Solve for the overall reliability of the system pictured in the attached diagram.

This diagram is presented without much context, but on what I believe is the usual interpretation, we can think of these as like "wiring", with the nodes that are wired "in series" being dependent on each other (if any fails, the system fails) while the nodes that are wired "in parallel" are redundant (as long as at least one does not fail, the system can continue functioning). Then, the numbers in each box are the reliability of each node, given as its probability of functioning correctly.

If that is indeed the intended interpretation, we would solve as follows.

Starting at the left, I'll label the nodes alphabetically. First we have 2 nodes in parallel, A is 0.99 and B is 0.98. The probability of at least one functioning properly is given by adding up the probability of each, then subtracting the probability of both (because it was double-counted):

`P(A or B) = P(A) + P(B) - P(A and B)`

`P (A or B) = 0.99 + 0.98 - (0.99)(0.98) = 1.97 - 0.9702 = 0.9998`
This first pair of nodes is 99.98% reliable.

The next 2 nodes in parallel are P(C) = 0.95 and P(D) = 0.90 respectively, and we do the same thing to find out how reliable they are as a redundant system:
`P(C or D) = 0.95 + 0.90 - (0.95)(0.90) = 1.85 - 0.855 = 0.995`
Then we want to combine these two probabilities; in order to make it through the first two layers, we need both A or B and C or D to work, so these probabilities are multiplied:
`P((A or B) and (C or D)) = (0.9998)(0.995) = 0.99301`

Now we have the next three nodes in parallel, which is a bit trickier but the same basic process; just remember that P(E or F or G) is the same as P((E or F) or G).

`P(E or F) = P(E) + P(F) - P(E)*P(F) `
`P((E or F) or G) = P(E or F) + P(G) - P(E or F)*P(G)`
`P(E or F or G) = P(E) + P(F) - P(E)*P(F) + P(G) - P(E)*P(G) - P(F)*P(G) + P(E)*P(F)*P(G)` `P(E or F or G) = 0.92 + 0.95 - (0.92)(0.95) + 0.98 - (0.92)(0.98) - (0.95)(0.98) + (0.92)(0.95)(0.98)` `P(E or F or G) = 1.87 - 0.874 + 0.98 - 0.9016 - 0.931 + 0.85652` `P(E or F or G) = 0.99992`

This third layer is 99.992% reliable. (Notice how redundancy dramatically improves reliability.)

Next we combine that with the first two layers, again multiplying because the first two layers must succeed and the third layer must succeed:

`P((A or B) and (C or D) and (E or F or G)) = (0.99301)(0.99992) = 0.9929305592`

Now all that remains is to multiply this by the reliability of the final node H, which is 0.92:

`P((A or B) and (C or D) and (E or F or G) and H) = (0.9929305592)(0.92) = 0.913496114464`

I kept all the decimals for the intermediate calculations, but it's still to put them in the final answer, so let's just do four significant figures.

The reliability of the whole system is 0.9135, which is 91.35%.