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Pulsified333

  • one year ago

Assume that you have 5 red balls and 4 blue balls from which randomly to select two without replacement. (1) What is the probability that exactly one red ball is chosen? (2) What is the probability that the second ball is blue given that at least one of the balls is blue?

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  1. anonymous
    • one year ago
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    there are a couple ways to do this perhaps the most intuitive way is to calculate the probability that the first ball is red and the second is not then the probability that the first is not red, and the second is do you know how to do that?

  2. Pulsified333
    • one year ago
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    i just got the first part but the second part I'm not sure how to do

  3. anonymous
    • one year ago
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    guess we need some notation

  4. anonymous
    • one year ago
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    actually a diagram would be bets lets see if i can draw one

  5. Pulsified333
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    actually no, lets just do it easy you have already computed the probability that exactly one ball is blue right?

  7. anonymous
    • one year ago
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    what did you get, and how did you do it?

  8. Pulsified333
    • one year ago
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    I got 5/9

  9. anonymous
    • one year ago
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    the probability that both balls are blue is \[\frac{4}{9}\times \frac{3}{8}=\frac{1}{6}\]

  10. anonymous
    • one year ago
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    the probability that the first is blue and the second is red is \[\frac{4}{9}\times \frac{5}{8}=\frac{5}{18}\]

  11. anonymous
    • one year ago
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    and the probability that the first is red and the second is blue is the same \[\frac{5}{18}\] add these up to get the probability that at least one is blue

  12. anonymous
    • one year ago
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    you need that for your denominator

  13. Pulsified333
    • one year ago
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    so 1/6+5/18

  14. anonymous
    • one year ago
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    no it is \[\frac{1}{6}+\frac{5}{18}+\frac{5}{18}\]

  15. Pulsified333
    • one year ago
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    oh

  16. Pulsified333
    • one year ago
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    the answer would be 13/18

  17. anonymous
    • one year ago
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    ok good so that is your denominator

  18. anonymous
    • one year ago
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    your numerator is the probability that the second ball is blue

  19. Pulsified333
    • one year ago
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    so 4/9

  20. anonymous
    • one year ago
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    yes

  21. Pulsified333
    • one year ago
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    8/13?

  22. anonymous
    • one year ago
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    that is what i get, yes

  23. Pulsified333
    • one year ago
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    :D it was right thanks man.

  24. anonymous
    • one year ago
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    yw nothing like instant gratification !

  25. Pulsified333
    • one year ago
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    btw if i want to send you a message how do i do that?

  26. anonymous
    • one year ago
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    you cannot

  27. Pulsified333
    • one year ago
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    ;(

  28. anonymous
    • one year ago
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    you can tag me in a question if you like @satellite73

  29. Pulsified333
    • one year ago
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    okay I'm going to be posting a lot of problems tonight because I have a midterm tomorrow that I have not even studied for yet and I pulled an all nighter last night to work on a presentation that I had to present today and study for a test that i took today.

  30. anonymous
    • one year ago
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    yikes

  31. Pulsified333
    • one year ago
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    @satellite73 I know. Right now I'm scared that I am going to have to stop doing my math hw and study for the test and take the fail on the hw

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