Bag I contains 4 white and 6 black balls.
Bag II contains 4 white and 3 black balls.
One ball is drawn at random from any one of the two bags and it is found to be a black ball. The probability that the black ball was drawn from Bag I is________ (rounded off to two decimal places).

The correct option is 0.58.

To find the probability that the black ball was drawn from Bag I, we can apply Bayes' Theorem.

Let us define the events involved in this problem:
Let $E_1$ be the event of choosing Bag I.
Let $E_2$ be the event of choosing Bag II.
Let $A$ be the event of drawing a black ball.

Since one bag is chosen at random from the two available bags, the probabilities of choosing Bag I or Bag II are equal:
$P\left(E_1\right)=\frac{1}{2}$
$P\left(E_2\right)=\frac{1}{2}$

Next, let us analyze the contents of the bags and determine the conditional probabilities of drawing a black ball from each bag:

Bag I: contains 4 white and 6 black balls. Total balls in Bag I = 4 + 6 = 10.
Therefore, the probability of drawing a black ball given that Bag I was chosen is:
$P\left(A\right|E_1)=\frac{6}{10}=0.6$

Bag II: contains 4 white and 3 black balls. Total balls in Bag II = 4 + 3 = 7.
Therefore, the probability of drawing a black ball given that Bag II was chosen is:
$P\left(A\right|E_2)=\frac{3}{7}$

We need to find the probability that the drawn black ball came from Bag I, which is $P\left(E_1\right|A)$. According to Bayes' Theorem, this is given by:

$P\left(E_1\right|A)=\frac{P\left(E_1\right)·P\left(A\right|E_1)}{P\left(E_1\right)·P\left(A\right|E_1)+P(E_2)·P(A\left|E_2\right)}$

Substituting the known values into the formula:
$P\left(E_1\right|A)=\frac{\frac{1}{2}·\frac{6}{10}}{\frac{1}{2}·\frac{6}{10}+\frac{1}{2}·\frac{3}{7}}$

We can cancel out the common factor of $\frac{1}{2}$ from the numerator and denominator:
$P\left(E_1\right|A)=\frac{\frac{6}{10}}{\frac{6}{10}+\frac{3}{7}}$

Convert the fractions to find a common denominator or calculate them directly:
$P\left(E_1\right|A)=\frac{0.6}{0.6+\frac{3}{7}}$

Let's simplify the denominator fraction:
$0.6+\frac{3}{7}=\frac{6}{10}+\frac{3}{7}=\frac{3}{5}+\frac{3}{7}=\frac{21+15}{35}=\frac{36}{35}$

Now, calculate the final probability value:

Dividing 7 by 12 yields:
$\frac{7}{12}\approx \mathrm{0.58333...}$

Rounding this result to two decimal places, we get 0.58.