Answer:
16.15% probability that exactly 3 of them would function
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
Probability of each system working:
4 components, which means that 
Each has a 30% probability of failing, so 
For the system to work, all 4 components have to work. This is P(X = 4).


0.2401 probability of a system working.
If you have 7 of these systems, what is the probability that exactly 3 of them would function?
Now 7 systems, so 
0.2401 probability of a system working.
We have to find P(X = 3).


16.15% probability that exactly 3 of them would function