Bayes Theorem

Question

Answer 1

Probability of Urn 3 given the balls are white and green
= P(Urn 3 and white + Urn 3 and green)/ P(white and green)
= P(Urn 3 and white + Urn 3 and green)/ [P(Urn 3 and white + Urn 3 and green) + P(Urn 2 and white + Urn 2 and green) + P(Urn 1 and white + Urn 1 and green)]
= 1/3 * (4/12 * 3/11  * 2C1) /  [1/3 * (4/12 * 3/11 * 2C1) + 1/3 * (2/4 * 1/3 * 2C1) + 1/3 * (1/6 * 3/5 * 2C1)]
= (2/33) / (2/33 + 1/9 + 1/15)
= (2/33) / ( 30 + 55 + 33)/495
= 2 * 15/118
= 15/59

(2C1 is multiplied because the 2 balls could be drawn in 2C1 ways)

Comments

Why we have considered without replacement here?

Can't this be like ......
1/3 * (4/12 * 3/12) /  [1/3 * (4/12 * 3/12 ) + 1/3 * (2/4 * 1/4 ) + 1/3 * (1/6 * 3/6 )]   ?

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After we pick 1 ball from 12, we have only 11 remaining. How can there be a repetition?

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Means the order of drawn matters.