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A discrete random variable is one that can only take on a finite number of different values.

\({\rm{N = 11}}\)

senior engineers, as well as

\({\rm{n = 4}}\)

job vacancies Six of the eleven are questioned on the first day, and the remaining five are interviewed the next day.

Assume that the population has M successes (S) and N鈥揗 failures (F). If X is a random variable,

X=number of successes in a random sample size of n,

\(\begin{array}{c}{\rm{h(x;n,M,N) = P(X = x)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{M}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{N - M}}}\\{{\rm{n - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\end{array}\)

It has a probability mass function after that.

\({\rm{max\{ 0,n - N + M\} }} \le {\rm{x}} \le {\rm{min\{ n,M\} }}\)

The probability distribution is known as the hypergeometric distribution.

In our situation, a success is when x of the top contenders for the position are interviewed on the first day, and there are no other candidates.

\({\rm{M = 6}}\)

On the first day, applicants were interviewed. Also,

\(\begin{array}{c}{\rm{N - M = 11 - 6}}\\{\rm{ = 5}}\end{array}\)

It is then from that:

\(\begin{array}{c}{\rm{max\{ 0,4 - 11 + 6\} }} \le {\rm{x}} \le {\rm{min\{ n,M\} }}\\{\rm{max\{ 0, - 1\} }} \le {\rm{x}} \le {\rm{min\{ 4,6\} }}\\{\rm{0}} \le {\rm{x}} \le {\rm{4}}\end{array}\)

With pmf, we may deduce that the random variable X has a hypergeometric distribution.

\(\begin{array}{c}{\rm{h(x;4,6,11) = P(X = x)}}\\{\rm{ = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{6}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{11 - 6}}}\\{{\rm{4 - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{11}}}\\{\rm{4}}\end{array}} \right)}}\end{array}\)

For all other values of x, use \({\rm{x}} \in {\rm{\{ 0,1,2,3,4\} }}\), and zero.

The probability we required were as follows:

Therefore, the probability is:

For random variable X with hypergeometric distribution and pmf h(x; n, M, N), the following is true.

\(\begin{array}{c}{\rm{E(X) = n \times }}\frac{{\rm{M}}}{{\rm{N}}}\\{\rm{V(X) = }}\left( {\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}} \right){\rm{ \times n \times }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{ \times (1 - }}\frac{{\rm{M}}}{{\rm{N}}}{\rm{)}}\end{array}\)

We have derived the following from the proposition:

\(\begin{aligned}E(X) &= n \times \frac{{\rm{M}}}{{\rm{N}}}\\ &= 4 \times \frac{{\rm{6}}}{{{\rm{11}}}}\\ &= 2{\rm{.18}}\end{aligned}\)

Or, to put it another way, we should expect two of the top four applicants to be interviewed on the first day.

Therefore, the required candidates are two.