Question

For a given process control chart, there are four rules for determining out-of-control state of the process which are being used simultaneously. The probability of Type-I error for the four rules are 0.005, 0.02, 0.03, and 0.05. Assuming independence of the rules, the probability of overall Type-I error when all the four rules are used simultaneously is

• A. 0.101
• B. 0.201
• C. 0.001
• D. 0.301

Hint:

When multiple independent rules are used, the overall error probability is calculated by multiplying the complement probabilities of no error for each rule.

Answer 1

The Correct Option is A

In this problem, we need to calculate the probability of Type-I error when multiple rules are applied simultaneously, assuming that the rules are independent. The probability of no Type-I error for each rule is the complement of the given probabilities. Thus, the probability of no Type-I error for each rule is: \[ P(\text{no error}) = 1 - P(\text{error}) \] For the four rules, the probabilities of no Type-I error are: \[ 1 - 0.005 = 0.995, \quad 1 - 0.02 = 0.98, \quad 1 - 0.03 = 0.97, \quad 1 - 0.05 = 0.95 \] The probability of no Type-I error for all four rules being applied simultaneously is the product of the probabilities for each rule: \[ P(\text{no error for all}) = 0.995 \times 0.98 \times 0.97 \times 0.95 \approx 0.899. \] Thus, the probability of overall Type-I error is the complement: \[ P(\text{overall error}) = 1 - 0.899 = 0.101. \] Therefore, the correct answer is (A).