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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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}}$ = $\large\frac{9}{9+4} = \frac{9}{13}$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...