To solve this problem, we need to find the probability that both marbles drawn are red when two marbles are selected at random from a jar containing 10 red marbles and 12 blue marbles.

Step 1: Total Number of Marbles

The total number of marbles in the jar is: $10 \, (\text{red}) + 12 \, (\text{blue}) = 22 \, \text{marbles}.$

Step 2: Total Number of Ways to Choose 2 Marbles

The number of ways to choose 2 marbles from a total of 22 marbles is given by the combination formula: $\binom{22}{2} = \frac{22 \times 21}{2} = 231.$ So, there are 231 possible ways to choose any 2 marbles from the jar.

Step 3: Number of Ways to Choose 2 Red Marbles

The number of ways to choose 2 red marbles from the 10 red marbles is: $\binom{10}{2} = \frac{10 \times 9}{2} = 45.$

Step 4: Probability of Drawing 2 Red Marbles

The probability of drawing 2 red marbles is the ratio of the number of favorable outcomes (choosing 2 red marbles) to the total number of outcomes (choosing any 2 marbles): $\text{Probability} = \frac{\binom{10}{2}}{\binom{22}{2}} = \frac{45}{231}.$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3: $\frac{45}{231} = \frac{15}{77}.$

Thus, the probability that both marbles drawn are red is: $\boxed{\frac{15}{77}}.$

Would you like further details or have any questions on this solution? Here are some related questions to explore:

1. How would the probability change if there were 12 red marbles and 10 blue marbles?
2. What is the probability of drawing at least one red marble?
3. How do you calculate the probability of drawing two marbles of different colors?
4. What happens if you draw 3 marbles instead of 2?
5. Can you generalize the formula for selecting "r" red marbles from "n" total marbles?

Tip: When calculating probabilities involving combinations, remember to use the combination formula, which is particularly useful when the order of selection doesn't matter.