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anonymous

  • one year ago

A bag contains 10 dimes and 4 quarters. You are going to select three coins at random, one at a time, without replacement. Find the probability of drawing a dime, then a quarter, then another quarter. 3/14 20/273 5/91 15/343

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  1. ybarrap
    • one year ago
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    How many coins are there total and how many dimes? Chance of getting a dime on the first then = #dimes/total number of coins Get this first

  2. anonymous
    • one year ago
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    Total 14, total dimes 10 so 10/14?

  3. ybarrap
    • one year ago
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    Yes. Now, we need a dime first AND quarter second AND lastly a quarter, since these are independent, we will be multiplying. After we get our dime, how many coins are left and how many quarters are available? This will give us #quarters/total , this is the next term

  4. ybarrap
    • one year ago
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    You should have 4 quarters and 13 coins available, since we took one out on the 1st draw: $$ \cfrac{10}{14}\times\cfrac{4}{13}\times ? $$ What's the last term?

  5. anonymous
    • one year ago
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    9/12?

  6. ybarrap
    • one year ago
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    Yes

  7. ybarrap
    • one year ago
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    Multiply and you get - http://www.wolframalpha.com/input/?i=%2810*4*9%29%2F%2814*13*12%29 Which is close to one of your options - maybe you left out a digit?

  8. calculusxy
    • one year ago
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    |dw:1434077226838:dw|

  9. anonymous
    • one year ago
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    Thank you!

  10. calculusxy
    • one year ago
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    These are all dependent variables. Thus you have the denominators decreasing every time you take it out ("without replacement").

  11. calculusxy
    • one year ago
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    @NANABE You're welcome :)

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