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) : \(\frac{56}{83}\).
Let:
Total probability of standard car:
\( P(S) = P(X) \cdot P(S|X) + P(Y) \cdot P(S|Y) = 0.7 \cdot 0.8 + 0.3 \cdot 0.9 = 0.56 + 0.27 = 0.83 \)
Using Bayes' Theorem:
\( P(X|S) = \frac{P(X) \cdot P(S|X)}{P(S)} = \frac{0.7 \cdot 0.8}{0.83} = \frac{0.56}{0.83} \)
Answer: \( \frac{56}{83} \)
A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant, and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage, and radish to be 25%, 35%, and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.
Given three identical bags each containing 10 balls, whose colours are as follows:
Bag I | 3 Red | 2 Blue | 5 Green |
Bag II | 4 Red | 3 Blue | 3 Green |
Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is: