A bag contains 5 balls. Two balls are drawn and are found to be white. What is the probability that all the balls are white?

Explanation

To solve the problem step by step, we will use Bayes' theorem and the concept of conditional probability.

### Step 1: Define Events
Let:
- \( E_1 \): Event that there are 2 white balls in the bag.
- \( E_2 \): Event that there are 3 white balls in the bag.
- \( E_3 \): Event that there are 4 white balls in the bag.
- \( E_4 \): Event that there are 5 white balls in the bag.
- \( A \): Event that 2 balls drawn are white.

We need to find \( P(E_4 | A) \), the probability that all balls are white given that 2 drawn balls are white.

### Step 2: Use Bayes' Theorem
According to Bayes' theorem:
\[ P(E_4 | A) = \frac{P(A | E_4) \cdot P(E_4)}{P(A)} \]

### Step 3: Calculate \( P(E_i) \)
Assuming there is no prior information, we can assume that each event \( E_1, E_2, E_3, E_4 \) is equally likely:
\[ P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4} \]

### Step 4: Calculate \( P(A | E_i) \)
Now we calculate \( P(A | E_i) \) for each event:

1. **For \( E_1 \)** (2 white balls):
- \( P(A | E_1) = \frac{2C2}{5C2} = \frac{1}{10} \)

2. **For \( E_2 \)** (3 white balls):
- \( P(A | E_2) = \frac{3C2}{5C2} = \frac{3}{10} \)

3. **For \( E_3 \)** (4 white balls):
- \( P(A | E_3) = \frac{4C2}{5C2} = \frac{6}{10} \)

4. **For \( E_4 \)** (5 white balls):
- \( P(A | E_4) = \frac{5C2}{5C2} = 1 \)

### Step 5: Calculate \( P(A) \)
Using the law of total probability:
\[ P(A) = P(A | E_1) P(E_1) + P(A | E_2) P(E_2) + P(A | E_3) P(E_3) + P(A | E_4) P(E_4) \]
Substituting the values:
\[ P(A) = \left(\frac{1}{10} \cdot \frac{1}{4}\right) + \left(\frac{3}{10} \cdot \frac{1}{4}\right) + \left(\frac{6}{10} \cdot \frac{1}{4}\right) + \left(1 \cdot \frac{1}{4}\right) \]
\[ = \frac{1}{40} + \frac{3}{40} + \frac{6}{40} + \frac{10}{40} = \frac{20}{40} = \frac{1}{2} \]

### Step 6: Substitute into Bayes' Theorem
Now substitute back into Bayes' theorem:
\[ P(E_4 | A) = \frac{P(A | E_4) \cdot P(E_4)}{P(A)} = \frac{1 \cdot \frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \cdot 2 = \frac{1}{2} \]

### Final Answer
Thus, the probability that all the balls are white given that the two drawn balls are white is:
\[ \boxed{\frac{1}{2}} \]