Understanding the Problem

This question asks for a conditional probability. We need to find the probability that a randomly selected ball came from Bag I, given that the selected ball is black. We can use Bayes' Theorem.

Defining Events and Probabilities

Let B1 be the event that Bag I is chosen, and B2 be the event that Bag II is chosen. Let B be the event that a black ball is drawn.

• Probability of choosing Bag I: $P(B1) = 1/2$
• Probability of choosing Bag II: $P(B2) = 1/2$
• Contents of Bag I: 4 white, 6 black balls (Total = 10).
• Contents of Bag II: 4 white, 3 black balls (Total = 7).
• Probability of drawing a black ball given Bag I was chosen: $P(B | B1) = 6 / 10$
• Probability of drawing a black ball given Bag II was chosen: $P(B | B2) = 3 / 7$

Calculating Total Probability of Drawing a Black Ball

We first find the overall probability of drawing a black ball, $P(B)$, using the law of total probability:

$P(B) = P(B | B1) * P(B1) + P(B | B2) * P(B2)$

Substituting the values:

$P(B) = (6 / 10) * (1 / 2) + (3 / 7) * (1 / 2)$

$P(B) = 3 / 10 + 3 / 14$

To add these fractions, find a common denominator (70):

$P(B) = (3 * 7) / 70 + (3 * 5) / 70$

$P(B) = 21 / 70 + 15 / 70$

$P(B) = 36 / 70$

Simplifying the fraction:

$P(B) = 18 / 35$

Applying Bayes' Theorem

Now we apply Bayes' Theorem to find the probability that the black ball was drawn from Bag I given that it is black, $P(B1 | B)$:

$P(B1 | B) = [ P(B | B1) * P(B1) ] / P(B)$

Substitute the calculated and given probabilities:

$P(B1 | B) = [ (6 / 10) * (1 / 2) ] / (18 / 35)$

$P(B1 | B) = (3 / 10) / (18 / 35)$

To divide by a fraction, multiply by its reciprocal:

$P(B1 | B) = (3 / 10) * (35 / 18)$

$P(B1 | B) = (3 * 35) / (10 * 18)$

$P(B1 | B) = 105 / 180$

Simplify the fraction by dividing numerator and denominator by their greatest common divisor (15):

$P(B1 | B) = 7 / 12$

Final Result and Rounding

Convert the fraction to a decimal:

$7 / 12 ≈ 0.58333...$

Rounding to two decimal places as required:

$P(B1 | B) ≈ 0.58$