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) = 1/3 x 1/4 = 1/2 = P(E1) {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’) = 2/3 x 3/4 = 1/2 = P(E2) {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|E2) = 1/20
And if their answer is correct obviously, they will get same answer.
∴ P(S|E1) = 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(E1|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-
Clearly our answer matches with option D.
∴ Option (D) is the only correct choice.
