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

Question

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

ybarrap · Answer

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 · Answer

Total 14, total dimes 10 so 10/14?

ybarrap · Answer

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

ybarrap · Answer

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

anonymous · Answer

9/12?

ybarrap · Answer

Yes

ybarrap · Answer

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 · Answer

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