The probability that a person who undergoes a kidney operation will recover is 0.7. If the six patients who undergoes similar operations,then the probability that at least half will recover is

Options

A. 0.4648
B. 0.04648
C. 0.9296
D. 0.09296

Answer

To solve the problem, we need to find the probability that at least half of the six patients who undergo a kidney operation will recover, given that the probability of recovery for each patient is \( p = 0.7 \).

### Step-by-Step Solution:

1. **Identify the total number of patients and the probability of recovery**:
- Let \( n = 6 \) (the number of patients).
- Let \( p = 0.7 \) (the probability of recovery).

2. **Define the event**:
- We need to find the probability that at least half of the patients recover. Since half of 6 is 3, we need to find \( P(X \geq 3) \).

3. **Use the complement rule**:
- Instead of calculating \( P(X \geq 3) \) directly, we can use the complement rule:
\[ P(X \geq 3) = 1 - P(X < 3) \]
- This means we need to calculate \( P(X < 3) \), which includes the probabilities of 0, 1, and 2 recoveries.

4. **Calculate \( P(X = k) \) using the binomial probability formula**:
- The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
- Where \( \binom{n}{k} \) is the binomial coefficient.

5. **Calculate \( P(X = 0) \)**:
\[ P(X = 0) = \binom{6}{0} (0.7)^0 (0.3)^6 = 1 \cdot 1 \cdot 0.000729 = 0.000729 \]

6. **Calculate \( P(X = 1) \)**:
\[ P(X = 1) = \binom{6}{1} (0.7)^1 (0.3)^5 = 6 \cdot 0.7 \cdot 0.00243 = 0.010626 \]

7. **Calculate \( P(X = 2) \)**:
\[ P(X = 2) = \binom{6}{2} (0.7)^2 (0.3)^4 = 15 \cdot 0.49 \cdot 0.0081 = 0.059535 \]

8. **Sum the probabilities for \( P(X < 3) \)**:
\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X < 3) = 0.000729 + 0.010626 + 0.059535 = 0.070890 \]

9. **Calculate \( P(X \geq 3) \)**:
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - 0.070890 = 0.929110 \]

### Final Answer:
The probability that at least half of the six patients will recover is approximately \( 0.9291 \).