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iAmCool

  • 2 years ago

Jose is paying for an 11$meal using bills in his wallet. He has four 1$ bills, two 5$ bills, and two 10 $bills. If he selects two bills at random, one at a time from his wallet, what is the probability that he will choose a 10$ bill and a 1$ bill to pay for the meal? Show your work.

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  1. iAmCool
    • 2 years ago
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    @tcarroll010 Please help:)

  2. graydarl
    • 2 years ago
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    The probability is equal to \[=\frac{ 4 }{ 8 }\frac{ 2 }{ 7 }=\frac{ 1 }{ 7 }=0.142857\]

  3. graydarl
    • 2 years ago
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    14.28%

  4. iAmCool
    • 2 years ago
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    o: Thanks.. :)

  5. iAmCool
    • 2 years ago
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    I'm ready for the answer.. because the answer it wrong.

  6. iAmCool
    • 2 years ago
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    is*

  7. graydarl
    • 2 years ago
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    7 is because after u take out the first bill out of the total of 8 it leaves in the wallet only 7 bills. 4 is the number of $1bills and 2 the number of $10 bills

  8. graydarl
    • 2 years ago
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    Experiment 1: A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack? Analysis: The probability that the first card is a queen is 4 out of 52. However, if the first card is not replaced, then the second card is chosen from only 51 cards. Accordingly, the probability that the second card is a jack given that the first card is a queen is 4 out of 51. Conclusion: The outcome of choosing the first card has affected the outcome of choosing the second card, making these events dependent. Definition: Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed. Now that we have accounted for the fact that there is no replacement, we can find the probability of the dependent events in Experiment 1 by multiplying the probabilities of each event. Experiment 1: A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack? Probabilities: P(queen on first pick) = 4 52 P(jack on 2nd pick given queen on 1st pick) = 4 51 P(queen and jack) = 4 · 4 = 16 = 4 52 51 2652 663

  9. tcarroll010
    • 2 years ago
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    It's (4/8)(2/7)(2) because you could pick the bills in either of 2 orders, so it's 2/7, not 1/7

  10. graydarl
    • 2 years ago
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    http://www.mathgoodies.com/lessons/vol6/dependent_events.html

  11. iAmCool
    • 2 years ago
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    ...I'm going with tcarrol010

  12. graydarl
    • 2 years ago
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    i am sorry for the mistake

  13. iAmCool
    • 2 years ago
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    It's okay:) Everyone makes mistakes. c:

  14. tcarroll010
    • 2 years ago
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    np. we all learn

  15. tcarroll010
    • 2 years ago
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    Here: in a little more detail: upcoming

  16. tcarroll010
    • 2 years ago
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    (4/8)(2/7) + (2/8)(4/7) = 2/7

  17. iAmCool
    • 2 years ago
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    After you help me with this problem.. I need help with another one.. if you don't mind.

  18. tcarroll010
    • 2 years ago
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    Have to go make pumpkin pie for tomorrow.

  19. iAmCool
    • 2 years ago
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    Oh..okay. Thanks for your help though:)

  20. tcarroll010
    • 2 years ago
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    ok

  21. graydarl
    • 2 years ago
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    i gave u a medal cause iamcool gave his mine and i didn t provide the correct answer :D

  22. iAmCool
    • 2 years ago
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    Your so nice.. lol ^

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