Probability of a particle penetrating through a shield Suppose that the probability that any particle is emitted by a
  radioactive material will penetrate a certain shield is 0.01. If 10
  particles are emitted, what is the probability that exactly one
  particle will penetrate the shield?


The answer is $10(0.01)(0.99)^9$, and I'm confused on why we need to multiply it by $10$ when the 9th power seems to constitute the $9$ failures and then $0.01$ for $1$ success.
You multiply by 10 because any of the 10 particles can be the penetrating particle. 

In the case in which exactly one particle penetrates, there are (1 choose 10) ways to organize 1 penetrating particle and 9 non-penetrating particles.
For each particle in turn getting to be the one that penetrates the shield, while none of the others do:

$\begin{align}
\#1 &amp;: 0.01 \times 0.99^9 \\
\#2 &amp;: 0.99 \times 0.01 \times 0.99^8 \\
\#3 &amp;: 0.99^2 \times 0.01 \times 0.99^7 \\
\#4 &amp;: 0.99^3 \times 0.01 \times 0.99^6 \\
\#5 &amp;: 0.99^4 \times 0.01 \times 0.99^5 \\
\#6 &amp;: 0.99^5 \times 0.01 \times 0.99^4 \\
\#7 &amp;: 0.99^6 \times 0.01 \times 0.99^3 \\
\#8 &amp;: 0.99^7 \times 0.01 \times 0.99^2 \\
\#9 &amp;: 0.99^8 \times 0.01 \times 0.99 \\
\#10 &amp;: 0.99^9 \times 0.01  \\
\end{align}$

...all exclusive (selecting one option excludes all the other options). Obviously in each case the resulting chance is the same. Total probability then is the sum of these, $10\times  0.01 \times 0.99^9$

If the question were asking the chance that exactly two particles penetrated the shield, the same exercise for every possible pair would be rather dull and instead we should turn to the combinatorics option, using the binomial coeeficient directly: $\binom {10}{2} = \frac {10!}{2!\,8!} = \frac {10\cdot 9 }{2} = 45$ possible pairings, multiplied by the identical chance of each considered pair, $0.01^2\cdot 0.99^8$