6. R1= .30R23= (.35)(.45) = .158R = 1 – [(1 - r1) (1 - (r2 x r3)) (1 – r4)] = 1 - (1 - .30) (1 - (.35 x .45)) (1 – .80) = 0.882Part 3: Human Reliability Analysis7.Assuming that all individual events are independent and must be completed successfully in orderto 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) = 0.5468. No, the probability of successfully completing the task was not as high as I expected. 0.546 isbasically 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 lowerthan theprobability 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.
Usability, 0%, Human reliability, Human Reliability Analysis
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