Step-by-Step Solution

Topic: Total Probability & Bayes’ Theorem

Probability Tree Diagram

The tree splits into 3 categories (A, B, C) with their respective failure rates (0.002, 0.02, 0.20).

1. Define Events & Probabilities

Total Students = 60.

Category Probabilities:

• • P(A) = 6/60
• • P(B) = 26/60
• • P(C) = (60 – 6 – 26)/60 = 28/60

Let E be the event “unable to get good marks”.

• • P(E|A) = 0.002
• • P(E|B) = 0.02
• • P(E|C) = 0.20

(I) TOTAL PROBABILITY P(E)

Using the Law of Total Probability:

Substitute the values (keeping 1/60 common factor):

P(E) = (1/60) [ 6(0.002) + 26(0.02) + 28(0.20) ]

P(E) = (1/60) [ 0.012 + 0.52 + 5.6 ]

P(E) = 6.132 / 60

(II) PROBABILITY (NOT CATEGORY A)

We need to find P(A’|E) = 1 – P(A|E).

First, find P(A|E) using Bayes’ Theorem:

Numerator: P(A)P(E|A) = (6/60)(0.002) = 0.012 / 60

Denominator: P(E) = 6.132 / 60

P(A|E) = 0.012 / 6.132 = 12 / 6132 = 1 / 511

Now, find the probability of NOT being category A:

P(Not A | E) = 1 – (1/511)