anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
depends entirely on whether you replace the marble or not after the first one is drawn
anonymous
  • anonymous
The first one is not replaced.
anonymous
  • anonymous
oh wait NOT replaced lets go slow

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More answers

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

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