Basic probability counting

I'm trying to self-study probability, and I'm trying to solve a problem (from Walpole) but, since don't have the solution inside don't know how to validate if my reasoning is fine.

Problem:

There are 16 army troops that need to protect 8 oil towers (numbered from A to H). They lost contact with each other. Supposing that each army troop has equal probability to protect any of the tower.

Event A: Which is the probability that at least 4 army troops end protecting tower A.

Solution:

Given that it doesn't matter the order (is the same to say alpha troop & beta troop that beta troop & alpha troop) then to count I can use Combinations. In particular $\binom{16}{8}$ (16 army troops combined within 8 towers) for the total of sample points.

Now, given that 4 army troops end up protecting tower A it means $\binom{4}{1}$ (4 army troops within 1 tower) and $\binom{12}{7}$ for the other towers (there remain 7 towers and 12 troops), so using simple probability:

$$P(A) = \frac{\binom{4}{1}\binom{12}{7}}{\binom{16}{8}}$$

Is this reasoning right? I'm new to this kind of stuff and I don't know if I pick the right book but, these counting thing is giving me some headaches.

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Your number of sample points is $8^{16}$ as each troop has eight places it can defend and these are multiplied. For event A, four of the troops have $1$ choice of where to defend and the other twelve have $7$ choices. This is another way to see my expression. The combinations only refer to selecting which set of four troops defend A, not which tower they defend. – Ross Millikan Aug 21 '13 at 23:38

I will take it that you want exactly four troops protecting tower A. If you mean at least four, you will need to adapt the analysis to cover five through eight and add them up. Then there are ${16 \choose 4}$ ways to pick the four troops that defend tower A. Each of the four has a probability of $\frac 18$ to defend A and the other twelve have a probability of $\frac 78$ of defending somewhere else. So the total probability is ${16 \choose 4}\left(\frac 18\right)^4\left(\frac 78\right)^{12}$ which Alpha gives as about $8.95\%$

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Thanks Ross for your answer. I have one question. Can you recommend me a book to learn how to count this things. The one I'm using is not good for counting. Thank you very much. – Randolf Rincón Fadul Aug 21 '13 at 23:40
Sorry, it has been a long time since I learned this and I don't remember what book I used. Can you see how to get the chance of five defending A? – Ross Millikan Aug 21 '13 at 23:42
This is a video from Kahn Academy related to this type of problem (which comes up a lot) and how approach it – Omnomnomnom Aug 21 '13 at 23:44
@RossMillikan It will be 16C5 * (1/8)^5 * (7/8)^11 ? – Randolf Rincón Fadul Aug 21 '13 at 23:45
@Omnomnomnom Thanks, I'll watch it right know! – Randolf Rincón Fadul Aug 21 '13 at 23:45

The problem asks for the probability of at least $4$ at the first tower, meaning $4,5,6,\dots,16$.

It is easier to find the probability $w$ that there are fewer than $4$. Then the answer to the original problem is $1-w$.

The probability there are exactly $k$ at the first tower is $\dbinom{16}{k}p^k(1-p)^{16-k}$, where $p=\dfrac{1}{8}$.

To find $w$, use $k=0,1,2,3$ and add up.

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