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:

The correct option is 2/7.

Let us define the events involved in this problem to solve it step-by-step using Bayes' Theorem:

Let $E_1$ be the event that the ball is drawn from Bag-1.
Let $E_2$ be the event that the ball is drawn from Bag-2.
Let $W$ be the event that the drawn ball is not black (which means it is a white ball).

First, we determine the probability of choosing Bag-1 or Bag-2 based on the roll of a die.
A standard six-sided die has outcomes {1, 2, 3, 4, 5, 6}.
The numbers divisible by 3 are {3, 6} (2 outcomes).
The numbers not divisible by 3 are {1, 2, 4, 5} (4 outcomes).

Therefore, the probabilities of selecting the bags are:
$P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$

$P\left(E_2\right)=\frac{4}{6}=\frac{2}{3}$

Next, we find the conditional probabilities of drawing a white ball from each bag:
Bag-1 contains 4 white and 6 black balls (total of 10 balls). Thus, the probability of drawing a white ball from Bag-1 is:
$P\left(W\right|E_1)=\frac{4}{10}=\frac{2}{5}$

Bag-2 contains 5 white and 5 black balls (total of 10 balls). Thus, the probability of drawing a white ball from Bag-2 is:
$P\left(W\right|E_2)=\frac{5}{10}=\frac{1}{2}$

We are asked to find the probability that the ball was not drawn from Bag-2, given that the ball drawn is not black (white).
Since there are only two bags, not being drawn from Bag-2 is equivalent to being drawn from Bag-1. Thus, we need to calculate $P\left(E_1\right|W)$.

By Bayes' Theorem, the formula is:
$P\left(E_1\right|W)=\frac{P\left(E_1\right)·P\left(W\right|E_1)}{P\left(E_1\right)·P\left(W\right|E_1)+P(E_2)·P(W\left|E_2\right)}$

Substitute the known probabilities into the expression:
Numerator:
$P\left(E_1\right)·P\left(W\right|E_1)=\frac{1}{3}·\frac{2}{5}=\frac{2}{15}$

Denominator:
$P\left(E_1\right)·P\left(W\right|E_1)+P(E_2)·P(W\left|E_2\right)=\frac{2}{15}+(\frac{2}{3}·\frac{1}{2})=\frac{2}{15}+\frac{1}{3}$

To add the terms in the denominator, find a common denominator:
$\frac{2}{15}+\frac{5}{15}=\frac{7}{15}$

Now divide the numerator by the denominator:
$P\left(E_1\right|W)=\frac{\frac{2}{15}}{\frac{7}{15}}=\frac{2}{7}$

Therefore, the probability that the ball was not drawn from Bag-2 is 2/7.