Question

Consider a sequence of independent Bernoulli trials, where \(\frac{3}{4}\) is the probability of success in each trial. Let X be a random variable defined as follows : If the first trial is a success, then X counts the number of failures before the next success. If the first trial is a failure, then X counts the number of successes before the next failure. Then 2E(X) equals __________

Answer 1

Correct Answer: 2

To solve the problem, we must understand the definition of the random variable \(X\). The variable \(X\) depends on whether the first Bernoulli trial is a success or a failure. Let's analyze both cases: 
1. **Case 1:** The first trial is a success. Here, \(X\) is the number of failures before the next success. This follows a geometric distribution with a success probability \(p = \frac{3}{4}\). The expected value for a geometric distribution, where \(X\) counts failures before the first success, is given by \(E(X) = \frac{1-p}{p} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}\). 

2. **Case 2:** The first trial is a failure. Here, \(X\) is the number of successes before the next failure. Again, this distribution follows a geometric pattern, but since we're counting successes, the success probability here is \(q = \frac{1}{4}\). The expected value in this context is \(E(X) = \frac{q}{1-q} = \frac{\frac{3}{4}}{\frac{1}{4}} = 3\).
Since the probability of the first trial being a success is \(\frac{3}{4}\) and being a failure is \(\frac{1}{4}\), the expected value of \(X\) can be calculated as: \[ E(X) = \left(\frac{3}{4}\right) \times \frac{1}{3} + \left(\frac{1}{4}\right) \times 3 = \frac{3}{12} + \frac{3}{4} = \frac{3}{12} + \frac{9}{12} = 1 \] Now, we need to find \(2E(X)\). Thus, \(2E(X) = 2 \times 1 = 2\).