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
To find the probability that the red ball was drawn from Bag 2, we can make use of Bayes' theorem.
Let's define the events:
A: Drawing a red ball
B1: Drawing from Bag 1
B2: Drawing from Bag 2
We want to find P(B2|A), the probability that the red ball was drawn from Bag 2, given that we drew a red ball.
According to Bayes' theorem:
P(B2|A) = (P(B2) * P(A|B2)) / P(A)
We know that P(B2) is the probability of drawing from Bag 2, which is 1/2, since there are two bags and we drew from one of them randomly.
To calculate P(A|B2), the probability of drawing a red ball given that we drew from Bag 2, we need to see how many red balls are there in Bag 2 and what is the total number of balls in Bag 2.
In Bag 2, there are 4 white balls and 5 red balls, making a total of 9 balls.
So, P(A|B2) = 5/9.
Now, let's calculate P(A), the probability of drawing a red ball.
To calculate P(A), we need to consider all possibilities of drawing a red ball, whether from Bag 1 or Bag 2.
The total number of red balls in both bags is 3 + 5 = 8.
The total number of balls in both bags is 5 + 7 = 12.
So, P(A) = 8/12 = 2/3.
Now, we have all the values to calculate P(B2|A):
P(B2|A) = (P(B2) * P(A|B2)) / P(A)
= (1/2 * 5/9) / (2/3)
= (5/18) / (2/3)
= (5/18) * (3/2)
= 15/36
= 5/12
Therefore, the probability that the red ball was drawn from Bag 2 is 5/12 or approximately 0.4167.