For k=1, 2, 3 the box \[{{B}_{k}}\] contains k red balls and \[(k+1)\] white balls. Let \[P({{B}_{1}})=\frac{1}{2},P({{B}_{2}})=\frac{1}{3}\]and \[P({{B}_{3}})=\frac{1}{6}.\] A box is selected at random and a ball is drawn from it, if a red ball is drawn, then the probability that it has come from box \[{{B}_{2}}\], is

For k=1, 2, 3 the box \[{{B}_{k}}\] contains k red balls and \[(k+1)\]

1.	For k=1, 2, 3 the box \[{{B}_{k}}\] contains k red balls and \[(k+1)\] white balls. Let \[P({{B}_{1}})=\frac{1}{2},P({{B}_{2}})=\frac{1}{3}\]and \[P({{B}_{3}})=\frac{1}{6}.\] A box is selected at random and a ball is drawn from it, if a red ball is drawn, then the probability that it has come from box \[{{B}_{2}}\], is
A.	\[\frac{35}{78}\]
B.	\[\frac{14}{39}\]
C.	\[\frac{10}{13}\]
D.	\[\frac{12}{13}\]
Answer» C. \[\frac{10}{13}\]

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