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anonymous

  • one year ago

An urn contains six red balls, five white balls, and four black balls. Four balls are drawn from the urn at random without replacement. For each red ball drawn, you win $10, and for each black ball drawn, you lose $15. Let X represent your net winnings. Compute E(X), your expected net winnings.

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  1. anonymous
    • one year ago
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    Hya

  2. anonymous
    • one year ago
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    hello

  3. anonymous
    • one year ago
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    I'm not good at this :P

  4. anonymous
    • one year ago
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    I'll give you a medal :D

  5. anonymous
    • one year ago
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    I need help not medal lol

  6. anonymous
    • one year ago
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    :P

  7. misty1212
    • one year ago
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    HI!!

  8. misty1212
    • one year ago
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    i cannot think of a snappy way to do this, this is going to take like forever

  9. misty1212
    • one year ago
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    first you have to think of all the possible outcomes when you pick 4 balls

  10. misty1212
    • one year ago
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    then you have to compute the amount of money you win or lose for each case then you have to find the probability of each possible outcome then you have to multiply and add !

  11. anonymous
    • one year ago
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    could you lead me through the problem step by step please.

  12. misty1212
    • one year ago
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    if i had like two hours i could lets at least begin

  13. anonymous
    • one year ago
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    sure thing thanks

  14. misty1212
    • one year ago
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    one possibility is you get all red and therefore win $40 the probability you get all red is \[\frac{\binom{6}{4}}{\binom{4}{15}}\]

  15. misty1212
    • one year ago
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    ok that is not quite right, the probabilty is \[\frac{\binom{6}{4}}{\binom{15}{4}}\]

  16. misty1212
    • one year ago
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    that is \(\frac{1}{91}\) http://www.wolframalpha.com/input/?i=%286+choose+4%29%2F%2815+choose+4%29

  17. misty1212
    • one year ago
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    then you could have 3 red, one white, win $30 that probability is \[\frac{\binom{6}{3}\times \binom{5}{1}}{\binom{15}{4}}\]

  18. misty1212
    • one year ago
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    this is really just a start there are lots of other possibilities i can't think of a quick way to do this though

  19. anonymous
    • one year ago
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    so if i get this right this is how it goes. P(1 red)=6C1×9C315C4 P(2 red)=6C2×9C215C4 P(3 red)=6C3×9C115C4 P(4 red)=6C415C4 P(1 red) = 0.369; P(2 red) = 0.396; P(3 red) = 0.132; P(4 red) = 0.011.

  20. anonymous
    • one year ago
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    the problem gets tough here on out i don't know what to do from here

  21. anonymous
    • one year ago
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    P(1 red)=6C1×9C3/15C4 P(2 red)=6C2×9C2/15C4 P(3 red)=6C3×9C1/15C4 P(4 red)=6C4/15C4 correct way

  22. anonymous
    • one year ago
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    @kropot need your help from this part

  23. anonymous
    • one year ago
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    i didn't understand anything from here on out \

  24. kropot72
    • one year ago
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    P(1 black) = 0.484; P(2 black) = 0.242; P(3 black) = 0.32; P(4 black) = 0.001 E(X) = 10(0.369 + 0.396 + 0.132 + 0.011) - 15(0.484 + 0.242 + 0.032 + 0.001)

  25. kropot72
    • one year ago
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    E(X) = a loss of $2.41.

  26. anonymous
    • one year ago
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    @kropot72 but we need to find the net winnings not loss

  27. kropot72
    • one year ago
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    Expected net winnings = -$2.41.

  28. kropot72
    • one year ago
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    "Net winnings" means (expected gains) - (expected losses). In this case the expected losses outweigh the expected gains.

  29. anonymous
    • one year ago
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    give me medal?

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