Find the probability that not all the targets in the group will be detected. 

There is group of $k$ targets, each of which independently of other targets, can be detected by a radar unit with probability $p$. Each of $m$ radar units detects the targets independently of other units. Find the probability that not all the targets in the group will be detected.


My Approach:
Let Event $A$ is "Not all the targets in group is detected"
And Event $B$ denotes "All the targets has been detected"
$\implies$$P(A)=1-P(B)$
According to question $p$ is probability that radar will detect targets $\implies$ $1-p$ will be Probability that radar will not detect target. So none of the radar will detect target is ${(1-p)^m}$. $\implies$ At least one radar will detect target is $1-(1-p)^m$.
$\implies$ At least one radar will detect the all the $k$ targets is $(1-(1-p)^m)^k$.
$\implies$ $P(B)=(1-(1-p)^m)^k$.
$\implies$ $P(A)=1-(1-(1-p)^m)^k$
My doubt:$(1)$ Is my Approach correct?
$(2)$ Can somebody suggest me All the other method to solve this Problem?

Every target will be aimed at $m$ times,
so P(a particular target is not hit) $= (1-p)^m$
P(a particular target is hit) $= 1 - (1-p)^m$
P(all targets are hit) $=(1-(1-p)^m)^k$
P(at least one target not hit) = $1 - (1-(1-p)^m)^k$
So your approach is correct.
