Who knows how to solve a similar problem on the basis of discrete mathematics and logic?

Question

Discrete Math

The numbers A, B and C, independently of each other, take two values ​​0 and 1, and they take the value 1 with probability p, q and r, respectively.
What is the probability that the expression:
(A + B) => (A&&!B||C) is equal to 1?
A + B - exclusive OR
A => B - implication operation
Answer options:

1. 1-(1-pq)(p+qr)
   
2. 1-(p+q)(p+qr)
   
3. (p+q-2pq)(p+qr-pqr)
   
4. 1-q(1-p)(1-r)
   

Answer 1

The arguments have 8 state options, let's look at those in which the second column has a one

ABC   A⊕B   A&&!B||C
--------------------
000    0       1     пофиг
001    0       1     пофиг
010    1       0     жопа       q(1-p)(1-r)
011    1       1      ок
100    1       1      ок
101    1       1      ок
110    0       0     пофиг
111    0       1     пофиг

It is necessary to sum their probabilities, and even easier - subtract the probability of the combination 1 => 0
from one. Thus, the 4th answer.