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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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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 (55.6k points)

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