Question

Bag I contains 3 red and 2 green balls and Bag II contains 5 red and 3 green balls. A ball is drawn from one of the bag at random and it is found to be green. Then the probability that it is drawn from Bag I is

• A. \(\frac{8}{31}\)
• B. \(\frac{12}{31}\)
• C. \(\frac{14}{31}\)
• D. \(\frac{16}{31}\)

Hint:

Apply Bayes' theorem with equal chance of choosing either bag.

Answer 1

The Correct Option is D

Step 1: Total probability
$P(G) = \dfrac{1}{2}\left(\dfrac{2}{5} + \dfrac{3}{8}\right) = \dfrac{1}{2}\cdot\dfrac{31}{40} = \dfrac{31}{80}$.

Step 2: Joint probability
$P(I \cap G) = \dfrac{1}{2}\cdot\dfrac{2}{5} = \dfrac{1}{5} = \dfrac{16}{80}$.

Step 3: Ratio
$\dfrac{16/80}{31/80} = \dfrac{16}{31}$.

Final Answer:
Option D, 16/31, is correct. \[ \boxed{\text{(D) }\dfrac{16}{31}} \]