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A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probabilities 0.77 and 0.23, respectively. The company tries to determine a candidate's qualifications by asking 20 true-false questions. A qualified candidate has probability 0.7 of answering a question correctly, while an unqualified candidate has a probability 0.7 of answering incorrectly. The answers to different questions are assumed to be independent. If the company considers anyone with at least 15 correct answers qualified, and everyone else unqualified, give the probability that the 20 questions will correctly identify someone to be qualified or unqualified.

Guest Feb 22, 2016

Best Answer 

 #3
avatar+2493 
+5

but it is really cool answer I have t thought of that 15 right answers can be in different arrangement 

Thank you very much !! :)

Solveit  Feb 24, 2016
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4+0 Answers

 #1
avatar+91001 
+10

Hi Solveit,

Is this your question Solveit or did you just want to know how to do it?

 

A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probabilities 0.77 and 0.23, respectively. The company tries to determine a candidate's qualifications by asking 20 true-false questions. A qualified candidate has probability 0.7 of answering a question correctly, while an unqualified candidate has a

 

probability 0.7 of answering incorrectly. The answers to different questions are assumed to be independent. If the company considers anyone with at least 15 correct answers qualified, and everyone else unqualified, give the probability that the 20 questions will correctly identify someone to be qualified or unqualified.

 

I think:

P(correct identification)=P(Q and pass)+P(U and F)

 

Say someone qualified is chosen then the probablility that they will pass will be

P(15 correct)+P(16 correct)+P(17 correct)+P(18correct)+P(19 correct)+P(20 correct)

nCr(20,15)*0.7^15*0.3^5+nCr(20,16)0.7^16*0.3^4+nCr(20,17)*0.7^17*0.3^3+nCr(20,18)*0.7^18*0.3^2+nCr(20,19)*0.7^19*0.3+0.7^20 = 0.4163708294474814

 

So the probability that someone IS qualified AND they pass is approximately

0.77*0.4163708294474814 = 0.320605538674560678

 

Now the probability that an unqualified person fails is exactly the same as the probablility that a qualified person passes so the probability that an individual is not qualified and fails is

0.23*0.4163708294474814 = 0.095765290772920722

 

So the probability that the person is corectly identified as qualified or unqualified is

0.320605538674560678+0.095765290772920722 = 0.4163708294474814

 

That is what i get  :/

Melody  Feb 24, 2016
 #2
avatar+2493 
0

it is not me :D

Solveit  Feb 24, 2016
 #3
avatar+2493 
+5
Best Answer

but it is really cool answer I have t thought of that 15 right answers can be in different arrangement 

Thank you very much !! :)

Solveit  Feb 24, 2016
 #4
avatar+91001 
0

You are always very welcome Solveit.

It is a real pleasure to have you on the forum :)

Melody  Feb 24, 2016

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