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 ${B\over A}={{3\choose 1}{21\choose 3}\over{24\choose 4}}={ {3!\over 1!2!}{21!\over3! 18! }\over {24!\over 4! 20! }} ={ {\color{darkgreen}3}\cdot{21\cdot\color{darkgreen}{20}\cdot19\over\color{darkgreen}{3!} }\over {\color{maroon}{24}\cdot23\cdot22\cdot21\over \color{maroon}{4!} }} ={ {21\cdot\color{darkgreen}{10}\cdot19 }\over {\color{maroon}{1}\cdot 23\cdot22\cdot21 }} ={{ 5\cdot19 }\over { 23\cdot11 }}={95\over253} \approx .375.$