A bag contains 5 balls of unknown colours. A ball is drawn at random from it and is found to be red. Then the probability that all tha balls in the bag are red, is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(2)/(5)`
D. `(1)/(3)`

Answer

To solve the problem step-by-step, we will use Bayes' theorem to find the probability that all the balls in the bag are red given that a red ball has been drawn.

### Step 1: Define Events
Let:
- \( E \): The event that all 5 balls in the bag are red.
- \( A \): The event that there is 1 red ball in the bag.
- \( B \): The event that there are 2 red balls in the bag.
- \( C \): The event that there are 3 red balls in the bag.
- \( D \): The event that there are 4 red balls in the bag.
- \( Q \): The event that a red ball is drawn.

### Step 2: Assign Probabilities
Assuming that each configuration of balls is equally likely, we have:
- \( P(E) = P(A) = P(B) = P(C) = P(D) = \frac{1}{5} \)

### Step 3: Calculate Conditional Probabilities
Now we calculate the probability of drawing a red ball given each event:

1. **For event \( A \)** (1 red ball):
\[ P(Q|A) = \frac{1}{5} \]

2. **For event \( B \)** (2 red balls):
\[ P(Q|B) = \frac{2}{5} \]

3. **For event \( C \)** (3 red balls):
\[ P(Q|C) = \frac{3}{5} \]

4. **For event \( D \)** (4 red balls):
\[ P(Q|D) = \frac{4}{5} \]

5. **For event \( E \)** (5 red balls):
\[ P(Q|E) = 1 \]

### Step 4: Apply Bayes' Theorem
We want to find \( P(E|Q) \):
\[ P(E|Q) = \frac{P(E) \cdot P(Q|E)}{P(A) \cdot P(Q|A) + P(B) \cdot P(Q|B) + P(C) \cdot P(Q|C) + P(D) \cdot P(Q|D) + P(E) \cdot P(Q|E)} \]

### Step 5: Substitute Values
Substituting the values we calculated:
\[ P(E|Q) = \frac{\left(\frac{1}{5}\right) \cdot 1}{\left(\frac{1}{5} \cdot \frac{1}{5}\right) + \left(\frac{1}{5} \cdot \frac{2}{5}\right) + \left(\frac{1}{5} \cdot \frac{3}{5}\right) + \left(\frac{1}{5} \cdot \frac{4}{5}\right) + \left(\frac{1}{5} \cdot 1\right)} \]

### Step 6: Simplify the Denominator
Calculating the denominator:
\[ = \frac{1}{5} \left( \frac{1}{5} + \frac{2}{5} + \frac{3}{5} + \frac{4}{5} + 1 \right) \]
\[ = \frac{1}{5} \left( \frac{1 + 2 + 3 + 4 + 5}{5} \right) \]
\[ = \frac{1}{5} \left( \frac{15}{5} \right) = \frac{1}{5} \cdot 3 = \frac{3}{5} \]

### Step 7: Final Calculation
Now substituting back into Bayes' theorem:
\[ P(E|Q) = \frac{\frac{1}{5}}{\frac{3}{5}} = \frac{1}{3} \]

### Conclusion
Thus, the probability that all the balls in the bag are red given that a red ball has been drawn is:
\[ \boxed{\frac{1}{3}} \]