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:
$$P = P(ABC) + P(AC) + P(AB) + P(BC)$$
$$P = 0.7 * (0.8455 + 0.1445 + 0.0045) + 0.3 * 0.8455 = 0.9498$$
$(a)$ and $(b)$ are correct.
Why do I have to sum those cases why can't I just have single
calculation such as $0.92\cdot0.92\cdot0.08$ ?
That is the probability of getting an acceptable component twice in a row followed by a failed component. However, we must account for the $3$ different ways to place a failed component, as your sample space correctly shows.
the result is obviously three times less, but the method works for
a-part and essentially also b-part
That is because there is only one way to place $3$ successful components and similarly for placing $3$ failed components.
For $(c)$ summing the event space probabilities is one acceptable way of doing it. Namely
$$P(AAF)+P(AFA)+P(FAA)=3\cdot 0.92^2 \cdot 0.08\approx 0.203$$
You may also use the binomial distribution. The probability of $k$ successes in $n$ trials is given by
$$\begin{align*}
P(X=k)
&={n \choose k}p^k (1-p)^{n-k}\\\\
&={3 \choose 2}0.92^2\cdot 0.08\\\\
&\approx 0.203
\end{align*}$$
Again, for $(d)$ taking
$$P(AAA)+P(AAF)+P(AFA)+P(FAA)=\left(0.92^3\right)+\left(3\cdot 0.92^2 \cdot 0.08\right)\approx 0.982$$
is perfectly acceptable.
You could also do
$$P(X\geq 2)=\sum_{k=2}^3 {n \choose k}p^k (1-p)^{n-k}$$
and come to the same conclusion.
In R statistical software,
> sum(dbinom(2:3,3,.92))
[1] 0.981824
Note: Using the binomial distribution did not save us much (if any) time in this situation but you can imagine writing out all the possible scenarios for larger $n$ and $k$ would be quite tedious.
Best Answer
Among $6,1$ can fail in $\binom 61$ ways etc.
So, the required probability will be $$\sum_{1\le r\le 6}\binom 6 r\left(\frac1{10}\right)^r=1-\left(1-\frac1{10}\right)^6$$
Alternatively, the probability that at least one fails $=1-$ the probability that all succeed $$=1-\left(1-\frac1{10}\right)^6$$