A & B are sharp shooters whose probabilities of hitting a target are \[\dfrac{9}{10}\] & \[\dfrac{14}{15}\] respectively. If it is known that exactly one of them has hit the target, then the probability that it was hit by A is equal to
(a) \[\dfrac{24}{55}\]
(b) \[\dfrac{27}{55}\]
(c) \[\dfrac{9}{23}\]
(d) \[\dfrac{10}{23}\]

Hint: In this question, we know the probability of success for each of the shooters to hit the target. Now, for A to hit the target B should always miss it then find the probability of B missing the target using the formula \[P\left( S \right)+P\left( F \right)=1\]. Now, use Baye's theorem for A to hit the target only when one of them hits using the formula \[P\left( {{E}_{i}}/A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}}\] and then simplify further to get the result.

Complete step-by-step answer:
BAYES THEOREM:
Let S be the sample space and let there be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with any of the events then the probability of occurrence of the event
\[P\left( {{E}_{i}}/A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( A/{{E}_{i}} \right)}}\]
As we already know that the sum of probabilities of two compliment events is 1
\[P\left( S \right)+P\left( F \right)=1\]
Now, given that the probability of success of B is
\[P\left( S \right)=\dfrac{14}{15}\]
Now, on substituting this in the above formula we get,
\[\Rightarrow \dfrac{14}{15}+P\left( F \right)=1\]
Now, this can be further written as
\[\Rightarrow P\left( F \right)=1-\dfrac{14}{15}\]
Now, on further simplification we get the probability of failure of B as,
\[\Rightarrow P\left( F \right)=\dfrac{1}{15}\]
Let us also find the probability of failure of A
\[\Rightarrow P\left( F \right)=1-\dfrac{9}{10}\]
Now, the probability of failure of A is given by
\[\Rightarrow P\left( F \right)=\dfrac{1}{10}\]
Let us assume that event 1 as only A hits the target, event 2 as only B hits the target and E as exactly one hits the target
Now, from the Baye's theorem we have
\[P\left( {{E}_{1}}/E \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( E/{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)P\left( E/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( E/{{E}_{2}} \right)}\]
Now, on substituting the respective values we get,
\[\Rightarrow P\left( {{E}_{1}}/E \right)=\dfrac{\dfrac{9}{10}\times \dfrac{1}{15}}{\dfrac{9}{10}\times \dfrac{1}{15}+\dfrac{1}{10}\times \dfrac{14}{15}}\]
Now, this can be further written in the simplified form as
\[\Rightarrow P\left( {{E}_{1}}/E \right)=\dfrac{\dfrac{9}{150}}{\dfrac{9}{150}+\dfrac{14}{150}}\]
Now, on further cancelling the common terms and simplifying further we get,
\[\Rightarrow P\left( {{E}_{1}}/E \right)=\dfrac{9}{9+14}\]
Now, on further simplification we get,
\[\therefore P\left( {{E}_{1}}/E \right)=\dfrac{9}{23}\]
So, the correct answer is “Option C”.

Note: Instead of considering the Bayes theorem if we use the Bernoulli trial to simplify this then the result will be completely incorrect because it is mentioned that only one of them hits the target and that target hit should be by A. So, if we consider that A should hit the target neglecting that only one hits then the answer will be completely incorrect.It is important to note that while substituting and simplifying we need to use the respective values because altering them changes the result completely. It is also to be noted that for A to hit B should not and same the other way.