Solution:

Sol. Probability of A's hitting the target, $=\frac{4}{5}$

Probability of B's hitting the target, $=\frac{3}{4}$

Probability of, $\mathrm{C}$ 's hitting the target, $=\frac{2}{3}$.

For at least two hits, we may have,

(i) $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$ all hit the target, the probability for which is,

$$ \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}=\frac{24}{60} \text {. } $$ (ii) $\mathrm{A}, \mathrm{B}$ hit the target and $\mathrm{C}$ misses it, the probability for which is, $$ \frac{4}{5} \times \frac{3}{4} \times\left(1-\frac{2}{3}\right)=\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}=\frac{12}{60} \text {. } $$

(iii) $\mathrm{A}$, $\mathrm{C}$ hit the target and $\mathrm{B}$ misses it, the probability for which is,

$$ \frac{4}{5} \times\left(1-\frac{3}{4}\right) \times \frac{2}{3}=\frac{4}{5} \times \frac{1}{4} \times \frac{2}{3}=\frac{8}{60} . $$ (iv) $\mathrm{B}$, $\mathrm{C}$ hit the target and $\mathrm{A}$ misses it, the probability for which is $$ \left(1-\frac{4}{5}\right) \times \frac{3}{4} \times \frac{2}{3}=\frac{1}{5} \times \frac{3}{4} \times \frac{2}{3}=\frac{6}{60} . $$

Since these are mutually exclusive events, required probability,

$$ =\frac{24}{60}+\frac{12}{60}+\frac{8}{60}+\frac{6}{60}=\frac{50}{60}=\frac{5}{6} \text {. } $$