# 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!!

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That's right. Redo your arithmetic... – David Mitra Apr 25 '12 at 2:44
For the probability to get exactly one dead battery (3C1 * 21C3)/(24C4) is correct, that is 37.5494%. – Byron Schmuland Apr 25 '12 at 2:50
I assumed it meant exactly one. But, it is ambiguous. In any case, $37.5\%$ would be approximately the probability of selecting exactly one, and the expression given in the post is correct. – David Mitra Apr 25 '12 at 2:51
goo.gl/Zu4jL – leonbloy Apr 25 '12 at 2:51

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

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