# In a test an examine either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he make a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it, is 1/8. Find the probability that he knew the answer to the question given that he correctly answered it.

Jitender Pal
7 years ago
Hello Student,
KEY CONCEPT :
Baye’s theorem: E1, E2, E­3, . . . . . . . . . . . . . . ,En are mutually exclusive and exhaustive events and E is an event which takes place in conjunction with any one of E1 then the probability of the event E1 happening when the event E has taken place is given by
P (E1 | E) = $\frac{P(E_{1})P(E|E_{1})}{\sum_{i=1}^{n}P(E_1)P(E|E_1)}$
Let us define the events :
A1 ≡ the examinee guesses the answer,
A2 ≡ the examinee copies the answer
A3 ≡ the examinee knows the answer,
A ≡ the examinee answers correctly
ATQ, P (A­1) = 1/3; P (A2) = 1/6
As any one happens out of A1, A2, A3, these are mutually exclusive and exhaustive events.
∴ P (A1) + P (A­2) + P (A3) = 1
⇒ P (A3) = 1 – 1/3 – 1/6 = 6 - 2 – 1/6 = 3/6 = 1/2
Also we have, P(A| A1) = 1/4
[∵ out of 4 choices only one is correct.] P (A| A2) = 1/8
(given) P (A| A3) = 1
[If examinee knows the ans., it is correct. i.e. true event]
To find P (A3| A). By Baye’s thm. P (A3| A)
= P (A3| A) P (A3)/P (A| A1) P(A) + P (A| A2) P(A) + P (A| A3) P(A3)
= $\frac{1\frac{1}{2}}{\frac{1}{4}.\frac{1}{3}+\frac{1}{8}.\frac{1}{6}+\frac{1.1}{2}}$= 1/2/29/48 = 1/2 x 18/29 = 24/29.

Thanks
Jitender Pal