Applied Mathematics for Electrical Engineering (3130908)

Question-4b

BE | Semester-3 Winter-2019 | 26-11-2019

Q4) (b)

Bayes’ theorem: Let

B_{1}, B_{2}, B_{3}, . . ., B_{n}

be an n-mutually exclusive and exhaustive events of a sample space S and let A be any event such that

P (A) \neq 0

, then

P (B_{i} / A) = \frac{P (B_{i}) \cdot P (A / B_{i})}{P (B_{1}) \cdot P (A / B_{1}) + P (B_{2}) \cdot P (A / B_{2}) + P (B_{3}) \cdot P (A / B_{3}) + . . . + P (B_{n}) \cdot P (A / B_{n})}

Let, D: Bolt is defective

P (A) = 25 % = 0.25

P (B) = 35 % = 0.35

P (C) = 40 % = 0.4

Given that,

P (D / A) = 5 % = 0.05

P (D / B) = 4 % = 0.04

P (D / C) = 2 % = 0.02

Then by Baye’s theorem,

P (A / D) = \frac{P (A) \cdot P (D / A)}{P (A) \cdot P (D / A) + P (B) \cdot P (D / B) + P (C) \cdot P (D / C)}

P (A / D) = \frac{(0.25) (0.05)}{(0.25) (0.05) + (0.35) (0.04) + (0.40) (0.02)}

P (A / D) = 0.3623

P (B / D) = \frac{P (B) \cdot P (D / B)}{P (A) \cdot P (D / A) + P (B) \cdot P (D / B) + P (C) \cdot P (D / C)}

P (A / D) = \frac{(0.35) (0.04)}{(0.25) (0.05) + (0.35) (0.04) + (0.40) (0.02)}

P (B / D) = 0.4058

P (C / D) = \frac{P (C) \cdot P (D / C)}{P (A) \cdot P (D / A) + P (B) \cdot P (D / B) + P (C) \cdot P (D / C)}

P (A / D) = \frac{(0.40) (0.02)}{(0.25) (0.05) + (0.35) (0.04) + (0.40) (0.02)}

P (C / D) = 0.2319