Question

A and B are two weak students in mathematics and their chances in solving the problem correctly are \[\dfrac{1}{8}\] and \[\dfrac{1}{12}\] respectively. They are given a problem and obtain the same answer. If the probability of a common mistake is \[\dfrac{x}{1001}\], then the probability that the answer was correct.

Answer 1

Hint: We are given with probabilities of A and B in solving the problem correctly. So firstly we must be multiplying the probabilities in order to obtain the total probability of the obtained answer to be correct. And then upon multiplying the probability of the correct answer with the probability of a common mistake, we obtain the probability of the correct answer.

Complete step by step answer:
Now let us learn about the probabilities. The probability of any event lies between \[0\] and \[1\]. It means that \[0\] indicates the impossibility of the event to occur and \[1\] indicates the certainty of the event. There are three types of probabilities. They are : theoretical probability, experimental probability and axiomatic probability.
Now let us find out the probability of the obtained answer to be correct.
Let us denote the event of \[\text{A and B}\] solve correctly as \[{{\text{E}}_{1}}\] and let us denote the event of \[\text{A and B}\] solve incorrectly as \[{{\text{E}}_{2}}\].
And let \[K\] be the event of obtaining the same answer.
Now let us find the probability of obtaining an answer correctly.
\[P\left( {{E}_{1}} \right)=\dfrac{1}{8}\times \dfrac{1}{12}=\dfrac{1}{96}\]
The probability of obtaining the same answer as well as the correct answer will be \[1\].
\[P\left( \dfrac{K}{{{E}_{1}}} \right)=1\]
The probability of obtaining answer incorrectly is \[P\left( {{E}_{2}} \right)=\dfrac{7}{8}\times \dfrac{11}{12}=\dfrac{77}{96}\]
Now the probability of obtaining common answer as well as incorrect answer would be,
\[P\left( \dfrac{K}{{{E}_{2}}} \right)=\dfrac{1}{1001}\]
Now, the probability of correct answer would be
\[\begin{align}
  & P\left( \dfrac{{{E}_{1}}}{K} \right)=P\left( {{E}_{1}} \right)\times P\left( \dfrac{K}{{{E}_{1}}} \right) \\
 & \Rightarrow \dfrac{1}{96}\times 1=\dfrac{1}{96} \\
\end{align}\]

\[\therefore \] The probability that the answer was correct is \[\dfrac{1}{96}\].

Note: We can see that the event with higher probability is likely to occur. The sum of the probabilities should be \[1\]. The probability of the incorrect answer is found by \[1-\] the probability of the event occurs. This is applicable to all of the cases in finding the probability of an event that does not occur.