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anonymous

  • 5 years ago

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 .

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  1. anonymous
    • 5 years ago
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    What does with replacement mean?

  2. anonymous
    • 5 years ago
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    putting back in the jar

  3. anonymous
    • 5 years ago
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    Oh. I dunno about that, but without replacement is pretty easy.

  4. anonymous
    • 5 years ago
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    i understand more of the with replacement but if you can help with the without replacement that would be good

  5. anonymous
    • 5 years ago
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    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: \[\frac{{7 \choose 1} * {5 \choose 2}}{12 \choose 3}\]

  6. anonymous
    • 5 years ago
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    so 12/3?

  7. anonymous
    • 5 years ago
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    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)?

  8. anonymous
    • 5 years ago
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    kind of

  9. anonymous
    • 5 years ago
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    was that without replacement?

  10. anonymous
    • 5 years ago
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    Yes

  11. anonymous
    • 5 years ago
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    do you know how to do d?

  12. anonymous
    • 5 years ago
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    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)

  13. anonymous
    • 5 years ago
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    140/1320....can be reduced though right?

  14. anonymous
    • 5 years ago
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    14/132 = 7/66

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