## 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

1. 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

2. anonymous

Total 14, total dimes 10 so 10/14?

3. 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

4. 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?

5. anonymous

9/12?

6. ybarrap

Yes

7. 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?

8. calculusxy

|dw:1434077226838:dw|

9. anonymous

Thank you!

10. calculusxy

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

11. calculusxy

@NANABE You're welcome :)