anonymous
  • anonymous
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
Mathematics
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anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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ybarrap
  • ybarrap
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
anonymous
  • anonymous
Total 14, total dimes 10 so 10/14?
ybarrap
  • ybarrap
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

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ybarrap
  • ybarrap
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?
anonymous
  • anonymous
9/12?
ybarrap
  • ybarrap
Yes
ybarrap
  • ybarrap
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?
calculusxy
  • calculusxy
|dw:1434077226838:dw|
anonymous
  • anonymous
Thank you!
calculusxy
  • calculusxy
These are all dependent variables. Thus you have the denominators decreasing every time you take it out ("without replacement").
calculusxy
  • calculusxy
@NANABE You're welcome :)

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