A bag contains 5 green marbles, 5 blue marbles, and 10 red marbles. You choose a marble, replace it, and choose again. What is P(green, then red)?
The probability of choosing a green marble on the first draw is 5/20 (since there are 5 green marbles out of a total of 20 marbles). After replacing the marble, the probability of choosing a red marble on the second draw is 10/20 (since there are 10 red marbles out of a total of 20 marbles).
Therefore, the probability of choosing a green marble and then a red marble is:
P(green, then red) = P(green) * P(red) = (5/20) * (10/20) = 0.125
So the probability of choosing a green marble on the first draw and a red marble on the second draw is 0.125 or 12.5%.
To find the probability of drawing a green marble and then a red marble, we need to calculate the probability of each event separately and then multiply these probabilities together.
First, let's calculate the probability of drawing a green marble.
The total number of marbles in the bag is 5 green + 5 blue + 10 red = 20 marbles.
So, the probability of drawing a green marble is 5/20 = 1/4.
Since we replace the marble each time, the probability of drawing a green marble again is also 1/4.
Now, let's calculate the probability of drawing a red marble.
We still have a total of 20 marbles in the bag, but now there are only 10 red marbles remaining.
So, the probability of drawing a red marble is 10/20 = 1/2.
Now, we can multiply the probabilities together to find the probability of drawing a green marble and then a red marble.
P(green, then red) = P(green) * P(red)
P(green, then red) = (1/4) * (1/2)
P(green, then red) = 1/8
Therefore, the probability of drawing a green marble and then a red marble is 1/8.