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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?

$\begin{array}{1 1} \large\frac{9}{13} \\ \large\frac{4}{13} \\ \large\frac{2}{13} \\ \large\frac{6}{13} \end{array} $

1 Answer

  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Let $E_1$: students residing in the hostel $\rightarrow$ P($E_1$) = 60% = $\large\frac{60}{100} = \frac{3}{5}$
Let $E_2$: day scholars $\rightarrow$ P($E_2$) = 40% = $\large\frac{40}{100} = \frac{2}{5}$
30% of hostel students get an A grade $\rightarrow$ P (E|$E_1$) = 30% = $\large\frac{30}{100} = \frac{3}{10}$
20% of day scholars get an A grade $\rightarrow$ P (E|$E_2$) = 20% = $\large\frac{20}{100} = \frac{1}{5}$
We need to find the probability that a student who is chosen from random that has an A grade is from the hostel.
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
Using Baye's theorem, P ($E_1$|A) =$ \Large\frac{\frac{3}{5}.\frac{3}{10}}{\frac{3}{5}.\frac{3}{10}+\frac{2}{5}.\frac{1}{5}}$ = $\large\frac{9}{9+4} = \frac{9}{13}$
answered Jun 19, 2013 by balaji.thirumalai

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