A machine has three parts. A. B and C, whose chances of being defective are 0.02, 0.10 and 0.05 respectively. The machine stops working if any one of the parts becomes defective. What is the probability that the machine will not stop working?

Options

A. 0.06
B. 0.16
C. 0.84
D. 0.94

Answer

To solve the problem, we need to find the probability that the machine will not stop working. The machine will not stop working if all three parts (A, B, and C) are functioning properly.

### Step-by-Step Solution:

1. **Identify the probabilities of each part being defective:**
- Probability of part A being defective, \( P(A) = 0.02 \)
- Probability of part B being defective, \( P(B) = 0.10 \)
- Probability of part C being defective, \( P(C) = 0.05 \)

2. **Calculate the probabilities of each part working properly:**
- Probability of part A working properly, \( P(A') = 1 - P(A) = 1 - 0.02 = 0.98 \)
- Probability of part B working properly, \( P(B') = 1 - P(B) = 1 - 0.10 = 0.90 \)
- Probability of part C working properly, \( P(C') = 1 - P(C) = 1 - 0.05 = 0.95 \)

3. **Determine the probability that the machine will not stop working:**
- The machine will not stop working if all parts are functioning properly. Since the parts are independent, we can multiply their probabilities:
\[ P(\text{Machine works}) = P(A') \times P(B') \times P(C') = 0.98 \times 0.90 \times 0.95 \]

4. **Calculate the multiplication:**
- First, calculate \( 0.98 \times 0.90 \):
\[ 0.98 \times 0.90 = 0.882 \]
- Next, multiply the result by \( 0.95 \):
\[ 0.882 \times 0.95 = 0.8361 \]

5. **Round the result:**
- Rounding \( 0.8361 \) gives approximately \( 0.84 \).

6. **Conclusion:**
- The probability that the machine will not stop working is approximately \( 0.84 \).

### Final Answer:
The probability that the machine will not stop working is \( 0.84 \).