# There are three urns containing 2 white and 3 black balls,3 white and 2 black balls,and 4 white and 1 black balls,respectively.There is an equal probability of each urn being chosen.A ball is drawn at random from the chosen urn and it is found to be white.find the probability that the ball drawn was from the second urn.

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

## 1 Answer

Toolbox:
• $$E_1=chosing\;urns\;1$$
• $$E_2=chosing\;urns\;2$$
• $$E_3=chosing\;urns\;3$$
• $$A\;getting\;white\;ball$$
• 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)=($$2 \;\;urn \;selected/it\;is\;a\;white\;ball$$)
$$\Large\frac{\frac{1}{3}\times\frac{3}{5}}{\frac{1}{3}\times\frac{2}{5}+\frac{1}{3}+\frac{3}{5}+\frac{1}{3}\times\frac{4}{5}}$$
=$$\Large\frac{3}{5}$$$$\times$$$$\frac{5}{9}$$
=$$\Large\frac{1}{3}$$

answered Mar 2, 2013
edited Jun 4, 2013

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