A bag contains 4 green marbles, 6 red marbles, 14 orange marbles, 5 brown marbles, and 8 blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
The probability of choosing a red marble on the first draw is 6/37 (6 out of 37 total marbles). Since we replace the marble before the second draw, the probability of choosing a blue marble on the second draw is also 8/37.
To find the probability of both events happening (choosing red and then blue), we multiply the probabilities:
P(red, then blue) = (6/37) * (8/37)
P(red, then blue) = 0.0366 or approximately 3.66%
3.66 as a fraction
We can write 3.66 as a fraction by dividing 3.66 by 1:
3.66/1
To get rid of the decimal, we can multiply both the numerator and denominator by 100:
3.66/1 x 100/100
This simplifies to:
366/100
We can further simplify by dividing both the numerator and denominator by their greatest common factor, which is 2:
183/50
Therefore, 3.66 as a fraction is 183/50.
To find the probability of choosing a red marble and then a blue marble, we need to multiply the probabilities of each event happening.
Step 1: Calculate the probability of choosing a red marble:
There are a total of 4 + 6 + 14 + 5 + 8 = 37 marbles in the bag. Out of these, there are 6 red marbles. Since we are replacing the marble after each selection, the probability of choosing a red marble on the first draw is 6/37.
Step 2: Calculate the probability of choosing a blue marble:
After replacing the first marble, there are still 37 marbles in the bag, but now there are 8 blue marbles. So the probability of choosing a blue marble on the second draw is 8/37.
Step 3: Multiply the probabilities:
To find the probability of both events happening, we multiply the probability of choosing a red marble and the probability of choosing a blue marble:
P(red, then blue) = (6/37) * (8/37) = 48/1369.
Therefore, the probability of choosing a red marble, replacing it, and then choosing a blue marble is 48/1369.