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

The probability mass function of the negative binomial random variable X with parameters r and p, for \({\rm{x}} \in {\rm{N}}\), is if there are r successes with a probability of success of p.

\({\rm{nb(x;r,p) = }}\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 1}}}\\{{\rm{r - 1}}}\end{array}} \right){{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\)

Assume that the chances of the offspring being male or female are equal.

\(\begin{array}{c}{\rm{P = P(F)}}\\{\rm{ = P(M)}}\\{\rm{ = 0}}{\rm{.5}}\end{array}\)

This scenario may be transformed into one in which a single family intends to produce children until they have a total of six female offspring.

The random variable X would have a negative binomial distribution, and the number of male children may potentially go from zero to infinity, therefore we get \({\rm{x}} \in {{\rm{N}}_{\rm{0}}}\).

\(\begin{array}{c}{\rm{nb(x;6,0}}{\rm{.5) = }}\left( {\begin{array}{*{20}{c}}{{\rm{x + 6 - 1}}}\\{{\rm{6 - 1}}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{5}}^{\rm{6}}}{{\rm{(1 - 0}}{\rm{.5)}}^{\rm{x}}}\\{\rm{ = }}\left( {\begin{array}{*{20}{c}}{{\rm{x + 5}}}\\{\rm{5}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{5}}^{\rm{6}}}{\rm{ \times 0}}{\rm{.}}{{\rm{5}}^{\rm{x}}}\end{array}\)

For random variable X with negative binomial distribution and pmf nb(x;r, p), the following is true.

\(\begin{array}{c}{\rm{E(X) = }}\frac{{{\rm{r(1 - p)}}}}{{\rm{p}}}\\{\rm{V(X) = }}\frac{{{\rm{r(1 - p)}}}}{{{{\rm{p}}^{\rm{2}}}}}\end{array}\)

As a result, the anticipated value is

\(\begin{array}{c}{\rm{E(X) = }}\frac{{{\rm{r(1 - p)}}}}{{\rm{p}}}\\{\rm{ = }}\frac{{{\rm{6(1 - 0}}{\rm{.5)}}}}{{{\rm{0}}{\rm{.5}}}}\\{\rm{ = 6}}\end{array}\)

The predicted number is six, and if we repeat the process for a family with parameter \({\rm{r = 2}}\) (for a family with a single brother), we will get

\(\begin{array}{c}{\rm{E(}}{{\rm{X}}_{\rm{1}}}{\rm{) = }}\frac{{{\rm{r(1 - p)}}}}{{\rm{p}}}\\{\rm{ = }}\frac{{{\rm{2(1 - 0}}{\rm{.5)}}}}{{{\rm{0}}{\rm{.5}}}}\\{\rm{ = 2}}\end{array}\)

As a result, we may deduce that the anticipated value of random variable X equals the total of the three brothers' expected values.

\({\rm{2 + 2 + 2 = 6}}\)

The expected value of X is the sum of the predicted number of male children born to each brother.

Therefore, the values are:

\(\begin{array}{c}{\rm{nb(x;6,0}}{\rm{.5) = }}\left( {\begin{array}{*{20}{c}}{{\rm{x + 5}}}\\{\rm{5}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{5}}^{\rm{6}}}{\rm{ \times 0}}{\rm{.}}{{\rm{5}}^{\rm{x}}}\\{\rm{E(X) = 6}}\end{array}\)