Hint: The probability that it works correctly is equal to $1$ minus the probability that it doesn't work correctly.
The only cases where it doesn't work correctly are (I) exactly 0 of the 4 components work correctly, and (II) exactly 1 of the 4 components work correctly. So compute $P(\text{Case I})$ and $P(\text{Case II})$; the answer is
$$
1 - P(\text{Case I}) - P(\text{Case II})
$$
To deal with Case II, note that there are four ways that exactly one of them can work - the first one can work, or the second one can work, or the third, or the fourth.
If the probabilities of the three components $A$, $B$, $C$ were independent of each other (in other words: uncorrelated), then it would indeed be correct to simply multiply the probabilities associated with $ABC$, $AB$, $AC$ and $BC$ working. Followed by adding up these four numbers, giving you the chance that least two components work.
However, as clearly stated in the text, the probabilities for components $B$ and $C$ are interdependent! This means that, before you can do a calculation as in the previous paragraph, you must first explore fully the interdependence of these probabilities.
Okay, let us do this. It is given that $C$ has a failure chance of $0.01$. Furthermore it is given that if $C$ fails, then the chance that $B$ fails is $0.55%$. This leads us to the conclusion that the probability that both $C$ and $B$ fail is equal to $0.01 * 0.55 = 0.0055$.
With a bit more effort we can derive that the probability that $C$ fails and $B$ works is equal to $0.0045$. The chance that $C$ works and $B$ fails is $0.1455$. The chance that both $C$ and $B$ work is $0.8455$. We now have the four values that determine the interdependence of $C$ and $B$.
Finally we can compute the overall probability that the system works. We get:
$$P = P(ABC) + P(AC) + P(AB) + P(BC)$$
$$P = 0.7 * (0.8455 + 0.1445 + 0.0045) + 0.3 * 0.8455 = 0.9498$$
Best Answer
All three components working happens with probability $p^3$.
Exactly two components working happens with probabiliy $3p^2(1-p)$.
So at least two components working happens with probability $p^3+3p^2(1-p)=3p^2-2p^3$.
Now, since $p$ itself is randomly distributed, our answer is just $\displaystyle \int_{0}^{1}(3p^2-2p^3)\,dp=0.5$.
In fact, this is kind of intuitive. There is the same chance that $2$ or $3$ components fail, as there are $2$ or $3$ components work.