Question

One urn contains 10 red balls and 10 white balls, a second urn contains 8 red...

One urn contains 10 red balls and 10 white balls, a second urn contains 8 red balls and 4 white balls, and a third urn contains 5 red balls and 10 white balls. An urn is selected at random, and a ball is chosen from the urn. If the chosen ball is white, what is the probability that it came from the third urn? Justify your answer.

let say Ai = choosen ball is from ith urn.

In question there are 3 urn so P(A1) = P(A2) = P(A3) = 1/3-----------------(i)

B = getting white ball

B|Ai  = getting white ball from ith Urn

P(B1) = 10/(10+10) = 1/2----------------(ii)

P(B2) = 4/(4+8) = 1/3-------------------(iii)

P(B3) = 10/(10+5) =2/3----------------(iv)

As per the question we are asked to calculate P(A3/B) means if choosen ball is white probability it came from 3rd urn.

As per Bayes Probability theorem we know that---->

P(A|B) = P(A) *P(B|A) /  P(B)

so P(A3|B) = P(A3) *P(B|A3) /  P(B) ---------------(v)

P(B) = P(A1)*P( B|A1) +  P(A2)*P( B|A3) +P(A3)*P( B|A3) (as per total probability theorem)

using results from (i) (ii) (iii) (iv)

we get P(B) = (1/3)*(1/2) + (1/3)*(1/3) + (1/3)*(2/3) = 1/2

so from equation (v)

P(A3|B) = (1/3)*(2/3)/(1/2) = 4/9 answer

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