Question

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).

Hint:

Use Bayes' theorem: weight each bag's black-ball chance by \(\frac{1}{2}\) and take Bag I's share of the total.

Answer 1

Correct Answer: 0.58

Step 1: Work with a common scale using the fractional chance from each bag.
Since the bag is picked with equal chance $\frac{1}{2}$ each, the weight contributed by Bag I toward getting a black ball is $\frac{1}{2} \times \frac{6}{10}$, and the weight contributed by Bag II is $\frac{1}{2} \times \frac{3}{7}$. The required probability is simply the Bag I weight divided by the sum of both weights, which is the same idea as Bayes' theorem but framed as a ratio of contributions rather than a formula.

Step 2: Compute the two weights using a common denominator.
Weight from Bag I: $\frac{1}{2}\times\frac{6}{10} = \frac{6}{20} = \frac{3}{10}$.
Weight from Bag II: $\frac{1}{2}\times\frac{3}{7} = \frac{3}{14}$.
To add these, use a common denominator of $70$: $\frac{3}{10} = \frac{21}{70}$ and $\frac{3}{14} = \frac{15}{70}$.

Step 3: Find the required probability as a ratio.
Total weight (probability of black overall) $= \frac{21}{70} + \frac{15}{70} = \frac{36}{70}$.
The share of this coming from Bag I is:
$$P(\text{Bag I} \mid \text{black}) = \frac{21/70}{36/70} = \frac{21}{36} = \frac{7}{12}$$

Step 4: Convert the fraction to a decimal and round.
$$\frac{7}{12} = 0.58333...$$
Rounded to two decimal places, this is $0.58$.
$$\boxed{P(\text{Bag I}\mid\text{black}) \approx 0.58}$$