# The machine works under the condition of simultaneous functioning of units A, B, C

**The machine works under the condition of simultaneous functioning of units A, B, C, which work independently of each other. The probability of breakage of these nodes is 0.2, 0.3, 0.1, respectively. what is the likelihood that the machine will fail?**

Probabilities of knot breakage: p1 = 0.2; p2 = 0.3; p3 = 0.1.

The probabilities that the nodes will work:

q1 = 1 – p1 = 1 – 0.2 = 0.8;

q2 = 1 – p2 = 1 – 0.3 = 0.7;

q3 = 1 – p3 = 1 – 0.1 = 0.9;

The probability that all nodes will work:

P (ABC) = q1 q2 q3 = 0.8 0.7 0.9 = 0.504.

The probability of an opposite event, such that at least one unit will fail, which will lead to a failure in the operation of the machine:

P (ABC) = 1 – P (ABC) = 1 – 0.504 = 0.496;

Answer: The probability that the machine will fail is 0.496.