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The probability of $A, B, C$ solving a problem are $\frac{1}{3}, \frac{2}{7}, \frac{3}{8}$ respectively. If all the three try to solve the problem simultaneously, the probability that exactly one of them will solve it.
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## 1 Answer

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Let $E_{1}, E_{2}, E_{3}$ be the eventsthat the problem issolved by $A, B, C$ respectively and let $\mathrm{p}_{1}, \mathrm{P}_{2}, \mathrm{p}_{3}$ be corresponding probabilities. Then,

$\mathrm{P}_{1}=P\left(\mathrm{E}_{1}\right)=\frac{1}{3}, \mathrm{P}_{2}=P\left(\mathrm{E}_{2}\right)=\frac{2}{7}, \mathrm{P}_{3}=\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{3}{8}, \mathrm{q}_{1}=\mathrm{P}\left(\overline{\boldsymbol{E}}_{1}\right)=1-\frac{1}{3}=\frac{2}{3}$,
$\mathrm{q}_{2}=\mathrm{P}\left(\bar{E}_{2}\right)=1-\frac{2}{7}=\frac{5}{7}, \mathrm{q}_{3}=\mathrm{P}\left(\overline{\boldsymbol{E}}_{3}\right)=1-\frac{3}{8}=\frac{5}{8} .$

The problem will be solved by exactly one of them if it happens in the following mutually exclusive ways:

(1) A solves and $B$, and $C$ do not solve;

(2) B solves and $A$, and $C$ do not solve;

(3) C solves and $A$, and $B$ do not solve;

Required probability $=p_{1} q_{2} q_{3}+q_{1} p_{2} q_{3}+q_{1} q_{2} p_{3}$

$=\frac{1}{3} \times \frac{5}{7} \times \frac{5}{8}+\frac{2}{3} \times \frac{2}{7} \times \frac{5}{8}+\frac{2}{3} \times \frac{5}{7} \times \frac{3}{8}=\frac{25}{168}+\frac{5}{42}+\frac{5}{28}=$
$\frac{25}{56}$.
by Diamond (77,939 points)

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