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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- 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

Total 14, total dimes 10 so 10/14?

- 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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## More answers

- 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

9/12?

- ybarrap

Yes

- 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

|dw:1434077226838:dw|

- anonymous

Thank you!

- calculusxy

These are all dependent variables. Thus you have the denominators decreasing every time you take it out ("without replacement").

- calculusxy

@NANABE You're welcome :)

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