In this system, the reliability of each component is the complement of its failure probability. Given that the failure probability of each component is $0.64$, the reliability (probability that each component works) is $1 - 0.64 = 0.36$.

Analyzing the Network:

The system has components arranged both in series and parallel. The reliability rules are:

1. Series components: If components are in series, all must work for the system to work. The overall reliability is the product of their individual reliabilities.

2. Parallel components: If components are in parallel, only one needs to work for the system to work. The overall reliability is $1 - \text{(failure probability of all)}$.

Step 1: Reliability of the two parallel branches

Each of the two branches between $A$ and $B$ consists of 2 components in series. The reliability of a series connection is: $R_{\text{series}} = 0.36 \times 0.36 = 0.1296$ This applies to both branches.

Step 2: Reliability of the entire system

Now, the two branches are in parallel. For parallel reliability: $R_{\text{parallel}} = 1 - (1 - R_{\text{series}}) \times (1 - R_{\text{series}})$ Substituting $R_{\text{series}} = 0.1296$: