To solve this problem, we use Bayes' theorem. Let's define the events:

• $$X$$: Selecting bag $$X$$
• $$Y$$: Selecting bag $$Y$$
• $$Z$$: Selecting bag $$Z$$
• $$A$$: Drawing a one-rupee coin

We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:

$$P(X) = P(Y) = P(Z) = \frac{1}{3}$$

Next, we need the probability of drawing a one-rupee coin from each bag:

• From bag $$X$$: $$P(A|X) = \frac{5}{5+4} = \frac{5}{9}$$
• From bag $$Y$$: $$P(A|Y) = \frac{4}{4+5} = \frac{4}{9}$$
• From bag $$Z$$: $$P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}$$

We need to find the probability that the coin came from bag $$Y$$ given that a one-rupee coin was drawn, i.e., we need $$P(Y|A)$$. Using Bayes' theorem:

$$P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}$$

To find $$P(A)$$, the total probability of drawing a one-rupee coin can be calculated as follows:

$$P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)$$

Substituting the values:

$$P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$$

Calculating the above, we get:

$$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}$$

Note that $$\frac{1}{9} = \frac{3}{27}$$, so:

$$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}$$

Now, substituting back into Bayes' theorem:

$$P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}$$

Hence, the probability that the coin came from bag $$Y$$ is:

Option B: $$\frac{1}{3}$$