Probability of a target being hit I don't have the answer of the following question so I wanted to cross check my solution.

$A$ can hit a target 3 times in 5 shots, $B$ 2 times in 5 shots and $C$ 3 times in 4 shots. Find the probability of the target being hit at all when all of them try.

my method
P(target being hit) = $$ \frac{3}{5}*\frac{3}{5}*\frac{1}{4} + \frac{3}{5}*\frac{3}{5}*\frac{3}{4} + \frac{3}{5}*\frac{2}{5}*\frac{1}{4} + \frac{3}{5}*\frac{2}{5}*\frac{3}{4} + \frac{2}{5}*\frac{3}{5}*\frac{3}{4} + \frac{2}{5}*\frac{2}{5}*\frac{1}{4} = 0.82$$
$$ = HMM + HMH + HHM + HHH + MMH + MHM$$
where, H = hit & M  = miss
 A: Let $a$ indicate that person $A$ hit the target, $b$ indicate that $B$ hit the target, and $c$ that $C$ hit the target.
We expect $a$ to occur $3/5$ of the time, $b$ to occur $2/5$ of the time and $c$ to occur $3/4$ of the time.
The event “at least one hits the target” is the complement of “none of them hits the target,” so
$$\begin{align}
\operatorname{P}(a\cup b\cup c) &= 1-\operatorname{P}\left([a\cup b\cup c]’\right) \\
&= 1-\operatorname{P}\left( a’\cap b’ \cap c’\right)\\
&=1- \operatorname{P}\left( a’ \right) \, \operatorname{P} \left( b’ \right) \, \operatorname{P} \left( c’ \right) \\
&= 1-\left( 1-\frac35 \right) \left( 1-\frac25 \right) \left( 1-\frac34 \right) \\
&= 1-\frac25 · \frac35 · \frac14 \\
&= 1-\frac{3}{50} \\
&= \frac{47}{50} \\
\end{align}$$
A: There is also a way we can use we take all the probable situations and subtract the ones that we don't succeed hitting;
For $A$ it is $\dfrac{2}{5}$
For $B$ it is $\dfrac{3}{5}$
And for $C$ it is $\dfrac{1}{4}$ 
$$1-\frac{2}{5}\frac{3}{5}\frac{1}{4}=\frac{47}{50}$$
I accidentally wrote this answer without seeing the first one, but I am not going to delete my effort. (Sorry:() 
