Answer to Question #125369 in Statistics and Probability for Rayyan

Question

In a university, 80% of the students are studying in various engineering disciplines and the rest are pursuing non-engineering degrees. 60% of the engineering students use IEEExplore for research and 45% of non-engineering students avail the IEEExplore service. A student is randomly selected from the university and asked whether he uses IEEExplore service or not. If the student turns out to be the user of the service, what is the probability that the randomly selected student is an engineering student? What is the probability that the randomly selected student is a non-engineering student?

Answer 1

Solution.

event $IEE$ - student is a user of the IEEExplore service;

event $nIEE$ - student is a user of the IEEExplore service;

event $EN$ - student is an engineering student;

event $nEN$ - student isn't an engineering student;

Bayes' formula says that $P(EN|IEE) = \dfrac{P(IEE|EN)\cdot P(EN)}{P(IEE)}\\[0.2cm]$

and $P(nEN|IEE) = \dfrac{P(IEE|nEN)\cdot P(nEN)}{P(IEE)}\\[0.2cm]$

Total probability formula says

that $P(IEE) = P(IEE|EN)\cdot P(EN) +P(IEE|nEN)\cdot P(nEN)$

Given probabilities are

$P(EN) = 0.8, \quad p(nEN)=0.2\\[0.2cm]$

$P(IEE|EN) = 0.6, \quad P(IEE|nEN) = 0.45\\[0.2cm]$

So $P(IEE) = 0.6\cdot0.8+0.45\cdot0.2=0.57$

The probability that the randomly selected student who is a user of the service is an engineering student is

$P(EN|IEE) = \frac{0.6\cdot0.8}{0.57} = \frac{16}{19} \approx 0.8421\\[0.2cm]$

The probability that the randomly selected student who is a user of the service is a non-engineering student is

$P(nEN|IEE) = \frac{0.45\cdot0.2}{0.57} = \frac{3}{19} \approx 0.1579\\[0.2cm]$

Answer

1. 16/19
2. 3/19