The teacher hasn't uploaded a solution for this question yet.
This is a Binomial Distribution problem where $n = 3$ (number of surgeries), $p = 0.9$ (probability of success), and $q = 1 - p = 0.1$ (probability of failure). The probability of $x$ successes is given by $P(X=x) = \binom{n}{x} p^x q^{n-x}$.
We need to find $P(X=1)$. $$P(X=1) = \binom{3}{1} (0.9)^1 (0.1)^2$$ $$P(X=1) = 3 \times 0.9 \times 0.01 = 0.027$$
We need to find $P(X \le 2)$, which is $1 - P(X=3)$. $$P(X=3) = \binom{3}{3} (0.9)^3 (0.1)^0 = 1 \times 0.729 \times 1 = 0.729$$ $$P(X \le 2) = 1 - 0.729 = 0.271$$
Final Answer: (i) 0.027, (ii) 0.271
AI generated content. Review strictly for academic accuracy.