There are two bags I and IL Bag I contains 3 white and 4 black balls, and bag II contains 5 white and 6 black balls. One ball is drawn at random from one of the bags and is found to be white. Find the probability that it was drawn from bag I.

Answer

To solve the problem, we will use Bayes' Theorem. We need to find the probability that the ball was drawn from Bag I given that the ball drawn is white.

### Step 1: Define Events
Let:
- \( A_1 \): The event that the ball is drawn from Bag I.
- \( A_2 \): The event that the ball is drawn from Bag II.
- \( B \): The event that the ball drawn is white.

### Step 2: Calculate Prior Probabilities
Since there are two bags, the probability of choosing either bag is:
- \( P(A_1) = \frac{1}{2} \)
- \( P(A_2) = \frac{1}{2} \)

### Step 3: Calculate Conditional Probabilities
Next, we need to find the probability of drawing a white ball from each bag:
- For Bag I: There are 3 white balls and 4 black balls, so the total number of balls is \( 3 + 4 = 7 \).
\[ P(B | A_1) = \frac{3}{7} \]

- For Bag II: There are 5 white balls and 6 black balls, so the total number of balls is \( 5 + 6 = 11 \).
\[ P(B | A_2) = \frac{5}{11} \]

### Step 4: Calculate Total Probability of Drawing a White Ball
Using the law of total probability, we can find \( P(B) \):
\[ P(B) = P(B | A_1) \cdot P(A_1) + P(B | A_2) \cdot P(A_2) \]
Substituting the values we calculated:
\[ P(B) = \left(\frac{3}{7} \cdot \frac{1}{2}\right) + \left(\frac{5}{11} \cdot \frac{1}{2}\right) \]
\[ P(B) = \frac{3}{14} + \frac{5}{22} \]

To add these fractions, we need a common denominator. The least common multiple of 14 and 22 is 154. Thus, we convert the fractions:
\[ P(B) = \frac{3 \cdot 11}{154} + \frac{5 \cdot 7}{154} = \frac{33 + 35}{154} = \frac{68}{154} \]

### Step 5: Apply Bayes' Theorem
Now we can apply Bayes' Theorem to find \( P(A_1 | B) \):
\[ P(A_1 | B) = \frac{P(B | A_1) \cdot P(A_1)}{P(B)} \]
Substituting the known values:
\[ P(A_1 | B) = \frac{\left(\frac{3}{7}\right) \cdot \left(\frac{1}{2}\right)}{\frac{68}{154}} \]
This simplifies to:
\[ P(A_1 | B) = \frac{\frac{3}{14}}{\frac{68}{154}} = \frac{3}{14} \cdot \frac{154}{68} = \frac{3 \cdot 11}{68} = \frac{33}{68} \]

### Final Answer
Thus, the probability that the ball was drawn from Bag I given that it is white is:
\[ \boxed{\frac{33}{68}} \]