The probability of obtaining a pair of tails in one throw of two fair coins is \(0.25\) (since there are four equally likely outcomes: HH, HT, TH, TT, and only TT is a pair of tails).

Let \(X\) be the number of times a pair of tails is obtained in 80 throws. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.25\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 80 \times 0.25 = 20\)

\(\sigma^2 = np(1-p) = 80 \times 0.25 \times 0.75 = 15\)

We need to find \(P(X > 25)\). Using the normal approximation,

\(P(X > 25) \approx P\left(Z > \frac{25.5 - 20}{\sqrt{15}}\right)\)

\(= P(Z > 1.42)\)

Using standard normal distribution tables, \(P(Z > 1.42) = 1 - P(Z \leq 1.42) = 1 - 0.9222 = 0.0778\).