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December 24th 2005, 12:01 AM
Could you please try to solve and explain this problem?
You are told that there are four urns containing the following Urn:1 1 red and 1 white balls, Urn:2 2 red and 4 white balls, Urn: 3 3 red and 5 white balls, Urn 4 , 4 red and 2 white balls . The probability of the ith urn being selected is proportional to the number on the urn. An urn is selected and ball is drawn from it. If the ball drawn is Red and you are told that Urn 4 was not selected, find the probability that the red ball came from Urn 1.
December 24th 2005, 01:08 AM
I don't know if you're familiar with it already (and if you aren't, I suggest you get very comfortable with it), but this problem is an application of Bayes' Theorem. It states that for two events A and B:
P("A happens given B occured")
=P(A|B)=P(A and B)/P(B) = P(B|A)P(A)/(P(B|A)P(A) + P(B|A')P(A'))
P(A')= complement of A = 1 - P(A)
The form in which I have written the theorem is somewhat simplistic and one where there are only two conditional probabilities of B occuring--one conditional on A and one conditional on A'.
A more general form for when there are multiple conditional probabilities is:
P(A|B)=P(B|A)P(A)/sum(P(B|A_i)*P(A_i)) (over all i)
This is what we have here. Let's define our events as follows:
A_1="event urn 1 is selected"
A_2="event urn 2 is selected"
A_3="event urn 3 is selected"
A_4="event urn 4 is selected"
R="event red ball is selected"
Thus, in terms of our definitions, "the probability that [the] red ball [we selected] came from Urn 1" = P(A_1|R).
From Bayes' Theorem,
P(A_1|R)=P(R|A_1)P(A_1)/[P(R|A_1)P(A_1) + P(R|A_2)P(A_2) + P(R|A_3)P(A_3) + P(R|A_4)P(A_4)]
The probability of the ith urn being selected is proportional to the number on the urn means that P(A_1)=1c, P(A_2)=2c, and so on, where c is some constant. But, as a general probability rule, these probabilities must add up to 1. Thus, we get 1c+2c+3c+4c=10c=1-->c=1/10, and from here, we have our probabilities of selecting individual urns.
We are also told that Urn 4 was not selected, so we must now replace P(A_4) with 0.
(1/2*1/10)/(1/2*1/10 + 2/6*2/10 + 3/8*3/10 + 0)
Does that help any? Really get Bayes' Theorem down . . . keep up the studying!
December 24th 2005, 10:23 PM
Thank you so much for making it so easy to understand. You are great!
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