A bag contains 7 green marbels 9 red marbels 10 orange marbels 5 brown marbels and 10 blue marbels 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 pick is 9/41 (since there are 9 red marbles out of 41 total marbles). After replacing the marble, the probability of choosing a blue marble on the second pick is 10/41 (since there are 10 blue marbles out of 41 total marbles).
Therefore, the probability of selecting a red marble then a blue marble is:
P(red, then blue) = P(red) x P(blue | red)
P(red, then blue) = (9/41) x (10/41)
P(red, then blue) = 90/1681 or approximately 0.0535
So the probability of selecting a red marble then a blue marble is approximately 0.0535.
To find the probability of selecting a red marble, replacing it, and then selecting a blue marble, we can use the formula:
P(red, then blue) = P(red) * P(blue | red)
First, let's find the probability of selecting a red marble:
P(red) = number of red marbles / total number of marbles
= 9 / (7 + 9 + 10 + 5 + 10)
= 9 / 41
Since the marble is replaced after each selection, the probability of selecting a blue marble given that a red marble was chosen is the same as the probability of selecting a blue marble initially:
P(blue | red) = number of blue marbles / total number of marbles
= 10 / (7 + 9 + 10 + 5 + 10)
= 10 / 41
Now, we can calculate the probability of selecting a red marble, then a blue marble:
P(red, then blue) = P(red) * P(blue | red)
= (9 / 41) * (10 / 41)
= 90 / 1681
Therefore, the probability of selecting a red marble, replacing it, and then selecting a blue marble is 90/1681.