The probability that a radar will detect an object in one cycle is $p$. The probability that the object will be detected in $n$ cycles is
(A) $1 - {p^n}$
(B) $1 - {\left( {1 - p} \right)^n}$
(C) ${p^n}$
(D) $p{\left( {1 - p} \right)^{n - 1}}$

Hint: This question is based on probability. If the probability of happening an event is $P(E)$, then the probability of an event that does not happen will be $P\left( {\bar E} \right) = 1 - P\left( E \right)$. This means that the sum of probability of an event that happens and the event that does not happen is zero.
Or, $P\left( E \right) + P\left( {\bar E} \right) = 1$

Complete step-by-step answer:
For the first cycle, there are two possibilities of events given below,
Let the probability of the event that the radar will “detect” an object in one cycle be $P\left( A \right)$.
Then, $P\left( A \right) = p{\rm{ }}\left( {{\rm{Given}}} \right)$.
Also, let the probability of the event that the radar will “not detect” the object in one cycle be $P\left( B \right)$.
Then, $P\left( B \right) = \left( {1 - p} \right)$.

If in the $1{\rm{st}}$ cycle the object was not detected, then $2{\rm{nd}}$ cycle is started.
Now for the $2{\rm{nd}}$ cycle, the probability would be,
$\begin{array}{c}
 = P\left( B \right) \cdot P\left( A \right)\\
 = \left( {1 - p} \right) \cdot p
\end{array}$

Similarly, if in the $2{\rm{nd}}$ cycle the object was not detected, then the $3{\rm{rd}}$ cycle is started.
Now for the $3{\rm{rd}}$ cycle, the probability would be,
$\begin{array}{c}
= P\left( B \right) \cdot P\left( B \right) \cdot P\left( A \right)\\
= \left( {1 - p} \right) \cdot \left( {1 - p} \right) \cdot p\\
= {\left( {1 - p} \right)^2} \cdot p
\end{array}$
For the $4{\rm{th}}$ cycle, the probability would be,
$\begin{array}{c}
 = P\left( B \right) \cdot P\left( B \right) \cdot P\left( B \right) \cdot P\left( A \right)\\
 = \left( {1 - p} \right) \cdot \left( {1 - p} \right) \cdot \left( {1 - p} \right) \cdot p\\
 = {\left( {1 - p} \right)^3} \cdot p
\end{array}$

From the successive terms, For the $n{\rm{th}}$ cycle, the probability would be,
$ = {\left( {1 - p} \right)^{n - 1}} \cdot p$

Therefore, the probability that the object will be detected in $n$ cycles is ${\left( {1 - p} \right)^{n - 1}} \cdot p$ and the correct option is (D)
So, the correct answer is “Option D”.

Note: In the question, the number of cycles is not specified, so the radar will keep increasing the number of cycles until the object is detected. Therefore, a series of events will be formed, the last term of this series would be the $n{\rm{th}}$ term and the value of $n{\rm{th}}$ term is determined by the succession.