Three identical boxes contain red and white balls. The first box contained 3 red and 2 white balls the second box 4 red and 5 white balls, and third box has 2 red and 4 white balls. A box chosen randomly and a ball is drawn from it, if the ball drawn out is red, what will be the probability of that second box is chosen? Use Bayesian theorem to find second box chosen?

To solve this problem, we will use Bayes' theorem, which is a way of finding a probability when we know certain other probabilities. The formula is:

P(A|B) = P(B|A) * P(A) / P(B)

Where:

• P(A|B) is the probability of event A given event B is true.
• P(B|A) is the probability of event B given event A is true.
• P(A) and P(B) are the probabilities of events A and B respectively.

In this case, we want to find the probability that the second box was chosen given that a red ball was drawn (P(Second Box|Red Ball)).

First, we need to find the following probabilities:

1. The probability of choosing any box, P(Box) which is 1/3 since there are three boxes.
2. The probability of drawing a red ball from each box, P(Red Ball|Box). For the first box, this is 3/5. For the second box, it's 4/9. And for the third box, it's 2/6 or 1/3.
3. The total probability of drawing a red ball, P(Red Ball). This is the sum of the probability of drawing a red ball from each box times the probability of choosing that box, so (1/3 * 3/5) + (1/3 * 4/9) + (1/3 * 1/3) = 0.2111.

Now we can substitute these values into Bayes' theorem to find P(Second Box|Red Ball):

P(Second Box|Red Ball) = P(Red Ball|Second Box) * P(Second Box) / P(Red Ball) = (4/9 * 1/3) / 0.2111 = 0.157

So, the probability that the second box was chosen given that a red ball was drawn is approximately 0.157 or 15.7%.