Hi everyone ! I'm new to this site and this is my first post.
This is a problem I found in my Stats textbook under the Bayes Theorem section.
Here's the problem:
Three urns of the same appearance have the following proportion of balls.
First urn:
Here's my attempt at the solution:
Let \(A\) be the event of selecting a white ball.
Let \(C\) be the event of selecting the second one.
Let \(E_i\) be the event of selecting urn \(i = 1,2,3\)
Now I suppose that that the white ball is already drawn from urn 1. So the probability of getting a second white ball given the first one is already drawn is:
\(P(C | A \wedge E_1) = \frac{1}{3}\cdot0 + \frac{1}{3}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{2} = \frac{7}{18}\)
Similarly selecting another white ball when a white ball was already selected from urn 2 is:
\(P(C | A \wedge E_2) = \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{1}{2} = \frac{4}{9}\)
Again, the prob of getting a white ball given a white was selected from third urn is:
\(P(C| A \wedge E_3) = \frac{4}{9}\)
I don't know how to proceed from here. I thought I would just add up the probabilities, but the answer I got was \(\frac{23}{18}\).
The answer given in the textbook is \(\frac{1}{3}\).
Can anyone please give me suggestions on how to approach/solve this problem ?
Also how do I approach such problems in the future ?
This is a problem I found in my Stats textbook under the Bayes Theorem section.
Here's the problem:
Three urns of the same appearance have the following proportion of balls.
First urn:
2 Black 1 White
Second urn:1 Black 2 White
Third Urn: 2 Black 2 White
One of the urns is selected and one ball is drawn. It turns out to be white. What is the probability of drawing a white ball again, the first one not having been returned ?Here's my attempt at the solution:
Let \(A\) be the event of selecting a white ball.
Let \(C\) be the event of selecting the second one.
Let \(E_i\) be the event of selecting urn \(i = 1,2,3\)
Now I suppose that that the white ball is already drawn from urn 1. So the probability of getting a second white ball given the first one is already drawn is:
\(P(C | A \wedge E_1) = \frac{1}{3}\cdot0 + \frac{1}{3}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{2} = \frac{7}{18}\)
Similarly selecting another white ball when a white ball was already selected from urn 2 is:
\(P(C | A \wedge E_2) = \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{1}{2} = \frac{4}{9}\)
Again, the prob of getting a white ball given a white was selected from third urn is:
\(P(C| A \wedge E_3) = \frac{4}{9}\)
I don't know how to proceed from here. I thought I would just add up the probabilities, but the answer I got was \(\frac{23}{18}\).
The answer given in the textbook is \(\frac{1}{3}\).
Can anyone please give me suggestions on how to approach/solve this problem ?
Also how do I approach such problems in the future ?
