Question 1210158: A basket contains 7 green marbles and 6 red marbles. Two marbles are taken from the basket at random without replacement (i.e., the first marble is not put back before the second marble is drawn). What is the probability that the first marble is green, and that the second marble is red?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break this problem down step by step:
**1. Probability of the First Marble Being Green**
* There are 7 green marbles and a total of 7 + 6 = 13 marbles in the basket.
* The probability of drawing a green marble first is the number of green marbles divided by the total number of marbles:
* P(Green first) = 7/13
**2. Probability of the Second Marble Being Red (Given the First Was Green)**
* After drawing a green marble, there are now 6 green marbles and 6 red marbles left in the basket, for a total of 12 marbles.
* The probability of drawing a red marble second, given that the first was green, is the number of red marbles divided by the remaining total number of marbles:
* P(Red second | Green first) = 6/12 = 1/2
**3. Probability of Both Events Occurring**
* To find the probability of both events happening (first green, then red), we multiply the probabilities of each individual event:
* P(Green first and Red second) = P(Green first) * P(Red second | Green first)
* P(Green first and Red second) = (7/13) * (1/2) = 7/26
**Therefore, the probability that the first marble is green and the second marble is red is 7/26.**
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