Rob wants to get certified in computer netwroking and he can make up to 3 attemp

Question

Rob wants to get certified in computer netwroking and he can make up to 3 attempts to pass the exam. The probability of Rob passing the exam in first attempt is 32%, in the second attempt is 65% and in the third attempt is 90%. If the Rob earns the certification, what is the probability that it took him at least 2 attempts to pass the exam? Rob wants to get certified in computer netwroking and he can make up to 3 attempts to pass the exam. The probability of Rob passing the exam in first attempt is 32%, in the second attempt is 65% and in the third attempt is 90%. If the Rob earns the certification, what is the probability that it took him at least 2 attempts to pass the exam?

Explanation / Answer

P(1 attempt) = 0.32 P(2 attempts) = (0.64)(0.65) = 0.416 P(3 attempts) = (0.64)(0.35)(0.90) = 0.2016 P(not passing) = (0.64)(0.35)(0.10) = 0.0224 P(2 or 3 attempts|pass) = [0.416+0.2016]/[1-0.0224] = 0.63175

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