Questions: A bag contains 10 marbles: 2 are green, 6 are red, and 2 are blue. Manuel chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that the first marble is green and the second is red? Write your answer as a fraction in simplest form.

Solution Steps

To solve this problem, we need to calculate the probability of two dependent events: drawing a green marble first and then a red marble.

1. Calculate the probability of drawing a green marble first.
2. Calculate the probability of drawing a red marble second, given that the first marble drawn was green.
3. Multiply these two probabilities to get the final answer.

The probability of drawing a green marble first is given by:

$$P(\text{Green First}) = \frac{\text{Number of Green Marbles}}{\text{Total Number of Marbles}} = \frac{2}{10} = 0.2$$

After drawing a green marble, the total number of marbles decreases to 9. The probability of drawing a red marble second is:

$$P(\text{Red Second} \mid \text{Green First}) = \frac{\text{Number of Red Marbles}}{\text{Total Number of Marbles After Green}} = \frac{6}{9} = 0.6667$$

The combined probability of both events occurring (drawing a green marble first and a red marble second) is:

$$P(\text{Green First and Red Second}) = P(\text{Green First}) \times P(\text{Red Second} \mid \text{Green First}) = 0.2 \times 0.6667 = 0.1333$$

The combined probability can be expressed as a fraction in simplest form:

$$P(\text{Green First and Red Second}) = \frac{2}{15}$$

Final Answer