There are two bags. Bag-1 contains $4$ white and $6$ black balls and Bag-2 contains $5$ white and $5$ black balls. A die is rolled, if it shows a number divisible by 3, a ball is drawn from Bag-1, else a ball is drawn from Bag-2. If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is :

Define our events:

• Event A: Ball is drawn from Bag-1
  • Event B: Ball is white (not black)

We need to find $P(A|B)$ - the probability that the ball came from Bag-1 given that it's white.

---

Using Bayes' Theorem: $P(A|B) = \dfrac{P(B|A) \times P(A)}{P(B)}$

---

Finding each component:

$P(A)$ = Probability of drawing from Bag-1 = Probability of rolling a number divisible by 3

Numbers divisible by 3 on a die: 3, 6 (2 numbers out of 6)

$P(A) = \dfrac{2}{6} = \dfrac{1}{3}$

---

$P(B|A)$ = Probability of drawing a white ball given we're drawing from Bag-1

Bag-1 has 4 white balls out of 10 total balls

$P(B|A) = \dfrac{4}{10} = \dfrac{2}{5}$

---

$P(B)$ = Total probability of drawing a white ball

$P(B) = P(B|A) \times P(A) + P(B|A^c) \times P(A^c)$

Where $P(A^c) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ (probability of drawing from Bag-2)

$P(B|A^c) = \dfrac{5}{10} = \dfrac{1}{2}$ (probability of white ball from Bag-2)

$P(B) = \dfrac{2}{5} \times \dfrac{1}{3} + \dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{2}{15} + \dfrac{1}{3} = \dfrac{2}{15} + \dfrac{5}{15} = \dfrac{7}{15}$

---

Apply Bayes' Theorem:

$P(A|B) = \dfrac{P(B|A) \times P(A)}{P(B)} = \dfrac{\dfrac{2}{5} \times \dfrac{1}{3}}{\dfrac{7}{15}} = \dfrac{\dfrac{2}{15}}{\dfrac{7}{15}} = \dfrac{2}{7}$

Therefore, the probability that the white ball was drawn from Bag-1 is $\dfrac{2}{7}$.