Question 1210157: A basket contains 7 green marbles and 6 red marbles. A marble is taken from the basket at random; its color is recorded, then the marble is returned to the basket. A second marble is then taken from the basket at random, and its color is recorded. What is the probability that the first marble is green, and the second marble is red?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step by step.
**1. Probability of the First Marble Being Green**
* There are 7 green marbles and 6 red marbles, for a total of 13 marbles.
* The probability of drawing a green marble is:
* P(Green first) = 7/13
**2. Probability of the Second Marble Being Red**
* Since the first marble is returned to the basket, the total number of marbles remains 13.
* The probability of drawing a red marble is:
* P(Red second) = 6/13
**3. Probability of Both Events Occurring**
* Since the events are independent (because the first marble is returned), we multiply the probabilities:
* P(Green first and Red second) = P(Green first) * P(Red second)
* P(Green first and Red second) = (7/13) * (6/13) = 42/169
**Therefore, the probability that the first marble is green and the second marble is red is 42/169.**
|
|
|
