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Home » Of the students in a college, it is known that reside in hostel and are day scholars (not residing in hostel). Previous year results report that of all students who reside in hostel attain A grade and of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?
Of the students in a college, it is known that 60 \% reside in hostel and 40 \% are day scholars (not residing in hostel). Previous year results report that 30 \% of all students who reside in hostel attain A grade and 20 \% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?
CBSE | Class 12 | Maths | NCERT | Probability
Of the students in a college, it is known that 60 \% reside in hostel and 40 \% are day scholars (not residing in hostel). Previous year results report that 30 \% of all students who reside in hostel attain A grade and 20 \% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Solution:

Let \mathrm{E}_{1} represent the event that the student is a hostler, \mathrm{E}_{2} represent the event that the student is a day scholar, and A represent the event that the student receives an A grade.

Then \mathrm{P}\left(\mathrm{E}_{1}\right)=60 \%=\frac{60}{100}=0.6

And \mathrm{P}\left(\mathrm{E}_{2}\right)=40 \%=\frac{40}{100}=0.4

Also \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}( students who attain A grade reside in hostel) =30 \%=0.3

And P\left(A \mid E_{2}\right)=P (students who attain A grade is day scholar) = 20 \%=0.2

The likelihood that students who live in hostels will receive an A grade is now \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right).

By using Bayes’ theorem, we have:

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}

We can acquire the values by substituting them.

=\frac{0.6 \times 0.3}{0.6 \times 0.3+0.4 \times 0.2}

=\frac{0.18}{0.18+0.08}

=\frac{0.18}{0.26}=\frac{18}{26}=\frac{9}{13}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{9}{13}

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