Question

Um 1 contains 3 red and 7 white balls. Um 2 contains 3 red and 6 white balls. A ball is drawn from urn 1 and placed in urn 2. Then a ball is drawn from um 2. If the ball drawn from um 2 is red, what is the probability that the ball drawn from um 1 was red?

Answer 1

To find the probability that the ball drawn from urn 1 was red given that the ball drawn from urn 2 is red, we can use Bayes' theorem. The probability that the ball drawn from urn 1 is red and the ball drawn from urn 2 is red is 1/3.

Step-by-step explanation:

To find the probability that the ball drawn from urn 1 was red given that the ball drawn from urn 2 is red, we can use Bayes' theorem. Let's denote A as the event that the ball drawn from urn 1 is red, and B as the event that the ball drawn from urn 2 is red. We are trying to find P(A|B), the probability of A given B.

Bayes' theorem states: P(A|B) = (P(B|A) * P(A)) / P(B)

In this case, P(B|A) is the probability of drawing a red ball from urn 2 given that the ball drawn from urn 1 is red. This probability is (3 red balls)/(10 total balls in urn 2) = 3/10.

P(A) is the probability of drawing a red ball from urn 1, which is (3 red balls)/(10 total balls in urn 1) = 3/10.

P(B) is the probability of drawing a red ball from urn 2, regardless of the ball drawn from urn 1. This probability is ((3 red balls from urn 1)/(10 total balls in urn 1)) * ((3 red balls)+(6 white balls) from urn 2)/(10 total balls in urn 2) = (3/10) * (9/10) = 27/100.

Using Bayes' theorem, we can calculate: P(A|B) = ((3/10) * (3/10)) / (27/100) = 9/27 = 1/3.