There are 6 red marbles, 9 blue marbles, and 10 green marbles in a bag. What is the theoretical probability of randomly drawing a red marble and then a green marble?

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There are 6 red marbles, 9 blue marbles, and 10 green marbles in a bag. What is the theoretical probability of randomly drawing a red marble and then a green marble?

Mathematics
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depends entirely on whether you replace the marble or not after the first one is drawn
The first one is not replaced.
oh wait NOT replaced lets go slow

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Other answers:

what is the probability the first one is red?
i hope that is more or less obvious it is the ratio of the number or red marbles to the total number of marbles
24%?
There are 4 answers: 10% 9.6% 16% 64%
not really interested in percents before we compute these numbers we need fractions first, then we can turn the final answer in to a percent
\[\frac{ 24 }{100}?\]
lets go slow how many red marbles are there?
6
yes, and how many marbles are there total?
25
right so the probability of picking a red marble first is \[\frac{6}{25}\]
now the red marble is chosen, and apparently not replaced how many green marbles are there ?
10
and how many marbles are left in the bag?
24
so the probability of picking a green marble once one red one is removes is?
\[\frac{ 10 }{ 24 }\]?
yes
to compute the probability that both things occur, multiply
\[\frac{6}{25}\times \frac{10}{24}\]
\[\frac{ 60 }{ 600 }\]
better known on planet earth as \(\frac{1}{10}\)
or \(0.1\) or even \(10\%\)
So that's the answer?

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