1. Define the events

Let:

• $A$ = event that Bag A is selected
• $B$ = event that Bag B is selected
• $W$ = event that the drawn ball is white

Since one bag is selected at random, $P(A)=P(B)=\frac{1}{2}.$

2. Find the probability of drawing a white ball from each bag

• Bag A has $3$ white and $7$ red balls, so total $10$ balls. $P(W\mid A)=\frac{3}{10}.$

• Bag B has $3$ white and $2$ red balls, so total $5$ balls. $P(W\mid B)=\frac{3}{5}.$

3. Use Bayes' theorem

We need: $P(A\mid W)=\frac{P(A)P(W\mid A)}{P(A)P(W\mid A)+P(B)P(W\mid B)}.$

Substitute the values: $P(A\mid W)=\frac{\frac{1}{2}\cdot\frac{3}{10}}{\frac{1}{2}\cdot\frac{3}{10}+\frac{1}{2}\cdot\frac{3}{5}}.$

4. Simplify

Numerator: $\frac{1}{2}\cdot\frac{3}{10}=\frac{3}{20}.$

Denominator: $\frac{3}{20}+\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{20}+\frac{3}{10}.$

Now, $\frac{3}{10}=\frac{6}{20},$ so denominator becomes $\frac{3}{20}+\frac{6}{20}=\frac{9}{20}.$

Thus, $P(A\mid W)=\frac{\frac{3}{20}}{\frac{9}{20}}=\frac{3}{9}=\frac{1}{3}.$

5. Match with the options

$\boxed{\frac{1}{3}}$ So the correct option is B.