anonymous
  • anonymous
Assume a jar has 7 red marbles and 5 black marbles. Draw out 3 marbles with and without replacement. Find the requested probabilities. (c) P(one red and two black marbles) With replacement , without replacement . (d) P(red on the first draw and black on the second draw and black on the third draw) With replacement , without replacement .
Mathematics
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anonymous
  • anonymous
Assume a jar has 7 red marbles and 5 black marbles. Draw out 3 marbles with and without replacement. Find the requested probabilities. (c) P(one red and two black marbles) With replacement , without replacement . (d) P(red on the first draw and black on the second draw and black on the third draw) With replacement , without replacement .
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
What does with replacement mean?
anonymous
  • anonymous
putting back in the jar
anonymous
  • anonymous
Oh. I dunno about that, but without replacement is pretty easy.

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anonymous
  • anonymous
i understand more of the with replacement but if you can help with the without replacement that would be good
anonymous
  • anonymous
For the first question of one red and two black we just want to know how many ways we can choose one red, and two black divided by how many ways we can choose 3 marbles from 12. So: \[\frac{{7 \choose 1} * {5 \choose 2}}{12 \choose 3}\]
anonymous
  • anonymous
so 12/3?
anonymous
  • anonymous
No, 7 choose 1 = 7 5 choose 2 = 10 12 choose 3 = 220 So it is 70/220. Are you familiar with the choose function (and it's notation)?
anonymous
  • anonymous
kind of
anonymous
  • anonymous
was that without replacement?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
do you know how to do d?
anonymous
  • anonymous
For the other one it's just (Probability of choosing a red)*(Probability of choosing a black)*(Probability of choosing a black) So (7/12) * (5*/11) * (4/10)
anonymous
  • anonymous
140/1320....can be reduced though right?
anonymous
  • anonymous
14/132 = 7/66

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