To determine the probability that a black ball is from the first set of bags, let's define the following:

B: Probability of drawing a black ball.

$B_1$: Probability associated with drawing from the first set of bags, where each bag contains 3 white and 5 black balls.

$B_2$: Probability associated with drawing from the second set of bags, where each bag contains 6 white and 4 black balls.

Given:

Probability of choosing a bag from the first set ($P(B_1)$) is $ \frac{1}{2} $.

Probability of choosing a bag from the second set ($P(B_2)$) is also $ \frac{1}{2} $ since there are two bags in the first set and four bags in the second set but we simplify by considering overall probabilities.

Probabilities of drawing a black ball from each type of bag:

$P(B \mid B_1) = \frac{5}{8}$ for Bag 1.

$P(B \mid B_2) = \frac{4}{10} = \frac{2}{5}$ for Bag 2.

Calculation:

The total probability of drawing a black ball ($P(B)$) is calculated by adding the probabilities of drawing a black ball from each type of bag, weighted by the probability of choosing each type of bag:

$ \begin{aligned} P(B) & = P(B_1) \cdot P(B \mid B_1) + P(B_2) \cdot P(B \mid B_2) \\ & = \frac{1}{2} \times \frac{5}{8} + \frac{1}{2} \times \frac{2}{5} \\ & = \frac{5}{16} + \frac{1}{5} \\ & = \frac{41}{80} \end{aligned} $

Final Probability:

Using Bayes' Theorem, we find the probability that a black ball is from the first set of bags:

$ \begin{aligned} P(B_1 \mid B) & = \frac{P(B \mid B_1) \cdot P(B_1)}{P(B)} \\ & = \frac{\frac{5}{8} \times \frac{1}{2}}{\frac{41}{80}} \\ & = \frac{5}{16} \times \frac{80}{41} \\ & = \frac{25}{41} \end{aligned} $

Thus, the probability that the black ball is from the first set of bags is $ \frac{25}{41} $.

Probability

Continue this paper

Similar Probability PYQs

More practice from this question