Question

A critical orthopedic surgery is performed on 3 patients. The probability of recovering a patient is 0.6. Then the probability that after surgery, exactly two of them will recover is

• A. 0.321
• B. 0.234
• C. 0.432
• D. 0.123

Hint:

The binomial distribution is applicable when there are a fixed number of independent trials with only two possible outcomes and a constant probability of success.

Answer 1

The Correct Option is C

This problem can be modeled using the binomial probability distribution. 
Step 1: Identify the parameters of the binomial distribution.
Number of trials (surgeries), \( n = 3 \).
Probability of success (recovery), \( p = 0.6 \).
Probability of failure (no recovery), \( q = 1 - p = 0.4 \).
Number of successful recoveries required, \( k = 2 \).
Step 2: Apply the binomial probability formula.
The probability of exactly \( k \) successes in \( n \) independent Bernoulli trials is given by: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] We want to find the probability of exactly 2 recoveries (\( k = 2 \)): \[ P(X = 2) = \binom{3}{2} (0.6)^2 (0.4)^{3-2} \] Step 3: Calculate the binomial coefficient.
\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3 \] Step 4: Calculate the probability.
\[ P(X = 2) = 3 \times (0.6)^2 \times (0.4)^1 = 3 \times (0.36) \times (0.4) = 3 \times 0.144 = 0.432 \] Thus, the probability that exactly two of them will recover is 0.432.