Answer :
Answer: 0.9259
Step-by-step explanation:
Given : The Industrial Statistics class consists of 50 DCIT students and 40 Engineering students.
Total students = 50+40=90
Probability of selecting a DCIT student= [tex]P(D)=\dfrac{50}{90}=\dfrac{5}{9}[/tex]
Probability of selecting a Engineering student= [tex]P(E)=\dfrac{40}{90}=\dfrac{4}{9}[/tex]
Also, The probability of a student failing from DCIT and Engineering are, P(D|F)=0.10 and P(E|F)=0.01 respectively.
Now, One of the students failed the course. Then by Bayes theorem, the probability that it is a student from DCIT will be :-
[tex]P(F|D)=\dfrac{P(D)\cdot P(D|F)}{P(D)\cdot P(D|F)+P(E)\cdot P(E|F)}\\\\=\dfrac{\dfrac{5}{9}\times0.10}{\dfrac{5}{9}\times0.10+\dfrac{4}{9}\times0.01}\\\\=\dfrac{0.0555555555556}{0.06}=0.925925925927\approx0.9259[/tex]
Hence, the probability that it is a student from DCIT = 0.9259