Let $$X$$ be a random variable such that the probability function of a distribution is given by $$P(X = 0) = \frac{1}{2}$$, $$P(X = j) = \frac{1}{3^j}$$ $$(j = 1, 2, 3, \ldots, \infty)$$. Then the mean of the distribution and $$P(X$$ is positive and even) respectively, are:

We have a discrete random variable $$X$$ whose probability mass function is given as follows:

$$P(X = 0) = \frac{1}{2}, \qquad P(X = j) = \frac{1}{3^{\,j}} \; \text{for} \; j = 1,2,3,\ldots$$

First we verify that the total probability is $$1$$. For the strictly positive values we must sum the geometric series

$$\sum_{j = 1}^{\infty} \frac{1}{3^{\,j}}.$$

We recall the infinite geometric‐series formula: if $$|r| < 1$$, then

$$\sum_{j = 1}^{\infty} r^{\,j} \;=\; \frac{r}{1 - r}.$$

Here $$r = \dfrac13$$, so

$$\sum_{j = 1}^{\infty} \frac1{3^{\,j}} = \frac{\tfrac13}{1 - \tfrac13} = \frac{\tfrac13}{\tfrac23} = \frac{1}{2}.$$

Adding the probability at $$j = 0$$ gives

$$P(X=0) + \sum_{j=1}^{\infty} P(X=j) = \frac12 + \frac12 = 1,$$

so the distribution is valid.

Now we find the mean (expected value) $$E[X]$$. By definition, for a discrete variable,

$$E[X] \;=\; \sum_{x} x\,P(X=x).$$

Because the term at $$x=0$$ contributes nothing, we have

$$E[X] = \sum_{j = 1}^{\infty} j \,\frac1{3^{\,j}}.$$

Again we use a standard result for an infinite geometric‐like sum. For $$|r| < 1$$,

$$\sum_{j = 1}^{\infty} j\,r^{\,j} \;=\; \frac{r}{(1 - r)^{2}}.$$

Setting $$r = \dfrac13$$ gives

$$E[X] = \frac{\tfrac13}{(1 - \tfrac13)^{2}} = \frac{\tfrac13}{\left(\tfrac23\right)^{2}} = \frac{\tfrac13}{\tfrac49} = \tfrac13 \times \tfrac94 = \frac34.$$

Next we need the probability that $$X$$ is both positive and even. Let us write the positive even values as $$2,4,6,\ldots$$, that is $$X = 2k$$ where $$k = 1,2,3,\ldots$$. Therefore

$$P(X \text{ is positive and even}) = \sum_{k = 1}^{\infty} P(X = 2k) = \sum_{k = 1}^{\infty} \frac1{3^{\,2k}} = \sum_{k = 1}^{\infty} \left(\frac1{3^{2}}\right)^{k} = \sum_{k = 1}^{\infty} \left(\frac1{9}\right)^{k}.$$

This is another geometric series with common ratio $$\dfrac19$$. Using the same formula for the sum, we get

$$\sum_{k = 1}^{\infty} \left(\frac19\right)^{k} = \frac{\tfrac19}{1 - \tfrac19} = \frac{\tfrac19}{\tfrac89} = \frac{1}{8}.$$

We have now obtained both required quantities: the mean $$E[X] = \dfrac34$$ and the probability that $$X$$ is positive and even $$= \dfrac18$$.

Hence, the correct answer is Option B.