20) At a birthday party, four toys are drawn by lot among the three boys present. The draws are independent, i.e., all the boys participate in all the draws. What is the probability $p$ that at least one boy ends up without a toy?
(A) $p=\frac{4}{9}$
(B) $\frac{4}{9}<p<\frac{1}{2}$
(C) $p=\frac{1}{2}$
(D) $\frac{1}{2}<p<\frac{5}{9}$
(E) $p=\frac{5}{9}$
20) The answer is (E)

A boy is left without toys if all four draws exclude him, and this happens with a probability of $\left(\frac{2}{3}\right)^{4}$. By summing the probabilities that each boy has of being left without toys, we get $3 \cdot\left(\frac{2}{3}\right)^{4}$, but in this way, cases where two boys are both left without toys (all toys went to the third) have been counted twice. These cases have a probability equal to $3 \cdot\left(\frac{1}{3}\right)^{4} 3$. The sought probability is therefore

$$
3 \cdot\left[\left(\frac{2}{3}\right)^{4}-\left(\frac{1}{3}\right)^{4}\right]=3 \cdot \frac{15}{81}=\frac{5}{9} \text {. }
$$