In answering a question on a multiple choice | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $E_{1}$ and $E_{2}$ be the respective events that the student knows the answer and he guesses the answer.Let $A$ be the event that the answer

Vedclass · Accepted Answer

$\frac{12}{13}$

Vedclass · Answer

(A) Let $E_{1}$ and $E_{2}$ be the respective events that the student knows the answer and he guesses the answer.Let $A$ be the event that the answer is correct.Given probabilities are $P(E_{1}) = \frac{3}{4}$ and $P(E_{2}) = \frac{1}{4}$.The probability that the student answers correctly given that he knows the answer is $P(A|E_{1}) = 1$.The probability that the student answers correctly given that he guesses is $P(A|E_{2}) = \frac{1}{4}$.We need to find $P(E_{1}|A)$. By Bayes' theorem:$P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$$P(E_{1}|A) = \frac{\frac{3}{4} 	imes 1}{(\frac{3}{4} 	imes 1) + (\frac{1}{4} 	imes \frac{1}{4})}$$P(E_{1}|A) = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} = \frac{\frac{3}{4}}{\frac{12+1}{16}} = \frac{\frac{3}{4}}{\frac{13}{16}}$$P(E_{1}|A) = \frac{3}{4} 	imes \frac{16}{13} = \frac{12}{13}$.

Similar Questions

$A$ bag $P$ contains $5$ white marbles and $3$ black marbles. Four marbles are drawn at random from $P$ and are put in an empty bag $Q$. If a marble drawn at random from $Q$ is found to be black,then the probability that all the three black marbles in $P$ were transferred to the bag $Q$ is:

From a certain population, the probability of choosing a colour blind man is $\frac{1}{20}$ and that of a colour blind woman is $\frac{1}{10}$. If a randomly chosen person is found to be colour blind, then the probability that the person is a man is

$A$ bag contains $6$ balls. If three balls are drawn at a time and all of them are found to be green,then the probability that exactly $5$ of the balls in the bag are green is:

$A$ bag contains $19$ unbiased coins and one coin with head on both sides. One coin is drawn at random and tossed,and a head turns up. If the probability that the drawn coin was unbiased is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $n^2 - m^2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module