Question

In answering a question on a multiple choice test , a student either knows the answer or guesses. Let $\dfrac{3}{4}$ be the probability and $\dfrac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\dfrac{1}{4}$.What is the probability that the student knows the answer given that he answered correctly?

Answer 1

Hint : First calculate the conditional probability$\;\;P\left( {A|{E_1}} \right)$. Then calculate the conditional probability$\;\;P\left( {A|{E_2}} \right)$.
At last use Bayes theorem $P({E_1}|A) = \dfrac{{P({E_1}).P(A|{E_1})}}{{P({E_1})P(A|{E_1}) + P({E_2}).P(A|{E_2})}}$ for the required probability.

Complete step-by-step answer :
Lets say student knows the answer so event is = E1
Student guesses the answer so event is = E2
Let's assume answer is correct so event is = A
$P\left( {{E_1}} \right) = \;\dfrac{3}{4}$
$P\left( {{E_2}} \right) = \;\dfrac{1}{4}$
Now student knows the answer so probability of getting answer correctly
$\;\;P\left( {A|{E_1}} \right) = 1$
Student guesses the answer is $\dfrac{1}{4}$ so probability of student answer it correctly
$P\left( {A|{E_2}} \right) = \dfrac{1}{4}$
Probability student knows the answer is $P\left( {{E_1}|A} \right)$
Now we will apply Bayes theorem which states that
$\Rightarrow P({E_1}|A) = \dfrac{{P({E_1}).P(A|{E_1})}}{{P({E_1})P(A|{E_1}) + P({E_2}).P(A|{E_2})}}$
$\Rightarrow P({E_1}|A)$ = $\dfrac{{\dfrac{3}{4}.1}}{{\dfrac{3}{4}.1 + \dfrac{1}{4}\dfrac{1}{4}}} = \dfrac{{\dfrac{3}{4}}}{{\dfrac{{13}}{{16}}}} = \dfrac{{3 \times 16}}{{4 \times 13}} = 0.92$

Note : Revise the Bay’s theorem of probability for the question and also revise some definitions of probability like conditional probability and the event and also remember that the probability is the possibility of getting an outcome.