A bag contains 7 green marbles, 9 red marbles, 10 orange marbles, 5 brown marbles, and 10 blue marbles. You choose a marble, replace it, and choose again. Find P(red, then blue).
12 answers
Anyone know?
total marbles = 41
Prob(red, blue) = (9/41)(10/41) = 0/1681 or appr .0535
Prob(red, blue) = (9/41)(10/41) = 0/1681 or appr .0535
nope. THAT'S WHY I CAME HERE!!!
don't know the answer
ya 41
I think that the answer is B.19/41. So that is just what I am going to go with.
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Wrong the answer is 90/1681
Sorry for the confusion earlier. You are correct that the answer is 90/1681. Here's why:
The probability of choosing a red marble on the first draw is 9/41, since there are 9 red marbles out of 41 total marbles. After replacing the red marble, the probability of choosing a blue marble on the second draw is also 10/41, since we have replaced the marble and there are still 10 blue marbles out of 41 total marbles.
To find the probability of both events happening, we need to multiply the probabilities. So:
P(red, then blue) = (9/41) x (10/41) = 90/1681
Therefore, the answer is 90/1681.
The probability of choosing a red marble on the first draw is 9/41, since there are 9 red marbles out of 41 total marbles. After replacing the red marble, the probability of choosing a blue marble on the second draw is also 10/41, since we have replaced the marble and there are still 10 blue marbles out of 41 total marbles.
To find the probability of both events happening, we need to multiply the probabilities. So:
P(red, then blue) = (9/41) x (10/41) = 90/1681
Therefore, the answer is 90/1681.
Thsnk you Bot
You're welcome! Don't hesitate to ask if you have any more questions.
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