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 of finding the probability that the sum of the numbers drawn from the bag is even, we can follow these steps:

### Step 1: Identify the possible outcomes
In the bag, we have three tickets numbered 1, 2, and 3. The numbers can be categorized as:
- Odd numbers: 1, 3
- Even number: 2

### Step 2: Determine the probability of drawing odd and even numbers
The probability of drawing an odd number (1 or 3) is:
\[ P(\text{odd}) = \frac{2}{3} \]
The probability of drawing an even number (2) is:
\[ P(\text{even}) = \frac{1}{3} \]

### Step 3: Consider the number of draws
We are drawing a ticket 4 times. We need to find the probability that the sum of the drawn numbers is even. This can happen in the following scenarios:
1. All four numbers are even.
2. Two numbers are odd and two numbers are even.
3. All four numbers are odd.

### Step 4: Calculate the probabilities for each scenario
1. **Probability of 0 odd numbers (all even)**:
\[ P(X = 0) = \left( P(\text{even}) \right)^4 = \left( \frac{1}{3} \right)^4 = \frac{1}{81} \]

2. **Probability of 2 odd numbers and 2 even numbers**:
We need to choose 2 draws to be odd out of 4, which can be calculated using combinations:
\[ P(X = 2) = \binom{4}{2} \left( P(\text{odd}) \right)^2 \left( P(\text{even}) \right)^2 \]
\[ = 6 \left( \frac{2}{3} \right)^2 \left( \frac{1}{3} \right)^2 = 6 \cdot \frac{4}{9} \cdot \frac{1}{9} = 6 \cdot \frac{4}{81} = \frac{24}{81} \]

3. **Probability of 4 odd numbers**:
\[ P(X = 4) = \left( P(\text{odd}) \right)^4 = \left( \frac{2}{3} \right)^4 = \frac{16}{81} \]

### Step 5: Combine the probabilities
Now, we add the probabilities of all scenarios 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
Thus, the probability that the sum of the numbers drawn is even is:
\[ \frac{41}{81} \]