The probability of success (p) = \(\frac{3}{5}\) and the probability of failure (q) = 1 - \(\frac{3}{5} = \frac{2}{5}\)
The probability that at least two of them failed is
P(x \(\geq\) 2) = 1 - p(x < 2) = 1 - [P(x = 0)] + [P(x = 1)]= 1 - [(\(^{10}_{0}\)) (\(\frac{2}{5}\))\(^0\) (\(\frac{3}{5}\))\(^{10}\) + (\(^{10}_1\)) (\(\frac{2}{5}\))\(^1\) (\(\frac{3}{5}\))\(^9\)]
Which simplified to 1 - 0.046357401 = 0.9536 correct to four decimal places

(b) The probability that exactly half of them passes is P(x = 6) = (\(^{10}_{5}\)) (\(\frac{3}{5}\))\(^5\) (\(\frac{2}{5}\))\(^{5}\)
Simplified to give; 252 x 0.7776 x 0.01024 = 0.2007 correct to four decimal places

(c) The probability that at most two of them failed is
P(x \(\leq\) 2) = P(x = 0) + P(x = 1) + P(x = 2). In this case p = \(\frac{2}{5}\) q = \(\frac{3}{5}\) so that
p(x \(\leq\) 2) = (\(^{10}_{0}\)) (\(\frac{2}{5}\))\(^0\) (\(\frac{3}{5}\))\(^{10}\) + (\(^{10}_1\)) (\(\frac{2}{5}\))\(^1\) (\(\frac{3}{5}\))\(^9\) + (\(^{10}_{2}\)) (\(\frac{2}{5}\))\(^2\) (\(\frac{3}{5}\))\(^{8}\)
Which simplies to 0.1673 correct to four decimal places.