Marble Selection Probability

Solution & Explanation

To solve this problem, we use Bayes' Theorem to find the probability that a red marble was drawn from Bucket #1, and then extend this to find the probability that two red marbles were consecutively drawn from Bucket #1.

Part 1: Probability of Drawing a Red Marble from Bucket #1

Given:

• Bucket #1: 30 red marbles, 10 black marbles.
• Bucket #2: 20 red marbles, 20 black marbles.

Objective: Find $P(Bucket1 | Marble = Red)$.

Using Bayes' Theorem:

$P(Bucket1 | Marble = Red) = \frac{P(Marble = Red | Bucket1) \cdot P(Bucket1)}{P(Marble = Red)}$

Calculations:

• $P(Marble = Red | Bucket1) = \frac{30}{40} = 0.75$
• $P(Bucket1) = \frac{1}{2}$
• $P(Marble = Red) = \left( \frac{30}{40} \cdot \frac{1}{2} \right) + \left( \frac{20}{40} \cdot \frac{1}{2} \right) = \frac{3}{8} + \frac{1}{4} = \frac{5}{8}$

Substituting into Bayes' Theorem:

$P(Bucket1 | Marble = Red) = \frac{0.75 \cdot 0.5}{0.625} = \frac{3}{5} = 0.6$

Part 2: Probability of Drawing Two Red Marbles from Bucket #1

Objective: Find the probability that both marbles are red and drawn from Bucket #1.

Scenario:

• The marble is replaced after each draw.
• The probability of drawing a red marble from Bucket #1 remains $0.75$ for each draw.

Calculating the Probability:

• Since the events are independent, the probability of drawing two red marbles from Bucket #1 is:

$P(Both Red from Bucket1) = P(Red from Bucket1) \cdot P(Red from Bucket1)$

• $P(Both Red from Bucket1) = 0.6 \cdot 0.6 = 0.36$

Conclusion:

• The probability that a red marble was drawn from Bucket #1 is $0.6$.
• The probability that two red marbles were drawn consecutively from Bucket #1 is $0.36$.