11. An urn contains 8 balls, on which the numbers from 1 to 8 are written. Federica draws two balls in succession, erases the number on the first and replaces it with its double, erases the number on the second ball and replaces it with its quadruple. She then reinserts the two balls into the urn (for example, if Federica has drawn the balls 3 and 7 in this order, she will reinsert into the urn two balls with the numbers 6 and 28). Finally, she draws a ball again: what is the probability that the ball drawn has the number 8?
(A) $\frac{1}{8}$
(B) $\frac{1}{7}$
(C) $\frac{7}{44}$
(D) $\frac{9}{56}$
(E) $\frac{1}{6}$
11. The answer is (A). Suppose the ball drawn at the end by Federica bears the number 8. This can happen for three different reasons (which are mutually exclusive):

- the ball was numbered 8 from the beginning, and was not drawn in the first two draws. This event has a probability of $\frac{1}{8} \times \frac{6}{8}$, which is the probability of drawing exactly the ball that initially bore the number $8\left(\frac{1}{8}\right)$, multiplied by the probability that this ball was not among the first two drawn ( $\frac{6}{8}$, since the balls not drawn are 6 out of a total of 8 );
- the ball initially bore the number 4, and was drawn on the first draw: with a similar reasoning to the previous one, we find that this event has a probability of $\frac{1}{8} \times \frac{1}{8}$;
- the ball initially bore the number 2, and was drawn on the second draw: this event again has a probability of $\frac{1}{8} \times \frac{1}{8}$.

In total, the probability that the ball drawn by Federica bears the number 8 is $\frac{1}{8}\left(\frac{6}{8}+\frac{1}{8}+\frac{1}{8}\right)=\frac{1}{8}$.