There are 3 white, 6 black balls in one bag and 6 white, 3 black balls in second bag. A bag is selected at random and a ball is drawn from it. Find the probability that the ball drawn is black.

Explanation

To solve the problem step by step, we will calculate the probability that a ball drawn from one of the two bags is black.

### Step 1: Define the events
Let:
- \( E_1 \): Event that Bag 1 is selected.
- \( E_2 \): Event that Bag 2 is selected.
- \( A \): Event that the ball drawn is black.

### Step 2: Determine the total number of balls in each bag
- **Bag 1**: 3 white balls + 6 black balls = 9 total balls.
- **Bag 2**: 6 white balls + 3 black balls = 9 total balls.

### Step 3: Calculate the probabilities of selecting each bag
Since there are two bags and selecting a bag is equally likely:
- \( P(E_1) = P(E_2) = \frac{1}{2} \)

### Step 4: Calculate the probability of drawing a black ball from Bag 1
Given that Bag 1 is selected, the probability of drawing a black ball is:
\[ P(A | E_1) = \frac{\text{Number of black balls in Bag 1}}{\text{Total number of balls in Bag 1}} = \frac{6}{9} = \frac{2}{3} \]

### Step 5: Calculate the probability of drawing a black ball from Bag 2
Given that Bag 2 is selected, the probability of drawing a black ball is:
\[ P(A | E_2) = \frac{\text{Number of black balls in Bag 2}}{\text{Total number of balls in Bag 2}} = \frac{3}{9} = \frac{1}{3} \]

### Step 6: Apply the Total Probability Theorem
Using the Total Probability Theorem, we can find the overall probability of drawing a black ball:
\[ P(A) = P(E_1) \cdot P(A | E_1) + P(E_2) \cdot P(A | E_2) \]
Substituting the values we calculated:
\[ P(A) = \left(\frac{1}{2} \cdot \frac{2}{3}\right) + \left(\frac{1}{2} \cdot \frac{1}{3}\right) \]
\[ P(A) = \frac{1}{3} + \frac{1}{6} \]
To add these fractions, we need a common denominator:
\[ P(A) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
The probability that the ball drawn is black is:
\[ \boxed{\frac{1}{2}} \]