combining probibilities I am working my way through Probability and Statistics for Computer Scientists 2nd ed.  And am stuck on how to answer the following question:
2.10 Three computer viruses arrived as an e-mail attachment. Virus A damages the system with probability 0.4. Independently of it, virus B damages the system with probability 0.5.  Independently of A and B, virus C damages the system with probability 0.2. 
What is the probability that the system gets damaged?
Since the events are not mutually exclusive I can not just add them but so to answer this don't I need more information?  i.e. Some values for the intersection of A,B & C so that the combination of these events adds up to less than 1?
P.S. I am not in any kind of course, and this is not an assignment.  I am doing it for my own interest.
Mark 
 A: "Since the events are not mutually exclusive I can not just add them but so to answer this don't I need more information? i.e. Some values for the intersection of A,B & C"
For independent events $U$ and $V$ you have $P(U \cap V) = P(U)P(V)$ 
The probability that the system gets damaged is 
$$
\begin{align}
P(A \cup B \cup C) & = P(A)+P(B)+P(C)  \\ 
 & - P(A \cap B)-P(A \cap C)-P(B \cap C) \\ 
 & + P(A \cap B \cap C) \\
 &  \\ 
 & =0.4+0.5+0.2-0.4 \cdot 0.5-0.4 \cdot 0.2-0.5 \cdot  0.2+0.4 \cdot 0.5 \cdot 0.2=0.76
\end{align}
$$

Here we are subtracting the double intersections. 
Maybe you are familiar with the formula
$P(U \cup V) = P(U) + P(V) - P(U \cap V)$
It is the same principle. Draw a Venn diagram if you need to convince yourself.
Finally we are adding $P(A \cap B \cap C)$ because we have subtracted too much.

This comes from the principle of inclusion-exclusion 
and is used for calculating the union of non-disjoint events.
A: Hint: Find first the probability the system does not get damaged. This is the probability that Virus A causes no damage and Virus B causes no damage and Virus C causes no damage.
Remark: You could find the intersection probabilities you refer to by using independence, but that is more work.
