Probability related math question Im having trouble with one of the problems on a review sheet for a test on thursday, the sheet provides an answer but not a process, and I would like to know how to solve it and why whatever your method is works. The problem is as follows: 

A box of 24 AAA-batteries contains 3 dead batteries. If you randomly select 4 from this box for use in your Wii controller, what is the probability that you select one dead battery?

The answer provided is 37.5% but I have no idea how to do this problem. I tried doing:
(3C1 * 21C3)/(24C4) but unfortunately that was not the correct answer. Thank you so much for taking the time to help me!!
 A: Assuming that we want the probability that exactly one dead battery is amongst the four chosen:
Since outcomes are equally likely, you need to 
$\ \ \ $1) Find the total number of ways to select four batteries; call this number $A$.
$\ \ \ $2) Find the total number of ways to select four batteries with exactly one of the four a dead battery; call this number $B$.
$\ \ \ $3) Calculate the probability as $B/A$.
Calculating $A$ is easy: you are choosing four objects from amongst 24 distinct objects without regard to order. So $A={24\choose 4}$ (recall ${n\choose k}={n!\over k!(n-k)!}$).
To calculate $B$, we will use the:
Multiplication Principle: if there are $n$ outcomes resulting from performing one task, and if for each of those outcomes there are $m$ outcomes resulting from performing another task, then the total number of outcomes resulting from performing the two tasks in succession is $n\cdot m$.
To count the number of ways to choose four batteries with exactly one dead, we first choose the dead battery and then choose three good batteries.  There are ${3\choose 1}$ ways to choose a dead battery. Once we've choosen a dead battery, there are still 21 good batteries, and thus there are ${21\choose 3}$ ways to choose the three good batteries. 
By the multiplication principle then, the number of ways to choose four batteries with exactly one dead is $B={3\choose 1}{21\choose 3}$.
And thus, the probability of choosing four batteries with exactly one dead is ${B\over A}=
{{3\choose 1}{21\choose 3}\over{24\choose 4}}$.
Computing the latter quantity gives 
$${B\over A}={{3\choose 1}{21\choose 3}\over{24\choose 4}}={ {3!\over 1!2!}{21!\over3! 18!  }\over {24!\over 4! 20! }} 
={ {\color{darkgreen}3}\cdot{21\cdot\color{darkgreen}{20}\cdot19\over\color{darkgreen}{3!}   }\over {\color{maroon}{24}\cdot23\cdot22\cdot21\over \color{maroon}{4!} }} 
={  {21\cdot\color{darkgreen}{10}\cdot19   }\over {\color{maroon}{1}\cdot  23\cdot22\cdot21  }} 
={{   5\cdot19   }\over {  23\cdot11  }}={95\over253}   \approx .375.$$
