Maths · Probability: Probability of an event, addition and multiplication theorems of probability
A machine has three parts, and whose chances of being defective are 0.02,0.10 an
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 arts becomes defective. What is the probability that the machine will not stop working?
- A. 0.06
- B. 0.16
- C. 0.84
- D. 0.94
Step-by-step solution
The machine does not stop working if all three parts are non-defective. Assuming independence, the probability is (1-0.02)*(1-0.10)*(1-0.05) = 0.98 * 0.90 * 0.95 = 0.8379 ≈ 0.84.
Related MCQs
- If and are independent events such that <\mathbf{1}, \mathbf{0}<\boldsymbol{P}(\boldsymbol{B})<\mathbf{1}, \) then This question has multipl…
- What is the probability of selecting two spade cards from a pack of 52 cards?…
- If =\boldsymbol{P}(\boldsymbol{B}), \) then the two events and are -…
- The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviation of 2 mon…
- The probability of an even happens in one trial of an experiment is Three independent trials of the experiments are performed. Find the prob…