Answer
The probability that a white appears on the second draw given that a white appeared on the first draw is 5/6.
Work Step by Step
An urn contains one white chip and a second chip that is equally likely to be white or black.
Let \[{{W}_{i}}\] be the event a white chip is selected on the ith draw, i = 1, 2 and B be the event that the urn contains one white and one black chip.
There are two ways a white chip could have been drawn from \[{{W}_{1}}{{W}_{2}}\] and \[{{W}_{1}}B\].
If both chips in urn are a white chip, then
\[P({{W}_{1}})=\frac{2}{2}=1\text{ otherwise }P({{W}_{1}})=\frac{1}{2}\]
The probability that both chips in the urn are white
\[\begin{align}
& =\left( \frac{1}{2} \right)(1) \\
& =\frac{1}{2} \\
\end{align}\]
The probability of that the urn has one white and one black, and the first drawing is white
\[\begin{align}
& =\left( \frac{1}{2} \right)\left( \frac{1}{2} \right) \\
& =\frac{1}{4} \\
\end{align}\]
So:
\[\begin{align}
& =\left( \frac{1}{2} \right)\left( \frac{1}{2} \right)\left( \frac{1}{2} \right) \\
& =\frac{1}{8} \\
\end{align}\]
The probability of getting two whites is
\[P({{W}_{1}}\cap {{W}_{2}})=\frac{1}{2}+\frac{1}{8}=\frac{5}{8}\]
The probability that the first drawing is white:
\[P({{W}_{1}})=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\]
The probability that a white appears on the second draw given that a white appeared on the first draw
\[\begin{align}
& P({{W}_{2}}|{{W}_{1}})=\frac{P({{W}_{1}}\cap {{W}_{2}})}{P({{W}_{1}})} \\
& P({{W}_{2}}|{{W}_{1}})=\frac{{}^{5}/{}_{8}}{{}^{3}/{}_{4}} \\
& P({{W}_{2}}|{{W}_{1}})=\frac{5}{6} \\
\end{align}\]
