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anonymous

  • one year ago

can someone help please Jose is paying for a six dollar meal using bills in his wallet. He has four one dollar bills, three five dollar bills, and two ten dollar bills. If he selects two bills at random, one at a time from his wallet, what is the probability that he will choose a one dollar bill and a five dollar bill to pay for the meal? Show your work.

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  1. anonymous
    • one year ago
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    @LegendarySadist

  2. anonymous
    • one year ago
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    can you guys help

  3. anonymous
    • one year ago
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    I'm here to see how someone else solves it - Probability was never my best subject

  4. anonymous
    • one year ago
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    agreed

  5. anonymous
    • one year ago
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    i could be totally wrong but ill take a guess

  6. anonymous
    • one year ago
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    i know this math subject sucks

  7. anonymous
    • one year ago
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    hate probability like i do not get my teacher explains and i am like uhhhh

  8. anonymous
    • one year ago
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    total is 4 + 3 + 2 = 9, first pull he has 4 / 9 chance to pull one dollar bill now total bills is 8 second pull he has 3/8 chance to to pull a five dollar bill so 4/9 * 3/8 = answer?

  9. anonymous
    • one year ago
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    its competition who will win

  10. anonymous
    • one year ago
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    that might be wrong i dont know

  11. anonymous
    • one year ago
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    But what if he gets a five dollar bill then a one dollar?

  12. anonymous
    • one year ago
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    ^^

  13. anonymous
    • one year ago
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    seems right i know you have to add first

  14. anonymous
    • one year ago
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    How many bills does he have in total? 4 + 2 + 2 = 8 The probability he picks a $10 bill first is therefore (2/8) = (1/4) The probability he picks a $10 bill second, given he didn't pick one first, is (2/7) because there are still two $10 bills in his wallet but only seven bills left Similarly, the probability he picks a $1 bill first is (4/8) = (1/2) The probability he picks a $1 bill second, given he didn't pick one first, is (4/7) because there are still four $1 bills in his wallet but only seven bills left So the probability he picks exactly $11 out with a $1 and $10 bill is the probability either he picks $10 then $1, or $1 then $10, which is: (1/4) * (4/7) + (1/2) * (2/7) = (1/7) + (1/7) = 2/7

  15. anonymous
    • one year ago
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    Isn't it 4+3+2=9?

  16. anonymous
    • one year ago
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    @nicxle.62 thats not what the question is asking

  17. anonymous
    • one year ago
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    uhhh okay um i think @LegendarySadist is right

  18. anonymous
    • one year ago
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    beause you do have to add

  19. anonymous
    • one year ago
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    woah woah, I'm not saying that's the solution

  20. anonymous
    • one year ago
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    no i know you have to add first then do the rest ya i knows but i suck at this

  21. anonymous
    • one year ago
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    So I think I have the question figured out.

  22. anonymous
    • one year ago
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    i know how to add this because i saw this in other questions like this but different XD

  23. anonymous
    • one year ago
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    There are two things that can happen. Either he draws a one and then a five, or he draws a five and then a one.

  24. anonymous
    • one year ago
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    So we have to test two things out. The chance of him getting a one then five and the chance of him getting a five then one.

  25. anonymous
    • one year ago
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    um kind of lost :/

  26. anonymous
    • one year ago
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    \[\frac{ 4 }{ 9 }*\frac{3}{8}\] will get you the probability of a one then five\[\frac{3}{9}*\frac{4}{8}\]will get you the probability of a five and then one.

  27. anonymous
    • one year ago
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    so basically what i said earlier

  28. anonymous
    • one year ago
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    ha XD

  29. anonymous
    • one year ago
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    Yeah, somewhat. The problem I'm thinking about is whether or not we're supposed to add these two probabilities. Since the question wants whether he gets a one and five, but doesn't give an order, it could be either of these.

  30. anonymous
    • one year ago
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    so will it be 0.16

  31. anonymous
    • one year ago
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    on the first one

  32. anonymous
    • one year ago
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    No need to convert to decimal form. Let's keep it in fraction form please.

  33. anonymous
    • one year ago
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    okay so will it be 1/6

  34. anonymous
    • one year ago
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    on both

  35. anonymous
    • one year ago
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    Yup.

  36. anonymous
    • one year ago
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    okay is that it or not

  37. anonymous
    • one year ago
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    thanks for your help guys

  38. anonymous
    • one year ago
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    @puppylife101 I did some testing to confirm my hypothesis. You ARE supposed to add the two probabilities.

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