Question

There are 2 bags each containing 3 white and 5 black balls and 4 bags each containing 6 white and 4 black balls. If a ball drawn randomly from a bag is found to be black, then the probability that this ball is from the first set of bags is:

• A. \(\frac{25}{57}\)
• B. \(\frac{25}{41}\)
• C. \(\frac{2}{5}\)
• D. \(\frac{3}{5}\)

Hint:

Use Bayes' Theorem to find conditional probabilities.

Answer 1

The Correct Option is B

Step 1: Define the events.
Let B_1 be the event that a bag is chosen from the first set (3 white, 5 black).
Let B_2 be the event that a bag is chosen from the second set (6 white, 4 black).
Let A be the event that a black ball is drawn.

Step 2: Calculate the probabilities of choosing a bag from each set.
There are 2 bags in the first set and 4 bags in the second set, for a total of 6 bags.
P(B_1) = \(\frac{2}{6} = \frac{1}{3}\)
P(B_2) = \(\frac{4}{6} = \frac{2}{3}\)

Step 3: Calculate the conditional probabilities of drawing a black ball from each set.
P(A|B_1) = \(\frac{5}{8}\) (5 black balls out of 8 total in the first set)
P(A|B_2) = \(\frac{4}{10} = \frac{2}{5}\) (4 black balls out of 10 total in the second set)

Step 4: Calculate the probability of drawing a black ball. Using the law of total probability:
P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2)
P(A) = \(\left(\frac{5}{8}\right)\left(\frac{1}{3}\right) + \left(\frac{2}{5}\right)\left(\frac{2}{3}\right)\)
P(A) = \(\frac{5}{24} + \frac{4}{15} = \frac{25 + 32}{120} = \frac{57}{120} = \frac{19}{40}\)

Step 5: Calculate the probability that the black ball came from the first set of bags. Using Bayes' Theorem:
P(B_1|A) = \(\frac{P(A|B_1)P(B_1)}{P(A)}\)
P(B_1|A) = \(\frac{\left(\frac{5}{8}\right)\left(\frac{1}{3}\right)}{\frac{19}{40}} = \frac{\frac{5}{24}}{\frac{19}{40}} = \frac{5}{24} \times \frac{40}{19} = \frac{200}{456} = \frac{25}{57}\)
However, this is not the answer given. Let's recalculate with the provided answer:
P(B_1|A) = \(\frac{\left(\frac{5}{8}\right)\left(\frac{1}{3}\right)}{\left(\frac{5}{8}\right)\left(\frac{1}{3}\right) + \left(\frac{4}{10}\right)\left(\frac{2}{3}\right)} = \frac{\frac{5}{24}}{\frac{5}{24} + \frac{8}{30}} = \frac{\frac{25}{120} + \frac{32}{120}} = \frac{\frac{5}{24}}{\frac{57}{120}} = \frac{5}{24} \times \frac{120}{57} = \frac{25}{57}\)
The answer given is \(\frac{25}{41}\). Let's see if we can get that:
P(B_1|A) = \(\frac{P(A|B_1)P(B_1)}{P(A)}\)
P(B_1|A) = \(\frac{\frac{5}{24}}{\frac{5}{24}+\frac{8}{30}} = \frac{\frac{25}{120}}{\frac{25}{120}+\frac{32}{120}} = \frac{25}{25+32} = \frac{25}{57}\)
However, we are given \(\frac{25}{41}\). Let's find the mistake.
P(A) = \(\frac{5}{8} \times \frac{1}{3} + \frac{4}{10} \times \frac{2}{3} = \frac{5}{24} + \frac{8}{30} = \frac{25}{120} + \frac{32}{120} = \frac{57}{120} = \frac{19}{40}\)
P(B_1|A) = \(\frac{\frac{5}{24}}{\frac{19}{40}} = \frac{5}{24} \times \frac{40}{19} = \frac{25}{57}\)
We have a mistake in the given answer. The correct answer is \(\frac{25}{57}\).