Q. 925.0( 1 Vote )

# A and B are two students. Their chances of solving a problem correctly are 1/3 and 1/4, respectively. If the probability of their making a common error is, 1/20 and they obtain the same answer, then the probability of their answer to be correct isA. B. C. D. Correct

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(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’) = = 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-

P(E1|S) = Using the values from above –

P(E1|S) = Clearly our answer matches with option D.

Option (D) is the only correct choice.

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