We have the formula to compute the probability of having exactly k successed over n trials, given a probability p of success (and implicitly a probability 1-p of failure), which is
[tex] P(\text{k successed on n trials}) = \binom{n}{k}p^k(1-p)^{n-k} [/tex]
Now, the probability of at least 3 successes is the union of the following event: exactly three successes,exactly four successes and exactly five successes.
We can compute their probability and sum them:[tex] \binom{5}{3}\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^2 + \binom{5}{4}\left(\frac{3}{7}\right)^4\left(\frac{4}{7}\right)^1 + \binom{5}{5}\left(\frac{3}{7}\right)^5 \approx 0.36788 [/tex]
So, the answer is about 36.79%
