(for explantion of Baye's rule ) three boxes contain 6 red , 4 black , 5 red , 5 black and 4 red , 6 black balls repectively one of the box is selected at random and a ball is drawn from it . If the ball drawn is red , find probability that it is drawn from the first bag .

Answer

To solve the problem using Bayes' theorem, we need to find the probability that the red ball drawn came from the first box. Let's denote the events as follows:

- Let A be the event that the first box is chosen.
- Let B be the event that the second box is chosen.
- Let C be the event that the third box is chosen.
- Let D be the event that a red ball is drawn.

We want to find \( P(A|D) \), the probability that the first box was chosen given that a red ball was drawn.

### Step 1: Calculate the prior probabilities
Since one of the boxes is selected at random, the prior probabilities for choosing each box are:
- \( P(A) = \frac{1}{3} \)
- \( P(B) = \frac{1}{3} \)
- \( P(C) = \frac{1}{3} \)

### Step 2: Calculate the probability of drawing a red ball from each box
Now we need to find the probability of drawing a red ball from each box:
- For Box A (6 red, 4 black):
\[ P(D|A) = \frac{6}{10} = \frac{3}{5} \]
- For Box B (5 red, 5 black):
\[ P(D|B) = \frac{5}{10} = \frac{1}{2} \]
- For Box C (4 red, 6 black):
\[ P(D|C) = \frac{4}{10} = \frac{2}{5} \]

### Step 3: Calculate the total probability of drawing a red ball
Using the law of total probability, we calculate \( P(D) \):
\[ P(D) = P(A) \cdot P(D|A) + P(B) \cdot P(D|B) + P(C) \cdot P(D|C) \]
Substituting the values:
\[ P(D) = \left(\frac{1}{3} \cdot \frac{3}{5}\right) + \left(\frac{1}{3} \cdot \frac{1}{2}\right) + \left(\frac{1}{3} \cdot \frac{2}{5}\right) \]
Calculating each term:
\[ = \frac{1}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{2}{5} \]
\[ = \frac{1}{5} + \frac{1}{6} + \frac{2}{15} \]
Finding a common denominator (30):
\[ = \frac{6}{30} + \frac{5}{30} + \frac{4}{30} = \frac{15}{30} = \frac{1}{2} \]

### Step 4: Apply Bayes' theorem
Now we can apply Bayes' theorem:
\[ P(A|D) = \frac{P(D|A) \cdot P(A)}{P(D)} \]
Substituting the values:
\[ P(A|D) = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{1}{3}\right)}{\frac{1}{2}} \]
Calculating:
\[ = \frac{\frac{3}{15}}{\frac{1}{2}} = \frac{3}{15} \cdot \frac{2}{1} = \frac{6}{15} = \frac{2}{5} \]

### Final Answer
Thus, the probability that the red ball drawn came from the first box is:
\[ \boxed{\frac{2}{5}} \]