Let the probability of passing the test be denoted by \(p = 0.3\) and the probability of failing be \(1 - p = 0.7\). We are looking for the probability that at least three out of five friends pass the test. This can be calculated using the binomial probability formula:

\(P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)\)

Using the binomial formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n = 5\):

\(P(X = 3) = \binom{5}{3} (0.3)^3 (0.7)^2 = 10 \times 0.027 \times 0.49 = 0.1323\)

\(P(X = 4) = \binom{5}{4} (0.3)^4 (0.7)^1 = 5 \times 0.0081 \times 0.7 = 0.02835\)

\(P(X = 5) = \binom{5}{5} (0.3)^5 (0.7)^0 = 1 \times 0.00243 \times 1 = 0.00243\)

Adding these probabilities gives:

\(P(X \geq 3) = 0.1323 + 0.02835 + 0.00243 = 0.16308\)