1. Define events

Let:

• $X, Y, Z$ be the events that bag $X, Y, Z$ is selected.
• $O$ be the event that the drawn coin is a one-rupee coin.

Since a bag is selected at random, $P(X)=P(Y)=P(Z)=\frac{1}{3}.$

2. Find conditional probabilities of drawing a one-rupee coin

• Bag $X$ has $5$ one-rupee coins and $4$ five-rupee coins, total $9$ coins. $P(O\mid X)=\frac{5}{9}.$

• Bag $Y$ has $4$ one-rupee coins and $5$ five-rupee coins, total $9$ coins. $P(O\mid Y)=\frac{4}{9}.$

• Bag $Z$ has $3$ one-rupee coins and $6$ five-rupee coins, total $9$ coins. $P(O\mid Z)=\frac{3}{9}=\frac{1}{3}.$

3. Use Bayes' theorem

We need: $P(Y\mid O)=\frac{P(Y)P(O\mid Y)}{P(X)P(O\mid X)+P(Y)P(O\mid Y)+P(Z)P(O\mid Z)}.$

Substitute values: $P(Y\mid O)=\frac{\frac{1}{3}\cdot\frac{4}{9}}{\frac{1}{3}\cdot\frac{5}{9}+\frac{1}{3}\cdot\frac{4}{9}+\frac{1}{3}\cdot\frac{3}{9}}.$

4. Simplify

Numerator: $\frac{1}{3}\cdot\frac{4}{9}=\frac{4}{27}.$

Denominator: $\frac{1}{3}\left(\frac{5}{9}+\frac{4}{9}+\frac{3}{9}\right)=\frac{1}{3}\cdot\frac{12}{9}=\frac{1}{3}\cdot\frac{4}{3}=\frac{4}{9}.$

Hence, $P(Y\mid O)=\frac{\frac{4}{27}}{\frac{4}{9}}=\frac{4}{27}\cdot\frac{9}{4}=\frac{1}{3}.$

5. Check options

• A: $\frac{1}{2}$ ❌
• B: $\frac{1}{3}$ ✅
• C: $\frac{5}{12}$ ❌
• D: $\frac{1}{4}$ ❌

Therefore, the required probability is $\boxed{\frac{1}{3}}.$