Let’s work through this step by step!

The question asks for the value of $c$ so that the given functions can serve as probability distributions for the discrete random variable $X$. For a function $f(x)$ to represent a valid probability distribution, the sum of the probabilities for all values of $X$ must equal 1. That is,

$\sum_{x} f(x) = 1.$

(a) $f(x) = c (x^2 + 4)$ for $x = 0, 1, 2, 3$:

1. Compute the function values for all $x = 0, 1, 2, 3$. $f(0) = c(0^2 + 4) = 4c, \quad f(1) = c(1^2 + 4) = 5c$ $f(2) = c(2^2 + 4) = 8c, \quad f(3) = c(3^2 + 4) = 13c$

2. Now sum all these values: $\sum_{x=0}^{3} f(x) = 4c + 5c + 8c + 13c = 30c.$

3. Since the sum of probabilities must be 1, we set: $30c = 1 \implies c = \frac{1}{30}.$

(b) $f(x) = c \binom{2}{x} \binom{3}{3-x}$ for $x = 0, 1, 2$:

This function uses binomial coefficients. Let’s compute the function values for all $x = 0, 1, 2$.

• For $x = 0$: $f(0) = c \binom{2}{0} \binom{3}{3} = c \cdot 1 \cdot 1 = c.$
• For $x = 1$: $f(1) = c \binom{2}{1} \binom{3}{2} = c \cdot 2 \cdot 3 = 6c.$
• For $x = 2$: $f(2) = c \binom{2}{2} \binom{3}{1} = c \cdot 1 \cdot 3 = 3c.$

2. Now sum these values: $\sum_{x=0}^{2} f(x) = c + 6c + 3c = 10c.$

3. Since the sum must be 1, we set: $10c = 1 \implies c = \frac{1}{10}.$

Final Answers:

• (a) $c = \frac{1}{30}$
• (b) $c = \frac{1}{10}$

Let me know if you need further clarification or details!

Here are 5 related questions for practice:

1. What properties must a valid probability distribution satisfy?
2. How do binomial coefficients work, and how are they used in probability?
3. What happens if the sum of probabilities exceeds or is less than 1?
4. How would the answer change in part (a) if the domain were $x = 0, 1, 2$ instead of $0, 1, 2, 3$?
5. Can the constant $c$ ever be negative for a valid probability distribution?

Tip: Always double-check that the probabilities sum to 1 when working with distributions!