6 r1 30 r23 3545 158 r 1 1 r 1 1 r 2 x r 3 1 r 4 1 1

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6. R1= .30 R23= (.35)(.45) = .158 R = 1 – [(1 - r 1 ) (1 - (r 2 x r 3 )) (1 – r 4 )] = 1 - (1 - .30) (1 - (.35 x .45)) (1 – .80) = 0.882 Part 3: Human Reliability Analysis 7. Assuming that all individual events are independent and must be completed successfully in order to complete the task, the probability the task will be completed successfully is: Probability (Successful) = (0.95)(0.95)(0.98)(0.99)(0.95)(0.95)(0.98)(0.99)(0.95)(0.85)(0.90) (0.98) = 0.546 8. No, the probability of successfully completing the task was not as high as I expected. 0.546 is basically 1 out of every 2 wont book the flight in the right way. I assume this is low because the probability of Entering the payment information and confirming all information are the lowest
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ones. This is understandable, since the majority of people don’t know all the credit card information and can make some mistakes. Two changes in the interface design that a human factors professional could to improve the probability of success is to make an easier version of “ entering payment information”, they could separate the numbers with dashes in groups of 4 just like credit cards. They could also add an option of “save payment information” for future transactions, so that probability can basically be 100%. 9. The probability of successful task completion without corrections is going to be lower than the probability with corrections. If we get the option to correct our mistakes, there is basically a 0% chance of screwing the transaction, it should go smooth.
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  • Fall '19
  • Usability, 0%, Human reliability, Human Reliability Analysis

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