Question

A school offers students the choice of three modes for attending classes:

• Mode A: Offline (in-person) – 40% of students 
• Mode B: Online (live virtual classes) – 35% of students 
• Mode C: Recorded lectures – 25% of students 

After a feedback survey:

• 20% of students from Mode A reported the class as “Excellent”.
• 30% from Mode B rated it as “Excellent”.
• 50% from Mode C rated it as “Excellent”.

A student is selected at random from the entire group, and it is found that they rated the class as “Excellent”.

(a) Represent the data in terms of probability. Define the events clearly.

(b) Using Bayes’ Theorem, find the probability that the student attended the Recorded lectures (Mode C), given that they rated the class as “Excellent”.

(c) Interpret your result. Which mode has the highest likelihood of being chosen if a student says “Excellent”?

Answer 1

(a) Let the events be:

A: The student chose Mode A (Offline – in person).

B: The student chose Mode B (Online – live virtual classes).

C: The student chose Mode C (Recorded lectures).

E: The student rated the class as “Excellent”.

From the information given:

P(A) = 0.40, P(B) = 0.35, P(C) = 0.25

`P(E/A)` = 0.20, `P(E/B)` = 0.30, `P(E/C)` = 0.50

(b) Find `P(C/E)` using Bayes’ theorem:

= `(P(C) xx P(E/C))/(P(A) xx P(E/A) + P(B) xx P(E/B) + P(C) xx P(E/C))`

= `(0.25 xx 0.50)/((0.40 xx 0.20) + (0.35 xx 0.30) + (0.25 xx 0.50))`

= `(0.25 xx 0.50)/((0.08) + (0.105) + (0.125))`

= `0.125/0.31`

= 0.403

(c) There’s about a 40.3% chance that a student who rated the class as “Excellent” attended Recorded lectures.

Numerator of `P(A/E) = P(A) xx P(E/A)`

= 0.40 × 0.20

= 0.08

Numerator of `P(B/E) = P(B) xx P(E/B)`

= 0.35 × 0.30

= 0.105

By checking the numerators of `P(C/E), P(B/E)` and `P(A/E)` we observed 0.125 > 0.105 > 0.08. Therefore, recorded lectures (Mode C) have the highest likelihood of being the chosen mode among students who gave an excellent rating.