NO LINKS OR ELSE YOU'LL BE REPORTED! Only answer if you're very good at Math.No guessing please.
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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
Answer:
You must survey 784 air passengers.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error is:
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
Assume that nothing is known about the percentage of passengers who prefer aisle seats.
This means that , which is when the largest sample size will be needed.
Within 3.5 percentage points of the true population percentage.
We have to find n for which M = 0.035. So
You must survey 784 air passengers.