In a bag there are three tickets numbered 1,2,3. A ticket is drawn at random and put back, and this is done four times. The probability that the sum of the numbers is even, is

Options

A. `41//81`
B. `39//81`
C. `40//81`
D. None of these

Answer

To solve the problem, we need to determine the probability that the sum of the numbers drawn from the tickets is even after drawing four times. The tickets are numbered 1, 2, and 3.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have three tickets numbered 1, 2, and 3.
- We draw a ticket four times, replacing it each time.
- We want to find the probability that the sum of the numbers drawn is even.

2. **Identifying Odd and Even Numbers**:
- The odd numbers are 1 and 3 (2 odd numbers).
- The even number is 2 (1 even number).

3. **Possible Outcomes**:
- When drawing four tickets, the sum will be even if we draw an even number of odd numbers (0, 2, or 4 odd numbers).

4. **Calculating the Probabilities**:
- Let \( X \) be the number of odd numbers drawn.
- We need to find \( P(X = 0) + P(X = 2) + P(X = 4) \).

5. **Calculating Each Probability**:
- The total number of outcomes when drawing four times is \( 3^4 = 81 \).

- **Case 1: \( X = 0 \)** (All draws are even):
- The only way this happens is if we draw the number 2 each time.
- Probability:
\[ P(X = 0) = \left(\frac{1}{3}\right)^4 = \frac{1}{81} \]

- **Case 2: \( X = 2 \)** (Two odd and two even):
- We can choose 2 draws to be odd from 4 draws. The number of ways to choose 2 draws from 4 is \( \binom{4}{2} = 6 \).
- Each odd draw can be either 1 or 3 (2 choices), and each even draw is 2 (1 choice).
- Probability:
\[ P(X = 2) = \binom{4}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^2 = 6 \cdot \frac{4}{9} \cdot \frac{1}{9} = \frac{24}{81} \]

- **Case 3: \( X = 4 \)** (All draws are odd):
- The only way this happens is if we draw either 1 or 3 each time.
- Probability:
\[ P(X = 4) = \left(\frac{2}{3}\right)^4 = \frac{16}{81} \]

6. **Summing the Probabilities**:
- Now, we sum the probabilities of the cases where the sum is even:
\[ P(\text{sum is even}) = P(X = 0) + P(X = 2) + P(X = 4) = \frac{1}{81} + \frac{24}{81} + \frac{16}{81} = \frac{41}{81} \]

### Final Answer:
The probability that the sum of the numbers drawn is even is \( \frac{41}{81} \).

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