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$\mathrm{A}, \mathrm{B}, \mathrm{C}$ try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are $\frac{3}{4}, \frac{1}{2}, \frac{5}{8}$. The probability that the target is hit by A or B but not by $\mathrm{C}$ is :
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The correct answer is:
$21 / 64$
$21 / 64$
$\mathrm{P}(\mathrm{A}$ or $\mathrm{B}$ but not by $\mathrm{C})=\mathrm{P}((\mathrm{A} \cup \mathrm{B}) \cap \overline{\mathrm{C}})$
$$
\begin{aligned}
& =\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \times \mathrm{P}(\overline{\mathrm{C}}) \\
& =[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \times \mathrm{P}(\overline{\mathrm{C}}) \\
& = \\
& {\left[\frac{3}{4}+\frac{1}{2}-\frac{3}{4} \times \frac{1}{2}\right] \times \frac{3}{8}=\left(\frac{6+4-3}{8}\right) \times \frac{3}{8}=\frac{21}{64}}
\end{aligned}
$$
$$
\begin{aligned}
& =\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \times \mathrm{P}(\overline{\mathrm{C}}) \\
& =[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \times \mathrm{P}(\overline{\mathrm{C}}) \\
& = \\
& {\left[\frac{3}{4}+\frac{1}{2}-\frac{3}{4} \times \frac{1}{2}\right] \times \frac{3}{8}=\left(\frac{6+4-3}{8}\right) \times \frac{3}{8}=\frac{21}{64}}
\end{aligned}
$$
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