Question

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 its other side is also back is

• A. $\frac{2}{9}$
• B. $\frac{4}{9}$
• C. $\frac{2}{3}$
• D. $\frac{2}{7}$

Answer 1

The correct option is B $\frac{4}{9}$

Let us consider the following events.
A : card shows up black
$B_1$: card with both sides black
$B_2$: card with both sides white
$B_3$: card with one side white and one black
$P\left(B_1\right)=\frac{2}{10},P\left(B_2\right)=\frac{3}{10}.P\left(B_3\right)=\frac{5}{10}P\left(\frac{A}{B_1}\right)=1,P\left(\frac{A}{B_2}\right)=0,P\left(\frac{A}{B_3}\right)=\frac{1}{2}P\left(\frac{B_1}{A}\right)=\frac{\frac{2}{10}\times 1}{\frac{2}{10}\times 1+\frac{3}{10}\times \left(0\right)+\frac{5}{10}\times \frac{1}{2}}=\frac{4}{4+5}=\frac{4}{9}$