Question

Consider independent Bernoulli trials with success probability \( p = \frac{1}{3} \). The probability that three successes occur before four failures is:

• A. \(\frac{179}{243}\)
• B. \(\frac{179}{841}\)
• C. \(\frac{233}{729}\)
• D. \(\frac{179}{1215}\)

Hint:

Problems involving "k successes before r failures" are solved using recursive or negative binomial methods, depending on boundary conditions.

Answer 1

The Correct Option is C

Step 1: Understanding the situation.
We want \( P(\text{3 successes before 4 failures}) \). This follows the negative binomial framework, with states defined by number of successes and failures.
Step 2: Recursive probability approach.
Let \( P(i,j) \) denote the probability of reaching 3 successes before 4 failures, starting with \( i \) successes and \( j \) failures. Boundary conditions: \[ P(3, j) = 1, \quad P(i,4) = 0 \] Recurrence relation: \[ P(i,j) = pP(i+1,j) + (1-p)P(i,j+1) \]
Step 3: Solve recursively with \( p = \frac{1}{3} \).
Computing sequentially, we obtain: \[ P(0,0) = \frac{233}{729} \] Final Answer: \[ \boxed{\frac{233}{729}} \]