The probability of landing on a 5 in one spin is \(P(5) = 0.2\) since the spinner is fair and has 5 sides.

We need to find the probability that it takes fewer than 7 spins to get a 5. This is the complement of not getting a 5 in all 6 spins.

The probability of not getting a 5 in one spin is \(1 - 0.2 = 0.8\).

The probability of not getting a 5 in 6 spins is \(0.8^6\).

Therefore, the probability of getting at least one 5 in fewer than 7 spins is:

\(P(X < 7) = 1 - 0.8^6\)

Calculating this gives:

\(1 - 0.8^6 = 1 - 0.262144 = 0.737856\)

Thus, the probability that it takes fewer than 7 spins for George to obtain a 5 is approximately 0.73856.