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, we will follow these steps:

### Step 1: Identify the total number of purses and their contents.
- 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 3 purses and each purse is selected at random, the probability of selecting any one purse is:
\[ P(X) = P(Y) = P(Z) = \frac{1}{3} \]

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

- From Purse Y:
\[ P(\text{White} | Y) = \frac{\text{Number of white balls in Y}}{\text{Total balls in Y}} = \frac{2}{6} = \frac{1}{3} \]

- From Purse Z:
\[ P(\text{White} | Z) = \frac{\text{Number of white balls in Z}}{\text{Total balls in Z}} = \frac{3}{4} \]

### Step 4: Use the law of total probability to find the overall probability of drawing a white ball.
The overall probability of drawing a white ball is given by:
\[ P(\text{White}) = P(X) \cdot P(\text{White} | X) + P(Y) \cdot P(\text{White} | Y) + P(Z) \cdot P(\text{White} | Z) \]

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

Calculating each term:
- From Purse X: \( \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \)
- From Purse Y: \( \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \)
- From Purse Z: \( \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4} \)

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

Now, adding 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 \( \frac{4}{9} \).

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