Probability of probability

TPFIN

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Apr 27, 2019
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Hi all!

Probability of being accepted for a new job is 15 %
An applicant makes 50 applications.
Probability being accepted to all of those jobs is ((15/100)^50)*100%
But what is the probability of being accepted to at least 5 % of those jobs?
Or what is the probability of being accepted to at least 95 % of those jobs?
 
The number of jobs the applicant is accepted to has a binomial distribution with parameters \(\displaystyle n=50,~p=0.15\)

\(\displaystyle P[\text{accepted to at least 5% of the jobs}] = P[k\geq 3]\)

\(\displaystyle P[\text{accepted to at least 95% of the jobs}] = P[k \geq 48]\)
 
Hi all!
Probability being accepted to all of those jobs is ((15/100)^50)*100%
That is not (exactly) correct. The probability of an even is a number between 0 and 1 while ((15/100)^50)*100% is a percentage.
The correct answer is Probability being accepted to all of those jobs is (15/100)^50
 
Probability of being accepted for a new job is 15 %
An applicant makes 50 applications. Probability being accepted to all of those jobs is ((15/100)^50)*100%
But what is the probability of being accepted to at least 5 % of those jobs?
Or what is the probability of being accepted to at least 95 % of those jobs?
An applicant makes 50 applications. Probability being accepted to all of those jobs is \(\displaystyle {\left( {0.15} \right)^{50}}\)
What is the probability of being accepted to at least 5 % of those jobs?
\(\displaystyle \sum\limits_{k = 3}^{50} {\binom{50}{k}{{(0.15)}^k}{{(0.85)}^{50 - k}}} \)
 
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