One purse (X) contains one white and three blue balls, another purse contains 2 white and 4 blue balls, and third purse (Z) contains 3 white and 1 blue balls. If a ball is drawn from one purse selected at random. Find the chance that it is white.

Answer

To solve the problem step by step, we will calculate the probability of drawing a white ball from each purse and then combine these probabilities.

### Step 1: Identify the contents of each purse
- Purse X: 1 white ball and 3 blue balls (Total = 4 balls)
- Purse Y: 2 white balls and 4 blue balls (Total = 6 balls)
- Purse Z: 3 white balls and 1 blue ball (Total = 4 balls)

### Step 2: Calculate the probability of selecting each purse
Since there are three purses and one is selected at random, the probability of selecting any one purse is:
\[ P(\text{selecting any purse}) = \frac{1}{3} \]

### Step 3: Calculate the probability of drawing a white ball from each purse
1. **From Purse X**:
- Probability of drawing a white ball from X:
\[ P(\text{white | X}) = \frac{\text{Number of white balls in X}}{\text{Total number of balls in X}} = \frac{1}{4} \]

2. **From Purse Y**:
- Probability of drawing a white ball from Y:
\[ P(\text{white | Y}) = \frac{\text{Number of white balls in Y}}{\text{Total number of balls in Y}} = \frac{2}{6} = \frac{1}{3} \]

3. **From Purse Z**:
- Probability of drawing a white ball from Z:
\[ P(\text{white | Z}) = \frac{\text{Number of white balls in Z}}{\text{Total number of balls in Z}} = \frac{3}{4} \]

### Step 4: Calculate the total probability of drawing a white ball
Using the law of total probability:
\[ P(\text{white}) = P(\text{selecting X}) \cdot P(\text{white | X}) + P(\text{selecting Y}) \cdot P(\text{white | Y}) + P(\text{selecting Z}) \cdot P(\text{white | Z}) \]

Substituting the values:
\[ P(\text{white}) = \left(\frac{1}{3} \cdot \frac{1}{4}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{3}{4}\right) \]

Calculating each term:
1. From Purse X:
\[ \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \]
2. From Purse Y:
\[ \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \]
3. From Purse Z:
\[ \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4} \]

### Step 5: Find a common denominator and sum the probabilities
The least common multiple of 12, 9, and 4 is 36. We convert each fraction:
1. \(\frac{1}{12} = \frac{3}{36}\)
2. \(\frac{1}{9} = \frac{4}{36}\)
3. \(\frac{1}{4} = \frac{9}{36}\)

Now, summing these:
\[ P(\text{white}) = \frac{3}{36} + \frac{4}{36} + \frac{9}{36} = \frac{16}{36} = \frac{4}{9} \]

### Final Answer
The probability that the ball drawn is white is:
\[ \boxed{\frac{4}{9}} \]