Question

frank’s bag of marbles contains: 4 red marbles, 10 blue marbles, 6 green marbles. a student draws one marble from the bag, does not replace it, then draws another marble from the bag. what is the probability that the first marble will be green and the second marble will be red?

Explanation:

Step1: Calculate total number of marbles

First, find the total number of marbles by adding the number of red, blue, and green marbles. There are 4 red, 10 blue, and 6 green marbles. So total marbles \( n = 4 + 10 + 6 = 20 \).

Step2: Probability of first marble being green

The number of green marbles is 6. So the probability of drawing a green marble first, \( P(G_1) = \frac{6}{20} \).

Step3: Probability of second marble being red (without replacement)

After drawing a green marble, the total number of marbles left is \( 20 - 1 = 19 \), and the number of red marbles is still 4. So the probability of drawing a red marble second, \( P(R_2|G_1) = \frac{4}{19} \).

Step4: Calculate the combined probability

Since these are dependent events, the probability of both events happening is the product of the two probabilities. So \( P(G_1 \cap R_2) = P(G_1) \times P(R_2|G_1) = \frac{6}{20} \times \frac{4}{19} \).
Simplify the fraction: \( \frac{6 \times 4}{20 \times 19} = \frac{24}{380} = \frac{6}{95} \).

Answer:

\(\frac{6}{95}\)