There are two bags out of which Bag 1 contains 2 white and 3 red balls and bag 2 contains 4 white and 5 red balls. 1 ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag 2
I know the probability of drawing red from the second bag = 5/9, but I don't know how to discount not drawing from the first bag. That is why I did not respond to you previously.
To find the probability that the red ball was drawn from bag 2, we can use Bayes' theorem. Let's define the following events:
A: the event that the ball is drawn from bag 2.
B: the event that a red ball is drawn.
We want to find P(A|B), the probability of event A given that event B has occurred. By Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability of drawing a red ball given that the ball is drawn from bag 2. We know that bag 2 contains 9 balls, with 5 of them being red. So, P(B|A) = 5/9.
P(A) is the probability of drawing a ball from bag 2. Since there are two bags and the choice of the bag is random, P(A) = 1/2.
P(B) is the probability of drawing a red ball, regardless of the bag it was drawn from. This can be calculated by considering the probabilities of drawing a red ball from each bag and then adding them up.
To find P(B), we need to calculate P(B|A') and P(B|A), where A' is the event that the ball is drawn from bag 1.
P(B|A') is the probability of drawing a red ball given that the ball is drawn from bag 1. We know that bag 1 contains 5 red balls out of a total of 5 balls, so P(B|A') = 5/5 = 1.
Now we can calculate P(B):
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
P(B) = (5/9) * (1/2) + (1) * (1/2)
P(B) = 5/18 + 1/2
P(B) = 15/18
P(B) = 5/6
Finally, we can calculate P(A|B) using Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (5/9 * 1/2) / (5/6)
P(A|B) = 5/18 / 5/6
P(A|B) = 5/18 * 6/5
P(A|B) = 1/3
Therefore, the probability that the red ball was drawn from bag 2 is 1/3.