Question

A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of cars and plant Y manufactures 30% of cars. 80% of cars at plant X and 90% of cars at plant Y are rated as standard quality. A car is chosen at random and is found to be standard quality. The probability that it has come from plant X is

• A. \(\frac{56}{73}\)
• B. \(\frac{56}{84}\)
• C. \(\frac{56}{83}\)
• D. \(\frac{56}{79}\)

Answer 1

The Correct Option is C

Let the events be defined as follows:
Let \( A_1 \) represent the event that the car came from plant X.
Let \( A_2 \) represent the event that the car came from plant Y.
Let \( B \) represent the event that the car is of standard quality.
We are given the following probabilities:
\( P(A_1) = 0.70 \),
\( P(A_2) = 0.30 \)
\( P(B|A_1) = 0.80 \) (80% of cars from plant X are of standard quality)
\( P(B|A_2) = 0.90 \) (90% of cars from plant Y are of standard quality)
We need to find \( P(A_1|B) \), the probability that the car came from plant X given that it is of standard quality. By Bayes' Theorem: \[ P(A_1|B) = \frac{P(B|A_1) \cdot P(A_1)}{P(B)} \] We first need to calculate \( P(B) \), the total probability that the car is of standard quality: \[ P(B) = P(B|A_1) \cdot P(A_1) + P(B|A_2) \cdot P(A_2) \] Substitute the given values: \[ P(B) = (0.80 \cdot 0.70) + (0.90 \cdot 0.30) = 0.56 + 0.27 = 0.83 \] Now, using Bayes' Theorem: \[ P(A_1|B) = \frac{(0.80 \cdot 0.70)}{0.83} = \frac{0.56}{0.83} \] Thus, the probability that the car came from plant X is \( \frac{56}{83} \).

The correct answer is (C) : \(\frac{56}{83}\).