Answer: 3/10

This is a conditional probability problem using Bayes' theorem. Let's define the events: - Let "1" be the event that the first bag is picked - Let "2" be the event that the second bag is picked - Let "3" be the event that the third bag is picked - Let "R" be the event that a round block is drawn We want P(2|R) = probability the block came from bag 2 given that a round block was selected. Using Bayes' theorem: P(2|R) = P(R|2) * P(2) / P(R) Where: - P(R|2) = probability of drawing a round block from bag 2 = 3/5 (3 round blocks out of 5 total) - P(2) = probability of selecting bag 2 = 1/3 (equal chance for each bag) - P(R) = total probability of drawing a round block To find P(R): P(R) = P(R|1)*P(1) + P(R|2)*P(2) + P(R|3)*P(3) - P(R|1) = 2/5 (2 round blocks out of 5 total in bag 1) - P(R|2) = 3/5 (3 round blocks out of 5 total in bag 2) - P(R|3) = 5/5 = 1 (all blocks are round in bag 3) - P(1) = P(2) = P(3) = 1/3 So: P(R) = (2/5)*(1/3) + (3/5)*(1/3) + (1)*(1/3) P(R) = 2/15 + 3/15 + 5/15 = 10/15 = 2/3 Now calculate P(2|R): P(2|R) = (3/5)*(1/3) / (2/3) = (3/15) / (2/3) = (1/5) / (2/3) = (1/5)*(3/2) = 3/10 Therefore, the probability that the round block came from the second bag is 3/10.

Author: Nikitesh Somanthe