Natalya put 14 blue marbles, 14 red marbles, and no other marbles into 3 empty cups. She put 2 blue...

Let's visualize this problem to make it crystal clear.

Phase 1: Owning the Dataset

Visualization

Let's create a simple table to organize our marble distribution:

We can verify our totals: \(2+6+6 = 14\) blue ✓ and \(6+4+4 = 14\) red ✓

Phase 2: Understanding the Question

Dmitry will randomly pick exactly one marble from each cup. We need to find:

1. 3 Blue: The probability that all three marbles picked are blue
2. 3 Red: The probability that all three marbles picked are red

Since he picks one marble from each cup independently, we'll multiply the individual probabilities.

Phase 3: Finding the Answer

Calculating P(3 Blue Marbles)

For Dmitry to pick 3 blue marbles, he needs:

• A blue marble from Cup A: \(\mathrm{P} = \frac{2}{8} = \frac{1}{4}\)
• A blue marble from Cup B: \(\mathrm{P} = \frac{6}{10} = \frac{3}{5}\)
• A blue marble from Cup C: \(\mathrm{P} = \frac{6}{10} = \frac{3}{5}\)

Combined probability:
\(\mathrm{P(3\,blue)} = \frac{1}{4} \times \frac{3}{5} \times \frac{3}{5} = \frac{9}{100} = 0.090\)

Calculating P(3 Red Marbles)

For Dmitry to pick 3 red marbles, he needs:

• A red marble from Cup A: \(\mathrm{P} = \frac{6}{8} = \frac{3}{4}\)
• A red marble from Cup B: \(\mathrm{P} = \frac{4}{10} = \frac{2}{5}\)
• A red marble from Cup C: \(\mathrm{P} = \frac{4}{10} = \frac{2}{5}\)

Combined probability:
\(\mathrm{P(3\,red)} = \frac{3}{4} \times \frac{2}{5} \times \frac{2}{5} = \frac{12}{100} = 0.120\)

Phase 4: Solution

Based on our calculations:

• 3 Blue: 0.090
• 3 Red: 0.120

These values match the answer choices provided, confirming our analysis.