Question

If $X \sim B(4, p)$ and $P(X = 0) = \frac{16}{81}$, then $P(X = 4) =$

• A. $\frac{1}{81}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $\frac{1}{27}$

Hint:

Notice the symmetry of the exponents in binomial boundaries! For $X = 0$, the value is simply $q^n$, and for $X = n$, the value is simply $p^n$. Since $q^4 = (2/3)^4$, its complement must be $p = 1/3$. Raising that to the fourth power gives $\frac{1}{81}$ in one quick step!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem presents a random variable $X$ that follows a binomial distribution with parameters $n = 4$ and success probability $p$. Given the probability value for zero successes, we need to evaluate the probability of getting exactly 4 successes.

Step 2: Key Formula or Approach:
The probability mass function of a binomial distribution is given by the formula: $$ P(X = x) = \binom{n}{x} p^x q^{n-x} $$ where $q = 1 - p$ represents the probability of failure.

Step 3: Detailed Explanation:
We are given that $n = 4$ and $P(X = 0) = \frac{16}{81}$. Let's write out this probability expression: $$ P(X = 0) = \binom{4}{0} p^0 q^{4-0} = \frac{16}{81} $$ Since $\binom{4}{0} = 1$ and $p^0 = 1$, the equation simplifies to: $$ q^4 = \frac{16}{81} $$ Taking the fourth root on both sides to find $q$: $$ q = \sqrt[4]{\frac{16}{81}} = \frac{2}{3} $$ Now, calculate the success probability parameter $p$: $$ p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3} $$ Next, let's find the required probability for $X = 4$ using our calculated values: $$ P(X = 4) = \binom{4}{4} p^4 q^{4-4} $$ Since $\binom{4}{4} = 1$ and $q^0 = 1$: $$ P(X = 4) = p^4 = \left(\frac{1}{3}\right)^4 = \frac{1}{81} $$

Step 4: Final Answer:
The probability $P(X = 4)$ is equal to $\frac{1}{81}$, which corresponds to option (A).