Question

# There are three categories of students in a class of 60 students: A : Very hardworking ; B : Regular but not so hardworking; C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C. It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.

Solution

## Let E denote the event that the student could not get good marks in the examination. Also, A : the event that student is very hardworking B : the event that student is regular but not so hardworking C : the event that student is careless and irregular ∴ P(A) = $\frac{10}{60}$, P(B) = $\frac{30}{60}$ and P(C) = $\frac{20}{60}$ Also,  $\mathrm{P}\left(\frac{E}{A}\right)$ = Pobability that the student of catagory A could not get good marks in the examination = 0.002 $\mathrm{P}\left(\frac{E}{B}\right)$ = Pobability that the student of catagory B could not get good marks in the examination = 0.02 $\mathrm{P}\left(\frac{E}{C}\right)$ = Pobability that the student of catagory C could not get good marks in the examination = 0.2 ∴ Required probability = $\mathrm{P}\left(\frac{C}{E}\right)=\frac{\mathrm{P}\left(C\right)\mathrm{P}\left(\frac{E}{C}\right)}{\mathrm{P}\left(A\right)\mathrm{P}\left(\frac{E}{A}\right)+\mathrm{P}\left(B\right)\mathrm{P}\left(\frac{E}{B}\right)+\mathrm{P}\left(C\right)\mathrm{P}\left(\frac{E}{C}\right)}=\frac{\frac{20}{60}×0.2}{\frac{10}{60}×0.002+\frac{30}{60}×0.02+\frac{20}{60}×0.2}=\frac{4}{4.62}=\frac{400}{462}=\frac{200}{231}$MathematicsRD Sharma XII Vol 2 (2019)All

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More

People also searched for
View More