Financial Risk Manager Part 1
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There are three different bags. The first bag contains 3 square blocks and 2 round blocks. The second bag contains 2 square blocks and 3 round blocks. The third bag contains 5 round blocks. In an experiment, a bag is randomly chosen and then a block picked. Given a round block was selected, what is the probability it came from the second bag?
Other
Community
NNikitesh
A
1/5
B
3/10
C
1/3
D
2/5
Explanation:
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.
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