question_answer

A hat contains a number of cards with 30% white on both sides, 50% black on one side and white on the other side, 20% black on both sides. The cards are mixed up, and a single card is drawn at random and placed on the table. Its upper side shows up black. The probability that it's other side is also black is

A) \[\frac{2}{9}\]

B) \[\frac{4}{9}\]

C) \[\frac{2}{3}\]

D) \[\frac{2}{7}\]

Correct Answer: B

Solution :

Let us consider the following events.

\[\operatorname{A}:\]Card shows up black

\[{{\operatorname{B}}_{1}}:\]Card with both sides black

\[{{\operatorname{B}}_{2}}:\]Card with both sides white

\[{{\operatorname{B}}_{3}}:\]Card with one side white and one black

\[P({{B}_{1}})=\frac{2}{10},P({{B}_{2}})=\frac{3}{10},P({{B}_{3}})=\frac{5}{10}\]

\[P\left( \frac{A}{{{B}_{1}}} \right)=1,P\left( \frac{A}{{{B}_{2}}} \right)=0,P\left( \frac{A}{{{B}_{3}}} \right)=\frac{1}{2}\]

\[\operatorname{P}\left( \frac{{{B}_{1}}}{{{A}_{1}}} \right)=\frac{\frac{2}{10}\times 1}{\frac{2}{10}\times 1+\frac{3}{10}\times (0)+\frac{5}{10}\times \frac{1}{2}}\]

\[=\frac{4}{4+5}=\frac{4}{9}\]