A five-member committee, whose members are $A, B, C, D, E$, decides every issue by voting, and resolutions are adopted by a simple majority. The smartest member of the committee is $A$, who makes a wrong decision only 5% of the time. (For simplicity, we assume that every decision is either correct or incorrect.) $B, C$, and $D$ make mistakes 10% of the time, and $E$ makes mistakes 20% of the time.

a) In what percentage of cases does the committee make an incorrect decision?

b) $E$, the "least capable" member of the company, has decided that from now on, he will always vote the same way as $A$. How will this affect the ratio of incorrect decisions?
a) Since decisions are made by a majority of words, a wrong decision is made if and only if at least 3 members decide incorrectly. This is only possible in the following cases: 1. everyone makes a wrong decision; 2. 4 people make a wrong decision, one makes a correct one; 3. three make wrong decisions, two make correct ones. Let $P_{1}, P_{2}$, and $P_{3}$ represent the probabilities of these events. Since these three events are pairwise disjoint, the desired probability is: $P_{1}+P_{2}+P_{3}$. Let's determine the individual probabilities.

1. $P_{1}=\frac{1}{20} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{5}=\frac{1}{10^{5}}$

since their decisions are independent.
2. $P_{2}=\frac{19}{20} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{5}+3 \cdot \frac{1}{20} \cdot \frac{9}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{5}+\frac{1}{20} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{4}{5}=\frac{50}{10^{5}}$.

Indeed, either $A$, or $B, \ldots$, or $E$ makes a correct decision. These five events are pairwise disjoint, and in each case, the decisions are independent.
3. $P_{3}=3 \cdot \frac{19}{20} \cdot \frac{9}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10}+3 \cdot \frac{1}{20} \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{1}{10} \cdot \frac{1}{5}+$

$+3 \cdot \frac{1}{20} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10} \cdot \frac{4}{5}+\frac{19}{20} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{4}{5}=\frac{940}{10^{5}}$

Thus, in case a), the desired probability is:

$$
P_{1}+P_{2}+P_{3}=\frac{991}{10^{5}}=0.00991
$$

b) Now we can distinguish two cases:

1. $A$ and $E$ make correct decisions; $B, C, D$ make wrong ones $\left(P_{1}\right)$.
2. $A$ and $E$ make wrong decisions; and at least one member makes a wrong decision $\left(P_{2}\right)$.
3. $P_{1}=\frac{19}{20} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10}=\frac{19}{2 \cdot 10^{4}}$
4. $P_{2}=\frac{1}{20} \cdot P$ where $P$ is the probability that at least one of $B, C$, and $D$ makes a wrong decision. But $1-P=\frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10}$, since it gives the probability of $B, C$, and $D$ making correct decisions. After this:

$$
P_{1}+P_{2}=\frac{19}{2 \cdot 10^{4}}+\frac{1}{20} \cdot \frac{10^{3}-9^{3}}{10^{3}}=\frac{290}{2 \cdot 10^{4}}=0.0145
$$

Spissich László (Pápa, Türr I. Gymnasium)

Note. It can be seen that in case b), there will be more wrong decisions, although at first glance, one would think that the number of wrong decisions should decrease with the "least capable" member adapting to the "smartest" one.

Tóth Csaba (Sopron, Széchenyi I. Gymnasium, IV. grade)