To solve the problem, we calculate the probability step by step.

Step 1: Understand the scenario

• Total number of marbles: $5 \text{ (blue)} + 3 \text{ (red)} = 8$.
• The first marble is drawn without replacement, meaning the total number of marbles decreases after the first draw.
• We are finding the probability that both selected marbles are red.

Step 2: Calculate the probability of the first red marble

The probability of drawing a red marble on the first draw is: $P(\text{first red}) = \frac{\text{Number of red marbles}}{\text{Total marbles}} = \frac{3}{8}.$

Step 3: Calculate the probability of the second red marble

After drawing one red marble, there are 2 red marbles left and 7 total marbles. The probability of drawing a second red marble is: $P(\text{second red | first red}) = \frac{\text{Remaining red marbles}}{\text{Remaining total marbles}} = \frac{2}{7}.$

Step 4: Multiply the probabilities

The probability of both events occurring (first and second red marbles) is the product of their individual probabilities: $P(\text{both red}) = P(\text{first red}) \cdot P(\text{second red | first red}) = \frac{3}{8} \cdot \frac{2}{7}.$

Simplify the product: $P(\text{both red}) = \frac{3 \cdot 2}{8 \cdot 7} = \frac{6}{56} = \frac{3}{28}.$

Final Answer:

The probability that both selected marbles are red is: $\boxed{\frac{3}{28}}$

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Do you want more details or further clarification?

Related Questions:

1. What is the probability of selecting one red and one blue marble in any order?
2. How would the probabilities change if the marbles were replaced after each draw?
3. What is the probability of selecting exactly one red marble in two draws?
4. How does the calculation change if there are more marbles in the jar?
5. What is the general formula for probabilities in problems involving draws without replacement?

Tip:

When solving probability problems involving "without replacement," always update both the numerator (favorable outcomes) and the denominator (total outcomes) after each draw.