Question

A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that :
(i) exactly one surgery is successful,
(ii) at most two surgeries are successful.

Hint:

Whenever an inequality asks for "at most \( n-1 \)" or "at least 1", checking the complement side (\( 1 - P(\text{opposite}) \)) almost always circumvents adding several tedious individual binomial combinations together.

Answer 1

Concept: Since the trials (surgeries) are independent and have a constant probability of success, this scenario follows a Binomial Distribution model, denoted by \( B(n, p) \). The probability of obtaining exactly \( r \) successes out of \( n \) independent trials is given by the formula: \[ P(X = r) = \binom{n}{r} p^r q^{n-r} \] where:
• \( n = 3 \) (total number of patients/surgeries)
• \( p = 90\% = 0.9 \) (probability of a successful surgery)
• \( q = 1 - p = 1 - 0.9 = 0.1 \) (probability of a failed surgery) Part (i): Probability that exactly one surgery is successful (\( r = 1 \)).
Substitute \( n = 3 \), \( r = 1 \), \( p = 0.9 \), and \( q = 0.1 \) into the binomial formula: \[ P(X = 1) = \binom{3}{1} (0.9)^1 (0.1)^{3-1} \] Since \( \binom{3}{1} = 3 \): \[ P(X = 1) = 3 \times 0.9 \times (0.1)^2 = 3 \times 0.9 \times 0.01 = 2.7 \times 0.01 = 0.027 \] Thus, the probability that exactly one surgery is successful is \( 0.027 \) or \( \frac{27}{1000} \). Part (ii): Probability that at most two surgeries are successful (\( X \le 2 \)).
"At most two" means the number of successful surgeries can be 0, 1, or 2. It is computationally quicker to solve this using the complement rule: \[ P(X \le 2) = 1 - P(X = 3) \] Let's first find \( P(X = 3) \), which represents the probability that all three surgeries are successful: \[ P(X = 3) = \binom{3}{3} (0.9)^3 (0.1)^{3-3} = 1 \times 0.729 \times 1 = 0.729 \] Now substitute this back into the complement equation: \[ P(X \le 2) = 1 - 0.729 = 0.271 \] Thus, the probability that at most two surgeries are successful is \( 0.271 \) or \( \frac{271}{1000} \).