Four persons can hit a target correctly with probabilities $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4}$ and $\dfrac{1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is?
A. $\dfrac{25}{192}$
B. $\dfrac{1}{192}$
C. $\dfrac{25}{32}$
D. $\dfrac{7}{32}$

Hint: First we will find out the probability of all the four persons not hitting the target, this will be done by using the formula: $P\left( \overline{t} \right)=1-P\left( t \right)$ and then we will apply the condition of independent events that is $P\left( M\bigcap N \right)=P\left( M \right)\times P\left( N \right)$ and then we will get the required probability.

Complete step-by-step answer:
Let the four persons be: A, B, C and D, now we are given the probabilities of hitting a target by all these four persons.
Let us take $P\left( A \right)$ to be the probability with which $A$ hits the target, therefore, according to the question we have:
$P\left( A \right)=\dfrac{1}{2}$.
Similarly, we will let $P\left( B \right)$ be the probability with which $B$ hits the target, therefore, according to the question we have:
$P\left( B \right)=\dfrac{1}{3}$
And again we will let $P\left( C \right)$ be the probability with which $C$ hits the target, therefore, according to the question we have:
$P\left( C \right)=\dfrac{1}{4}$
Finally, let us take $P\left( D \right)$ to be the probability with which $D$ hits the target, therefore, according to the question we have:
$P\left( D \right)=\dfrac{1}{8}$
So we have all the probabilities as: $P\left( A \right)=\dfrac{1}{2}$ , $P\left( B \right)=\dfrac{1}{3}$ , $P\left( C \right)=\dfrac{1}{4}$, $P\left( D \right)=\dfrac{1}{8}$
Now we know that if probability of an event to happen is $P\left( t \right)$ , then probability of the event to not happen will be $P\left( \overline{t} \right)=1-P\left( t \right)$
Now, we will apply this condition to our obtained probabilities for the persons:
Therefore probability of $A$ not hitting the target will be: $P\left( \overline{A} \right)=1-P\left( A \right)\Rightarrow P\left( \overline{A} \right)=\left( 1-\dfrac{1}{2} \right)\Rightarrow P\left( \overline{A} \right)=\dfrac{1}{2}$
Similarly we will find out the probabilities of other three persons:
$\begin{align}
  & P\left( \overline{B} \right)=1-P\left( B \right)\Rightarrow P\left( \overline{B} \right)=\left( 1-\dfrac{1}{3} \right)\Rightarrow P\left( \overline{B} \right)=\dfrac{2}{3} \\
 & P\left( \overline{C} \right)=1-P\left( C \right)\Rightarrow P\left( \overline{C} \right)=\left( 1-\dfrac{1}{4} \right)\Rightarrow P\left( \overline{C} \right)=\dfrac{3}{4} \\
 & P\left( \overline{D} \right)=1-P\left( D \right)\Rightarrow P\left( \overline{D} \right)=\left( 1-\dfrac{1}{8} \right)\Rightarrow P\left( \overline{D} \right)=\dfrac{7}{8} \\
\end{align}$
We already know that if $M$ and $N$ are two events and their probability of happening is $P\left( M \right)\text{ and }P\left( N \right)$ ,then condition of independent events is: $P\left( M\bigcap N \right)=P\left( M \right)\times P\left( N \right)$ .
Therefore the probability of not hitting the target of all the four persons is:
\[P\left( \overline{A}\bigcap \overline{B}\bigcap \overline{C}\bigcap \overline{D} \right)=P\left( \overline{A} \right)\times P\left( \overline{B} \right)\times P\left( \overline{C} \right)\times P\left( \overline{D} \right)\]
Putting the values in RHS: \[P\left( \overline{A}\bigcap \overline{B}\bigcap \overline{C}\bigcap \overline{D} \right)=\left( \dfrac{1}{2} \right)\times \left( \dfrac{2}{3} \right)\times \left( \dfrac{3}{4} \right)\times \left( \dfrac{7}{8} \right)\Rightarrow \dfrac{7}{32}\text{ }.......\text{Equation 1}\text{.}\]
Therefore, the required probability of hitting the target = $1-$ probability of not hitting the target:
Putting the value from equation 1 $\Rightarrow 1-\dfrac{7}{32}=\dfrac{32-7}{32}=\dfrac{25}{32}$ .
Hence, the probability that the target would be hit is $\dfrac{25}{32}$ .
Therefore Option C is correct.

Note: In the subjective questions related to probabilities, try to describe the steps clearly as there is not much calculation or formulas, the examiner may find it difficult to understand what you are trying to do in the solutions. Also probability of any event will always be equal to or more than 0 but less than 1.