Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In an entrance test, there are multiple choice questions. There are four possible answers in each question, of which one is correct. The probability that a student knows the answer to a question is \[\dfrac{9}{{10}}\]. If he gets the correct answer to a question, then what is the probability that he was guessing?
(a). \[\dfrac{{37}}{{40}}\]
(b). \[\dfrac{1}{{37}}\]
(c). \[\dfrac{{36}}{{37}}\]
(d). \[\dfrac{1}{9}\]

seo-qna
Last updated date: 27th Mar 2024
Total views: 414.3k
Views today: 6.14k
MVSAT 2024
Answer
VerifiedVerified
414.3k+ views
Hint: Determine the probability that the student doesn’t know that answer and the probability that he chooses the correct answer given he knows the answer and he doesn’t know the answer. Then use Bayes theorem to find the answer.

Complete step by step answer:
Let \[{E_1}\] be the event that the student knows the answer. It is given that this event occurs with the probability \[\dfrac{9}{{10}}\].

\[P({E_1}) = \dfrac{9}{{10}}..........(1)\]

Let \[{E_2}\] be the event that the student doesn’t know the answer. Then, the probability of this event is given as follows:

\[P({E_2}) = 1 - \dfrac{9}{{10}}\]

\[P({E_2}) = \dfrac{1}{{10}}..........(2)\]

Let A be the event that the student gets the correct answer.

The probability of the student getting the correct answer, given that, he knows the correct answer is 1.

\[P(A|{E_1}) = 1..........(3)\]

The probability of the student getting the correct answer, given that he doesn’t know the correct answer is \[\dfrac{1}{4}\], since there are four choices among which one is correct.
\[P(A|{E_2}) = \dfrac{1}{4}..........(4)\]

Let the probability of the student doesn’t know the answer, given that, he gets the correct answer be \[P({E_2}|A)\]. Then, by Bayes theorem, we have:

\[P({E_2}|A) = \dfrac{{P({E_2})P(A|{E_2})}}{{P({E_1})P(A|{E_1}) + P({E_2})P(A|{E_2})}}........(5)\]

Substituting equations (1), (2), (3) and (4) in equation (5), we have:

\[P({E_2}|A) = \dfrac{{\dfrac{1}{{10}}.\dfrac{1}{4}}}{{\dfrac{9}{{10}}.1 + \dfrac{1}{{10}}.\dfrac{1}{4}}}\]

\[P({E_2}|A) = \dfrac{1}{{9 \times 4 + 1}}\]

\[P({E_2}|A) = \dfrac{1}{{36 + 1}}\]

\[P({E_2}|A) = \dfrac{1}{{37}}\]

Hence, the correct answer is option (b).


Note: Guessing here means that the student doesn’t know the correct answer and chooses one option out of the four. If you do not use the Bayes formula correctly, then you will get option (c) as the answer, which is wrong.

Recently Updated Pages