Answer :

Let E denotes the event that student ‘A’ solves the problem correctly.

**∴** **P(E) = 1/3**

Similarly, if we denote the event of ’B’ solving the problem correctly with F

We can write that:

**P(F) = 1/4**

**Note:** Observe that both the events are independent.

∴ Probability that both the students solve the question correctly can be represented as-

**P (E** **∩** **F) =** **= P(E _{1}) {say}** {we can multiply because events are independent}

∴ Probability that both the students could not solve the question correctly can be represented as-

**P(E’** **∩** **F’) =** **= P(E _{2}) {say}**

Now we are given with some more data and they can be interpreted as:

Given: probability of making a common error and both getting same answer.

**Note:** If they are making an error, we can be sure that answer coming out is wrong.

Let S denote the event of getting same answer.

∴ above situation can be represented using conditional probability.

**P(S|E _{2}) = 1/20**

And if their answer is correct obviously, they will get same answer.

**∴** **P(S|E _{1}) = 1**

We need to find the probability of getting a correct answer if they committed a common error and got the same answer.

Mathematically,

i.e **P(E _{1}|S) = ?**

By observing our requirement and availability of equations, we can make guess that Bayes theorem is going to help us.

**∴** **Using Bayes theorem, we get-**

**P(E _{1}|S) =**

Using the values from above –

**P(E _{1}|S) =**

Clearly our answer matches with option D.

**∴** **Option (D) is the only correct choice.**

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