There are 6 tickets with number 1,1,2,2,3,3 in a box if 2 tickets are selected find the probability distribution and expected value of the total on the numbers on digits.

Let's solve this step by step.

### Step 1: Understand the scenario

We have 6 tickets numbered:  
\[ 1, 1, 2, 2, 3, 3 \]  
We are randomly selecting 2 tickets and the sum of their numbers is considered as the outcome.

### Step 2: List possible outcomes

The possible pairs of tickets (with their sums) are:

- \( (1, 1) \): Sum = 2
- \( (1, 2) \): Sum = 3
- \( (1, 3) \): Sum = 4
- \( (2, 2) \): Sum = 4
- \( (2, 3) \): Sum = 5
- \( (3, 3) \): Sum = 6

### Step 3: Find the probabilities of each sum

We calculate the probabilities by considering how many ways each pair of tickets can be drawn.

- For \( (1, 1) \): there are \( \binom{2}{2} = 1 \) way to select the two 1's.
- For \( (1, 2) \): there are \( 2 \times 2 = 4 \) ways to select a 1 and a 2.
- For \( (1, 3) \): there are \( 2 \times 2 = 4 \) ways to select a 1 and a 3.
- For \( (2, 2) \): there is \( \binom{2}{2} = 1 \) way to select the two 2's.
- For \( (2, 3) \): there are \( 2 \times 2 = 4 \) ways to select a 2 and a 3.
- For \( (3, 3) \): there is \( \binom{2}{2} = 1 \) way to select the two 3's.

The total number of ways to select two tickets is \( \binom{6}{2} = 15 \).

Now, the probability of each sum is:

- \( P(\text{Sum} = 2) = \frac{1}{15} \)
- \( P(\text{Sum} = 3) = \frac{4}{15} \)
- \( P(\text{Sum} = 4) = \frac{5}{15} \)
- \( P(\text{Sum} = 5) = \frac{4}{15} \)
- \( P(\text{Sum} = 6) = \frac{1}{15} \)

### Step 4: Probability distribution table

\[
\begin{array}{|c|c|}
\hline
\text{Sum (X)} & \text{Probability (P(X))} \\
\hline
2 & \frac{1}{15} \\
3 & \frac{4}{15} \\
4 & \frac{5}{15} \\
5 & \frac{4}{15} \\
6 & \frac{1}{15} \\
\hline
\end{array}
\]

### Step 5: Expected value

The expected value \( E(X) \) is calculated as:
\[
E(X) = \sum X \cdot P(X)
\]
Substituting the values:
\[
E(X) = 2 \times \frac{1}{15} + 3 \times \frac{4}{15} + 4 \times \frac{5}{15} + 5 \times \frac{4}{15} + 6 \times \frac{1}{15}
\]
\[
E(X) = \frac{2}{15} + \frac{12}{15} + \frac{20}{15} + \frac{20}{15} + \frac{6}{15}
\]
\[
E(X) = \frac{60}{15} = 4
\]

### Final Answer:

- **Probability Distribution**: The table above shows the distribution.
- **Expected Value**: The expected value of the sum is 4.

Would you like more details or have any other questions?

Here are 5 related questions you could explore:
1. What is the variance of the sum of the numbers on the tickets?
2. How would the probability distribution change if there were 3 tickets numbered 1, 2, and 3 each?
3. How can we generalize the expected value if we increase the number of tickets?
4. What is the probability that the sum is greater than 4?
5. How does the expected value change if the tickets have different weights?

**Tip**: The expected value gives the long-run average outcome but does not describe variability or extremes!

Explore the probability distribution and expected value of the sum when selecting 2 tickets from a set of 6 (1, 1, 2, 2, 3, 3). This detailed solution covers combinatorics and expected value, making it ideal for students in Grades 10-12.

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