If the first marble is replaced before the second marble is drawn, what is the probability of selecting a white marble then a red marble? there are 4 red marbles 5 white and 3 blue
prob(white, then red)
= (5/12)(4/12) = 5/36
Well, let's roll out the red carpet for this probability problem! We're looking for the likelihood of selecting a white marble first, followed by a red one.
Now, there are a total of 12 marbles (4 red + 5 white + 3 blue). If we assume the first marble is replaced before the second is drawn, then the total number of marbles remains the same for both draws.
The probability of selecting a white marble on the first draw is 5/12, since there are 5 white marbles out of the total 12.
After the first white marble is replaced, we still have 12 marbles to choose from, but now the count has changed slightly: 4 red marbles, 5 white marbles, and 3 blue marbles. So, the probability of selecting a red marble on the second draw is 4/12 or 1/3.
To find the probability of both events occurring together, we multiply their individual probabilities. Therefore, the probability of selecting a white marble and then a red marble is (5/12) * (1/3) = 5/36.
So, the odds are 5 out of 36, or approximately 0.138. Good luck on your marble adventures!
To calculate the probability of selecting a white marble and then a red marble, assuming the first marble is replaced before the second marble is drawn, we need to calculate the individual probabilities of selecting a white marble and a red marble.
Step 1: Calculate the probability of selecting a white marble.
Total number of marbles = 4 (red) + 5 (white) + 3 (blue) = 12 marbles
Number of white marbles = 5 white marbles
Probability of selecting a white marble = Number of white marbles / Total number of marbles
P(white marble) = 5/12
Step 2: Calculate the probability of selecting a red marble.
Total number of marbles = 4 (red) + 5 (white) + 3 (blue) = 12 marbles
Number of red marbles = 4 red marbles
Probability of selecting a red marble = Number of red marbles / Total number of marbles
P(red marble) = 4/12 = 1/3
Step 3: Calculate the probability of selecting a white marble and then a red marble.
Since the marbles are replaced before the second draw, the events are independent, and we can multiply the individual probabilities together.
P(white marble and then red marble) = P(white marble) * P(red marble)
P(white marble and then red marble) = (5/12) * (1/3)
P(white marble and then red marble) = 5/36
Therefore, the probability of selecting a white marble and then a red marble, with replacement, is 5/36.
To determine the probability of selecting a white marble and then a red marble, we can break down the problem into two separate events: the probability of selecting a white marble on the first draw, and the probability of selecting a red marble on the second draw.
First, let's determine the probability of selecting a white marble on the first draw.
The total number of marbles is 4 (red) + 5 (white) + 3 (blue) = 12 marbles.
The probability of selecting a white marble on the first draw is:
P(White marble on first draw) = Number of white marbles / Total number of marbles
= 5 / 12
Now, let's determine the probability of selecting a red marble on the second draw.
Since the marble from the first draw is being replaced before the second draw, the number of marbles remains the same for the second draw.
The total number of marbles is still 12.
The probability of selecting a red marble on the second draw is:
P(Red marble on the second draw) = Number of red marbles / Total number of marbles
= 4 / 12
= 1 / 3
To find the probability of both events occurring (white marble on the first draw and red marble on the second draw), we multiply the individual probabilities together:
P(White marble then red marble) = P(White marble on first draw) * P(Red marble on second draw)
= (5 / 12) * (1 / 3)
= 5 / 36
Therefore, the probability of selecting a white marble then a red marble, with replacement, is 5/36.