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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. (a) P(3 red marbles) With replacement , without replacement . (b) P(3 black marbles) With replacement , without replacement . (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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    a) with replacement: P(3 red) = P(1st red). P(2nd red). P(3rd red)=(7/10)(7/10)(7/10)=343/1000

  2. anonymous
    • 5 years ago
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    these are the answers I got I just wanted to get them checked and see if they are right? a) with 343/128 without 21/132 b)125/1728 without 60/132 c) I did not know how to do this one D) with 175/1728 without 14/132

  3. anonymous
    • 5 years ago
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    wouldnt it be 7/12? not 7/10?

  4. anonymous
    • 5 years ago
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    Oh sorry over 12 :(

  5. anonymous
    • 5 years ago
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    ok could you possibly tell me if my answers are right? and possibly how to do C?

  6. anonymous
    • 5 years ago
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    for a with replacement: P(3 red)=(7/12)^3

  7. anonymous
    • 5 years ago
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    without replacement: P(3 red)=(7/12)(6/11)(5/10)=....

  8. anonymous
    • 5 years ago
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    Use the same method to solve B.

  9. anonymous
    • 5 years ago
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    ya I did that, but I just wanted to make sure i got the right answers and simplified right?

  10. anonymous
    • 5 years ago
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    a) with replacement: P(3 red)=343/1728 = 0.1984

  11. anonymous
    • 5 years ago
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    without replacement: P(3 red)=7/44 = 0.159

  12. anonymous
    • 5 years ago
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    what about the others?

  13. anonymous
    • 5 years ago
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    B) with replacement: P(3 black)=(5/12)^3=125/1728

  14. anonymous
    • 5 years ago
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    without replacement: P(3 black)=(5/12)(4/11)(3/10)=1/22

  15. anonymous
    • 5 years ago
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    how did you get 1/22?

  16. anonymous
    • 5 years ago
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    Calculator.

  17. anonymous
    • 5 years ago
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    You can do it by hand though.

  18. anonymous
    • 5 years ago
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    do you know how to do c?

  19. anonymous
    • 5 years ago
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    Sorry, I had to go. In part (c), we want to get one red and two black marbles. There are three possibilities for this, red comes first, second or last. if R denotes red marbles and B denotes black marbles, then: i) with replacement: P(one red and two black)=P(R B B)+P(B R B)+P(B B R) =(7/12)(5/12)(5/12)+(5/12)(7/12)(5/12)+(5/12)(5/12)(7/12) =3(7/12)(7/12)(5/12)= (use a calculator).

  20. anonymous
    • 5 years ago
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    Try the same way and do the "without replacement" part.

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