Question

A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probabilities q and 1 − q, respectively. The company tries to determine a candidate’s qualifications by asking 20 true-false questions. A qualified candidate has probability p of answering a question correctly, while an unqualified candidate has a probability p 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 a formula for the probability that the 20 questions will correctly identify someone to be qualified or unqualified.

Answer 1

`P(C=1|T=1)=q(\sum_(i=15)^(20)\binom{20}{i} p^i(1-p)^(20-i))( \sum_(i=15)^(20)\binom{20}{i}[qp^i(1-p)^(20-i) + (1-q)p^(20-i)(1-p)^i])^(-1)`

Explanation:

Hi!

Lets define:

C = 1 if candidate is qualified

C = 0 if candidate is not qualified

A = 1 correct answer

A = 0 wrong answer

T = 1 test passed

T = 0 test failed

We know that:

`P(C=1)=q\\P(A=1 | C=1) = p\\P(A=0 | C=0) = p`

The test consist of 20 questions. The answers are indpendent, then the number of correct answers X has a binomial distribution (conditional on the candidate qualification):

`P(X=x | C=1)=f_1(x)=\binom{20}{x}p^x(1-p)^(20-x)\\P(X=x | C=0)=f_0(x)=\binom{20}{x}(1-p)^xp^(20-x)`

The probability of at least 15 (P(T=1))correct answers is:

`P(X\geq 15|C=1)=\sum_(i=15)^(20)f_1(i)\\P(X\geq 15|C=0)=\sum_(i=15)^(20)f_0(i)\\`

We need to calculate the conditional probabiliy P(C=1 |T=1). We use Bayes theorem:

`P(C=1|T=1)=(P(T=1|C=1)P(C=1))/(P(T=1))\\P(T=1) = qP(T=1|C=1) + (1-q)P(T=1|C=0)`

`P(T=1)=q\sum_(i=15)^(20)f_1(i) + (1-q)\sum_(i=15)^(20)f_0(i)\\P(T=1)=\sum_(i=15)^(20)\binom{20}{i}[qp^i(1-p)^(20-i) + (1-q)p^(20-i)(1-p)^i)]`

`P(C=1|T=1)=q(\sum_(i=15)^(20)\binom{20}{i} p^i(1-p)^(20-i))( \sum_(i=15)^(20)\binom{20}{i}[qp^i(1-p)^(20-i) + (1-q)p^(20-i)(1-p)^i])^(-1)`