To solve this problem, we calculate the probability of selecting two red marbles in succession without replacement.

Step 1: Understand the problem

• Total marbles: $7 \text{ blue} + 3 \text{ red} = 10 \text{ marbles}$
• Event: Both selected marbles are red.
• Selection is without replacement, so the probabilities change after the first marble is drawn.

Step 2: Calculate the probabilities

1. Probability of the first marble being red:
   Since there are 3 red marbles and 10 total marbles:
   $P(\text{First red}) = \frac{3}{10}.$

2. Probability of the second marble being red given the first is red:
   After one red marble is removed, there are 2 red marbles left and 9 total marbles remaining:
   $P(\text{Second red | First red}) = \frac{2}{9}.$

Step 3: Multiply the probabilities

The probability of both events occurring is the product of the probabilities:
$P(\text{Both red}) = P(\text{First red}) \cdot P(\text{Second red | First red}) = \frac{3}{10} \cdot \frac{2}{9}.$

$P(\text{Both red}) = \frac{6}{90} = \frac{1}{15}.$

Step 4: Convert to decimal and round

$P(\text{Both red}) = 0.067 \quad \text{(rounded to three decimal places)}.$

Final Answer:

The probability that both of the selected marbles are red is 0.067.

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Would you like further details or help with a similar problem?
Here are 5 related questions to deepen your understanding:

1. What is the probability of selecting one red marble and one blue marble in any order?
2. How would the probability change if the marbles were replaced after each draw?
3. What is the probability of selecting two blue marbles in a row without replacement?
4. If the jar had 10 blue and 5 red marbles, how would the probability of selecting two red marbles change?
5. Can you explain how conditional probability applies to this situation?

Tip: Always double-check whether the problem involves "with replacement" or "without replacement," as it significantly changes the probabilities.