A bag contains three tickets numbered 1, 2 and 3. A ticket is drawn at random its number is noted and is put back in the bag. This is done 4 times. Let p denote the probability that sum of the numbers on tickets is odd, then `2/p` is equal to _____

Answer

To solve the problem, we need to determine the probability \( p \) that the sum of the numbers on the tickets drawn is odd when a ticket is drawn 4 times from a bag containing tickets numbered 1, 2, and 3. We will then calculate \( \frac{2}{p} \).

### Step-by-step Solution:

1. **Understanding the Problem**:
- The tickets are numbered 1, 2, and 3.
- The sum of the numbers drawn can be odd or even.
- We need to find the probability that the sum is odd after drawing 4 tickets.

2. **Identifying Odd and Even Numbers**:
- The odd numbers are 1 and 3.
- The even number is 2.
- The sum of numbers is odd if there is an odd count of odd numbers drawn.

3. **Possible Cases**:
- In 4 draws, we can have:
- 1 odd and 3 even (not possible since we only have one even number).
- 3 odd and 1 even.
- 2 odd and 2 even (not possible since 2 even numbers are not available).
- Thus, the only valid case for an odd sum is having 3 odd numbers and 1 even number.

4. **Calculating the Favorable Outcomes**:
- We need to choose 3 draws to be odd (1 or 3) and 1 draw to be even (2).
- The arrangements of 3 odd and 1 even can be calculated using combinations:
- The number of ways to choose which of the 4 draws will be even is \( \binom{4}{1} = 4 \).
- For the odd numbers, we have 2 choices (1 or 3) for each of the 3 odd draws:
- Total odd combinations = \( 2^3 = 8 \).

5. **Total Favorable Outcomes**:
- Total favorable outcomes for the case of 3 odd and 1 even:
\[ \text{Total favorable outcomes} = 4 \text{ (ways to choose the even draw)} \times 8 \text{ (odd combinations)} = 32. \]

6. **Total Possible Outcomes**:
- Each draw has 3 options (1, 2, or 3), and since we draw 4 times, the total possible outcomes are:
\[ 3^4 = 81. \]

7. **Calculating the Probability \( p \)**:
- The probability \( p \) that the sum is odd:
\[ p = \frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{32}{81}. \]

8. **Calculating \( \frac{2}{p} \)**:
- Now, we need to find \( \frac{2}{p} \):
\[ \frac{2}{p} = \frac{2}{\frac{32}{81}} = \frac{2 \times 81}{32} = \frac{162}{32} = \frac{81}{16}. \]

### Final Answer:
Thus, \( \frac{2}{p} = \frac{81}{16} \).