Person A can hit the target with probability $\frac{4}{5}$

Person B with probability $\frac{3}{4}$

Person C with probability $\frac{2}{3}$

They fire together.

We need:

Probability(target is hit at least two times)

This means: exactly 2 hits OR exactly 3 hits.

Let $p_A=\frac{4}{5}$, $p_B=\frac{3}{4}$, $p_C=\frac{2}{3}$ and miss probabilities:

$q_A=\frac{1}{5}$, $q_B=\frac{1}{4}$, $q_C=\frac{1}{3}$.

1.Exactly 2 hits:

-A and B hit, C misses: $p_A p_B q_C$

- A and C hit, B misses: $p_A q_B p_C$

- B and C hit, A misses: $q_A p_B p_C$

So, $P(\text{exactly 2}) = p_A p_B q_C + p_A q_B p_C + q_A p_B p_C$

$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \;+\; \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} \;+\; \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$

$= \frac{1}{5} + \frac{2}{15} + \frac{2}{20}$

$= \frac{4}{20} + \frac{8}{60} + \frac{6}{60}$

$= \frac{12}{60} + \frac{14}{60}$

$= \frac{26}{60} = \frac{13}{30}$

2. Exactly 3 hits: $P(\text{all hit}) = p_A p_B p_C$

$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$

$= \frac{2}{5}$

Therefore,

$P(\text{at least 2 hits}) = P(\text{exactly 2}) + P(\text{exactly 3})$

$= \frac{13}{30} + \frac{2}{5}$

$= \frac{13}{30} + \frac{12}{30}$

$= \frac{25}{30}$

$= \frac{5}{6}$

Final Answer: $\frac{5}{6}$

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