A bag contains 2 red marbles and 4 green marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then a second marble is drawn. What is the probability that the first marble is green and the second one is red?
There are two ways to approach this problem: using conditional probability or using permutations.
Using conditional probability:
The probability of drawing a green marble on the first draw is 4/6 (or 2/3), since there are 4 green marbles out of 6 marbles total. If a green marble is drawn on the first draw and not replaced, then there are 5 marbles left in the bag, including 2 red marbles and 3 green marbles. Therefore, the probability of drawing a red marble on the second draw, given that a green marble was drawn on the first draw and not replaced, is 2/5.
Using the formula for conditional probability:
P(Red on second draw | Green on first draw) = P(Red on second draw and Green on first draw) / P(Green on first draw)
The probability of drawing a green marble on the first draw and a red marble on the second draw is (4/6) * (2/5) = 4/15.
Therefore, the probability of drawing a green marble on the first draw and a red marble on the second draw is:
P(Green on first draw and Red on second draw) = P(Red on second draw | Green on first draw) * P(Green on first draw) = (4/15) * (2/3) = 8/45
Using permutations:
There are 6 marbles total, so there are 6 ways to choose the first marble. If a green marble is chosen on the first draw and not replaced, there are 5 marbles left, including 2 red marbles. Therefore, there are 2 ways to choose a red marble on the second draw.
Therefore, the total number of ways to choose a green marble on the first draw and a red marble on the second draw is:
6 * 2 = 12
The probability is then:
12 / (6 choose 2) = 12 / 15 = 4/5
Therefore, the probability of drawing a green marble on the first draw and a red marble on the second draw is 4/5 or 8/45, depending on the method used.
4/6 green
2/5 red
Thank you for the correction.
Using conditional probability:
P(Red on second draw | Green on first draw) = P(Red on second draw and Green on first draw) / P(Green on first draw)
The probability of drawing a green marble on the first draw and a red marble on the second draw is (4/6) * (2/4) = 1/3.
Therefore, the probability of drawing a green marble on the first draw and a red marble on the second draw is:
P(Green on first draw and Red on second draw) = P(Red on second draw | Green on first draw) * P(Green on first draw) = (1/3) * (4/6) = 2/9
Using permutations:
There are 6 marbles total, so there are 6 ways to choose the first marble. If a green marble is chosen on the first draw and not replaced, there are 5 marbles left, including 2 red marbles. Therefore, there are 2 ways to choose a red marble on the second draw.
Therefore, the total number of ways to choose a green marble on the first draw and a red marble on the second draw is:
6 * 2 = 12
The probability is then:
12 / (6 choose 2) = 12 / 15 = 4/5
However, this method assumes that the order in which the marbles are drawn is significant (i.e. the first marble is drawn and then the second marble). In this particular problem, the order is not significant, as the question only asks for the probability of drawing a green marble and then a red marble, regardless of which marble is drawn first. Therefore, the probability should be calculated using combinations, as opposed to permutations.
The number of ways to choose a green marble and then a red marble (without replacement) is:
(4 choose 1) * (2 choose 1) = 8
The number of total possible outcomes (without replacement) is:
6 choose 2 = 15
Therefore, the probability of drawing a green marble and then a red marble is:
8/15.
To find the probability that the first marble is green and the second one is red, we need to consider the two draws separately and multiply their probabilities together.
First, let's calculate the probability of drawing a green marble on the first draw. There are a total of 6 marbles in the bag (2 red + 4 green), so the probability of drawing a green marble on the first draw is 4/6 or 2/3.
Since the first marble is not replaced, there are a total of 5 marbles left in the bag for the second draw, consisting of 1 red marble and 4 green marbles. Therefore, the probability of drawing a red marble on the second draw is 1/5.
To find the probability of both events occurring, we multiply the probabilities together: (2/3) * (1/5) = 2/15.
So, the probability that the first marble drawn is green and the second one is red is 2/15.