Answer:
Step-by-step explanation:
Let the first unknown integer be x
Let the second unknown integer be y
A very weak university is on and off probationary status with the accrediting agency. A different procedure is used in July and
1 answer:
Answer:
0.3425 = 34.25% probability it will be off probation in February 2020
Step-by-step explanation:
We have these desired outcomes:
Off probation in July 2019, with 0.25 probability, then continuing off in January, with 1 - 0.08 = 0.92 probability.
Still in probation in July 2019, with 1 - 0.25 = 0.75 probability, then coming off in January, with 0.15 probability.
What is the probability it will be off probation in February 2020?
0.3425 = 34.25% probability it will be off probation in February 2020
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Answer:
The probability is
Step-by-step explanation:
If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:
If k=1
Because, in her first try she has n possibles options and just one give her a successful login.
If k=2
Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.
If k=3
Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.
Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n