Paganellash
2021-11-16
Whouldess
Beginner2021-11-17Added 11 answers
Geometric Distribution: Let there is a sequence of trials, each having two outcomes, say Success and Failure, trials being independent and the probability of success, say p, remaining constant. Then the distribution of the number of failures before the first success follows geometric.
Let us assume X be a random variable denoting the number of failures before the first success. Then the probability mass function of X is given by,
It is denoted by,
(First Solution)
Let us assume X is a random variable denoting the number of people will be tested before the first false-positive occurs
The probability of false-positive is
Here,
1) each person's test is independent of the other
2) there are only two possibilities in each person's test viz whether the test result is false-positive or not
3) the number of tests before the false-positive can go up to infinity
iv) the probability of false-positive remains constant for all the tests
In these circumstances, we can state that,
with
Now, the event that the first false-positive will occur when the third person is tested implies that there will be two persons tested before the first false-positive occurs.
Hence, the probability that there will be two persons tested before the first false-positive occurs is
(rounded to 4 decimal places)
Result: The probability that the first false-positive will occur when the third person is tested is 0.1084.
(Second Solution)
The probability that less than 4 people are tested before the first false-positive occurs is
(rounded to 4 decimal places)
Answer: The probability that less than 4 people are tested before the first false-positive occurs is 0.4780.
(Third Solution)
The probability that more than 3 people are tested before the first false-positive occurs is
(rounded to 4 decimal places)
Answer: The probability that more than 3 people are tested before the first false-positive occurs is 0.5220.
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