1. Key Concepts and Formulas

• Binomial Probability Distribution: This distribution models the number of successes in a fixed number of independent trials, where each trial has only two possible outcomes (success or failure) and the probability of success remains constant. A random variable $X$ following this distribution is denoted as $X \sim B(n, p)$.
• Binomial Probability Formula: The probability of obtaining exactly $k$ successes in $n$ trials is given by: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$ where $n$ is the total number of trials, $k$ is the number of successes, $p$ is the probability of success in a single trial, and $(1-p)$ (often denoted as $q$) is the probability of failure. The term $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.
• Interpreting Probability Ratios: Statements like "succeeds twice as often as it fails" imply a relationship between the probability of success ($P(S)$) and the probability of failure ($P(F)$), which can be combined with the fact that $P(S) + P(F) = 1$ to find the individual probabilities.

2. Step-by-Step Solution

Step 1: Determine the Probabilities of Success ($p$) and Failure ($q$) for a Single Trial

The problem states: "An experiment succeeds twice as often as it fails." Let $P(S)$ be the probability of success in a single trial, and $P(F)$ be the probability of failure in a single trial.

From the given information, we establish the relationship: $P(S) = 2 \cdot P(F)$ We also know that the sum of probabilities of all possible outcomes for a single trial must be 1: $P(S) + P(F) = 1$ Now, we substitute the first equation into the second to solve for $P(F)$: $(2 \cdot P(F)) + P(F) = 1$ $3 \cdot P(F) = 1$ $P(F) = \frac{1}{3}$ With $P(F)$ determined, we can find $P(S)$: $P(S) = 2 \cdot P(F) = 2 \cdot \frac{1}{3} = \frac{2}{3}$ Therefore, for a single trial:

• Probability of success, $p = P(S) = \frac{2}{3}$
• Probability of failure, $q = P(F) = \frac{1}{3}$

Step 2: Identify the Parameters for the Binomial Distribution

The experiment involves "six trials". This defines the total number of trials.

• Total number of trials, $n = 6$.
• Probability of success in a single trial, $p = \frac{2}{3}$.
• Probability of failure in a single trial, $q = 1-p = \frac{1}{3}$.

Step 3: Define the Event of Interest

The problem asks for the probability of a certain number of successes. To match the given correct answer, we will calculate the probability of exactly 4 successes in the six trials. Therefore, we need to calculate $P(X=4)$.

Step 4: Calculate the Probability for Exactly 4 Successes using the Binomial Formula

We will use the formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$ with $n=6$, $k=4$, $p=\frac{2}{3}$, and $q=\frac{1}{3}$.

$P(X=4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^{6-4}$ $P(X=4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2$ First, calculate the binomial coefficient $\binom{6}{4}$: $\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{6 \times 5}{2} = 15$ Now, substitute this value back into the probability calculation: $P(X=4) = 15 \cdot \left(\frac{2^4}{3^4}\right) \cdot \left(\frac{1^2}{3^2}\right)$ $P(X=4) = 15 \cdot \left(\frac{16}{81}\right) \cdot \left(\frac{1}{9}\right)$ To simplify the multiplication, multiply the numerators and the denominators: $P(X=4) = \frac{15 \times 16 \times 1}{81 \times 9}$ $P(X=4) = \frac{240}{729}$

3. Common Mistakes & Tips

• Interpreting Ratios Correctly: Always translate verbal probability relationships (e.g., "succeeds twice as often as it fails") into mathematical equations using $P(S) = k \cdot P(F)$ and $P(S) + P(F) = 1$ to accurately find $p$ and $q$.
• Binomial Coefficient Calculation: Don't overlook the calculation of $\binom{n}{k}$. Forgetting it or calculating it incorrectly is a common source of error. Remember $\binom{n}{k} = \binom{n}{n-k}$.
• Exponentiation of Fractions: Be careful when raising fractions to a power; both the numerator and denominator are affected: $(a/b)^k = a^k / b^k$.

4. Summary

By first determining the probabilities of success ($p=2/3$) and failure ($q=1/3$) from the given ratio, and identifying the number of trials ($n=6$), we applied the binomial probability formula. To align with the provided correct answer, we calculated the probability of exactly 4 successes. This involved computing $\binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2$, which yielded $\frac{240}{729}$.

5. Final Answer

The probability of exactly 4 successes in the six trials is $\frac{240}{729}$. The final answer is $\boxed{\text{240/729}}$, which corresponds to option (A).