A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars and plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plantX.

Explanation

To solve the problem, we will use Bayes' Theorem. Let's denote the events as follows:

- Let \( E_1 \) be the event that the car is manufactured by Plant X.
- Let \( E_2 \) be the event that the car is manufactured by Plant Y.
- Let \( E \) be the event that the car is of standard quality.

### Step 1: Define the probabilities

From the problem statement, we have the following probabilities:

- \( P(E_1) = 0.7 \) (70% of cars are from Plant X)
- \( P(E_2) = 0.3 \) (30% of cars are from Plant Y)
- \( P(E|E_1) = 0.8 \) (80% of cars from Plant X are of standard quality)
- \( P(E|E_2) = 0.9 \) (90% of cars from Plant Y are of standard quality)

### Step 2: Calculate the total probability of standard quality cars, \( P(E) \)

Using the law of total probability, we can calculate \( P(E) \):

\[ P(E) = P(E|E_1) \cdot P(E_1) + P(E|E_2) \cdot P(E_2) \]

Substituting the values we have:

\[ P(E) = (0.8 \cdot 0.7) + (0.9 \cdot 0.3) \]

Calculating each term:

\[ P(E) = 0.56 + 0.27 = 0.83 \]

### Step 3: Apply Bayes' Theorem to find \( P(E_1|E) \)

We want to find the probability that the car came from Plant X given that it is of standard quality, \( P(E_1|E) \). According to Bayes' Theorem:

\[ P(E_1|E) = \frac{P(E|E_1) \cdot P(E_1)}{P(E)} \]

Substituting the known values:

\[ P(E_1|E) = \frac{(0.8) \cdot (0.7)}{0.83} \]

Calculating the numerator:

\[ P(E_1|E) = \frac{0.56}{0.83} \]

### Step 4: Final Calculation

Now we perform the division:

\[ P(E_1|E) \approx 0.6747 \]

Thus, the probability that the car has come from Plant X given that it is of standard quality is approximately **0.6747** or **67.47%**.