** define a valid probability distribution.

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**Summary of Results:**
- (a) Does not define a probability distribution because the sum of probabilities is not equal to 1.
- (b) Does not define a probability distribution because it violates the non-negativity condition.
- (c) Does not define a probability distribution because the sum of probabilities is not equal to 1.

Would you like more details or clarification on any of these steps? 

Here are some related questions to explore further:
1. What adjustments would make function (a) valid as a probability distribution?
2. How do you handle negative probabilities in a distribution?
3. What is the significance of normalization in probability distributions?
4. Can a probability distribution have a sum greater than 1?
5. How does the concept of a probability mass function apply to discrete variables?

**Tip**: Always check both non-negativity and normalization when evaluating if a function can define a probability distribution!

Check whether the following can define a probability distribution and explain:
(a) f(x) = x/14, x = 0, 1, 2, 3, 4
(b) f(x) = (3 - x^2)/4, x = 0, 1, 2
(c) f(x) = (2x + 1)/50, x = 1, 2, 3, 4, 5

Learn how to check whether functions define valid probability distributions by verifying non-negativity and normalization conditions. This detailed solution includes examples of f(x) = x/14, f(x) = (3 - x^2)/4, and f(x) = (2x + 1)/50, suitable for students in Grades 10-12.

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