Animated Solution for Mathematics - Probability: Comprehension Passage

Imagine standing before three boxes, each a treasure chest of numbered cards. Box 1 holds cards $\{1,2,3\}$, Box 2 holds $\{1,2,3,4,5\}$, and Box 3 holds $\{1,2,3,4,5,6,7\}$.

When you draw a card from each, you are navigating a sample space of $3\times 5\times 7=105$ total possibilities. This is the foundation of our journey.

The sum of three integers is odd only under specific conditions. To get an odd sum, you need an odd number of odd cards.

This leads us to four mutually exclusive scenarios: 1. (Odd, Odd, Odd) 2. (Odd, Even, Even) 3. (Even, Odd, Even) 4. (Even, Even, Odd)

By calculating the probability of each and summing them, we find the answer is:

This demonstrates the beauty of breaking down a complex problem into manageable, mutually exclusive parts.

Now, the plot thickens. We want $x_1,x_2,x_3$ to be in an arithmetic progression. This condition implies:

Because $2x_2$ is always even, the sum $x_1+x_3$ must be even. This forces $x_1$ and $x_3$ to have the same parity (both odd or both even).

We explore the odd-odd and even-even cases, ensuring $x_2$ always lands within the valid range of Box 2.

For the odd-odd case: - Box 1 has $\{1,3\}$ (2 choices). - Box 3 has $\{1,3,5,7\}$ (4 choices). - For each pair $(x_1,x_3)$, the sum $x_1+x_3$ is even, and $x_2=\frac{x_1+x_3}{2}$ must be in $\{1,2,3,4,5\}$. All resulting $x_2$ values satisfy this. - Total pairs = $2\times 4=8$.

For the even-even case: - Box 1 has $\{2\}$ (1 choice). - Box 3 has $\{2,4,6\}$ (3 choices). - For each pair $(x_1,x_3)$, the sum $x_1+x_3$ is even, and $x_2=\frac{x_1+x_3}{2}$ must be in $\{1,2,3,4,5\}$. All resulting $x_2$ values satisfy this. - Total pairs = $1\times 3=3$.

The total number of favorable outcomes is $8+3=11$.

Thus, the probability that the drawn numbers form an arithmetic progression is:

You have mastered the logic. Keep pushing forward, for every problem is just a puzzle waiting for your insight.