$\begin{array}{1 1} \frac{1}{3} \\ \frac{1}{2} \\ \frac{1}{5} \\ \frac{1}{6} \end{array} $

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- \(E_1=chosing\;urns\;1\)
- \(E_2=chosing\;urns\;2\)
- \(E_3=chosing\;urns\;3\)
- \(A\;getting\;white\;ball\)
- since each urn has equal chanceof being selected
- \(P(E_1)=P(E_2)=P(E_3)=\Large\frac{1}{3}\)
- P(E\(_2\)/A)=\(\Large\frac{{P(E_2)}{P(A/E_2)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}\)

\(P(A/E_1\))=\(P(white\; ball \;chosen/\;1 \;\;urn \;selected\))

=\(\Large\frac{2}{5}\)

\(P(A/E_2\))=\(P(white\; ball \;chosen/\;2 \;\;urn \;selected\))

=\(\Large\frac{3}{5}\)

\(P(A/E_3\))=\(P(white\; ball \;chosen/\;3 \;\;urn \;selected\))

=\(\Large\frac{4}{5}\)

P(E\(_2\)/A)=(\(2 \;\;urn \;selected/it\;is\;a\;white\;ball\))

\(\Large\frac{\frac{1}{3}\times\frac{3}{5}}{\frac{1}{3}\times\frac{2}{5}+\frac{1}{3}+\frac{3}{5}+\frac{1}{3}\times\frac{4}{5}}\)

=\(\Large\frac{3}{5}\)\(\times\)\(\frac{5}{9}\)

=\(\Large\frac{1}{3}\)

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