Step 1: Understanding the Concept:
A binomial random variable $X \sim B(n, p)$ represents the number of successes in $n$ independent trials, each with probability of success $p$.
The range of $X$ is $\{0, 1, 2, \dots, n\}$. Given the range is $\{0, 1, 2, 3, 4\}$, we identify $n = 4$.
Step 2: Key Formula or Approach:
The probability mass function of a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where $q = 1 - p$.
Substitute the given values into the equation $P(X = 3) = 3P(X = 4)$ and solve for $p$.
Step 3: Detailed Explanation:
Given $n = 4$ and $P(X = 3) = 3P(X = 4)$.
Let's express both probabilities using the binomial formula:
\[ P(X = 3) = \binom{4}{3} p^3 q^{4-3} = 4 p^3 q \]
\[ P(X = 4) = \binom{4}{4} p^4 q^{4-4} = 1 \cdot p^4 \cdot 1 = p^4 \]
Substitute these into the given equation:
\[ 4 p^3 q = 3 p^4 \]
Since $p$ is a probability parameter for a distribution taking values up to 4, we assume $p>0$. We can divide both sides by $p^3$:
\[ 4q = 3p \]
We know that $q = 1 - p$. Substitute this:
\[ 4(1 - p) = 3p \]
\[ 4 - 4p = 3p \]
\[ 4 = 7p \]
\[ p = \frac{4}{7} \]
The value derived mathematically is $\frac{4}{7}$, which is not among the given options (A: 1/4, B: 3/4, C: 1/3, D: 2/5).
Note: It is highly probable there is a typographical error in the original question. A common variant is $P(X=2) = 3P(X=3)$, which yields $6p^2q^2 = 3(4p^3q) \Rightarrow 6q = 12p \Rightarrow q = 2p \Rightarrow 1-p = 2p \Rightarrow p = 1/3$ (Option C). However, solving the text exactly as presented yields $4/7$.
Step 4: Final Answer:
Based on the exact text, the parameter $p$ is $\frac{4}{7}$.