What is the probability of selecting a red marble, replacing it, and then selecting a blue marble?
The probability of selecting a red marble on the first draw is the same as the probability of selecting a blue marble on the second draw, since we replace the first marble. Let's assume that the probability of selecting a red marble is P(red) = 0.4 and the probability of selecting a blue marble is P(blue) = 0.3. Then:
P(red, blue) = P(red) x P(blue) = 0.4 x 0.3 = 0.12
So the probability of selecting a red marble, replacing it, and then selecting a blue marble is 0.12 or 12%.
To calculate the probability of selecting a red marble, replacing it, and then selecting a blue marble, we need to consider the total number of marbles and the number of red and blue marbles.
Let's assume we have a bag of marbles that contains 10 marbles in total, of which 4 are red and 6 are blue.
Step 1: Probability of selecting a red marble:
The probability of selecting a red marble on the first draw is the number of red marbles divided by the total number of marbles:
P(Red) = Number of Red Marbles / Total Number of Marbles
= 4 / 10
= 2/5
Step 2: Probability of selecting a blue marble:
Since we are replacing the red marble back into the bag, the total number of marbles remains the same. Therefore, the probability of selecting a blue marble on the second draw is still:
P(Blue) = Number of Blue Marbles / Total Number of Marbles
= 6 / 10
= 3/5
Step 3: Probability of both events happening together (selecting a red marble and then a blue marble):
To calculate the probability of both events happening together, we multiply the probabilities of the individual events:
P(Red and Blue) = P(Red) * P(Blue)
= (2/5) * (3/5)
= 6/25
Therefore, the probability of selecting a red marble, replacing it, and then selecting a blue marble is 6/25 or approximately 0.24 or 24%.
Note: If the marbles are not replaced after each draw, the probability would change for the second draw, but in this case, we assumed that the marbles are replaced.