. An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is

Options

A. `(1)/(10)`
B. `(3)/(10)`
C. `(3)/(5)`
D. `(1)/(2)`

Answer

To solve the problem, we need to determine the probability that all the balls in the urn are white given that we have drawn two white balls. We can use Bayes' theorem for this purpose.

### Step-by-step Solution:

1. **Define Events**:
- Let \( A_1 \): The event that there are 2 white balls in the urn.
- Let \( A_2 \): The event that there are 3 white balls in the urn.
- Let \( A_3 \): The event that there are 4 white balls in the urn.
- Let \( A_4 \): The event that there are 5 white balls in the urn.
- Let \( B \): The event that 2 balls drawn are white.

2. **Prior Probabilities**:
Since we have no prior information, we can assume that each event \( A_1, A_2, A_3, A_4 \) is equally likely. Thus, the prior probabilities are:
\[ P(A_1) = P(A_2) = P(A_3) = P(A_4) = \frac{1}{4} \]

3. **Calculate \( P(B|A_i) \)**:
- For \( A_1 \) (2 white balls):
\[ P(B|A_1) = 1 \quad \text{(since both drawn balls must be white)} \]
- For \( A_2 \) (3 white balls):
\[ P(B|A_2) = \frac{2}{3} \quad \text{(choose 2 white from 3)} \]
- For \( A_3 \) (4 white balls):
\[ P(B|A_3) = \frac{6}{10} = \frac{3}{5} \quad \text{(choose 2 white from 4)} \]
- For \( A_4 \) (5 white balls):
\[ P(B|A_4) = 1 \quad \text{(since all balls are white)} \]

4. **Calculate \( P(B) \)**:
Using the law of total probability:
\[ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) + P(B|A_4)P(A_4) \]
\[ P(B) = 1 \cdot \frac{1}{4} + \frac{2}{3} \cdot \frac{1}{4} + \frac{3}{5} \cdot \frac{1}{4} + 1 \cdot \frac{1}{4} \]
\[ = \frac{1}{4} + \frac{2}{12} + \frac{3}{20} + \frac{1}{4} \]
To compute this, we need a common denominator. The least common multiple of 4, 12, and 20 is 60:
\[ = \frac{15}{60} + \frac{10}{60} + \frac{9}{60} + \frac{15}{60} = \frac{49}{60} \]

5. **Apply Bayes' Theorem**:
We want to find \( P(A_4|B) \):
\[ P(A_4|B) = \frac{P(B|A_4)P(A_4)}{P(B)} \]
\[ = \frac{1 \cdot \frac{1}{4}}{\frac{49}{60}} = \frac{60}{196} = \frac{15}{49} \]

### Final Answer:
The probability that all the balls are white given that two drawn balls are white is \( \frac{15}{49} \).