Solution:

\(\text {For these types of question, use the Total Probability Theorem/Formula.}\\ \mathbb{P}(B_1) =\mathbb{P}(B_2) =\mathbb{P}(B_3) = \dfrac{1}{3}\\ \text {Probability of mint from Bag} \tiny \text {#} \\ \mathbb{P}(M|B_1) = \dfrac{1}{2}\\ \mathbb{P}(M|B_2) = \dfrac{3}{5}\\ \mathbb{P}(M|B_3) = \dfrac{2}{3}\\ \;\\ \mathbb{P}(B_1|M) = \Big (\mathbb{P}(B_1) \cdot \mathbb{P}(M|B_1)\Big)\div \Big(\mathbb{P}(B_1) \mathbb{P}(M|B1) + \mathbb{P}(B_2)\mathbb{P}(M|B_2) +\mathbb{P} P(B_3) \mathbb{P}(M|B_3)\Big)\)

\(\text {The probability that Jane chose bag 1, given that she selects a mint is equal to the probability}\\ \text {Jane chose a mint from bag #1 divided by the probability that she selected a mint.}\\\)

\(\mathbb{P}(B_1|M) = \dfrac{\Big(\dfrac{1}{3} \cdot \dfrac{1}{2}\Big) }{\Big(\dfrac{1}{3} \cdot \dfrac{1}{2}\Big) + \Big(\dfrac{1}{3} \cdot \dfrac{3}{5}\Big) +\Big( \dfrac{1}{3} \cdot \dfrac{2}{3}\Big)}\\ \;\\ \hspace{1.8cm}=\dfrac {15}{53}\\ \;\\ \hspace{1.8cm} \approx 0.28301\\ \)

GA