91Ó°ÊÓ

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

Given: \({\rm{p = 0}}{\rm{.2}}\)

Let \({\rm{X}}\)be the number of failures that must occur before the \({\rm{r = 2}}\) success (a success is obtaining a prize).

The negative binomial distribution is a probability distribution for a variable that quantifies the number of failures required to achieve the \({\rm{r}}\)success (with independent trials and a constant probability of success), therefore \({\rm{X}}\) has a negative binomial distribution with \({\rm{r = 2}}\)and \({\rm{p = 0}}{\rm{.2}}\)

X~nb(x;2,0.2)

Given: \({\rm{p = 0}}{\rm{.2}}\)

Let \({\rm{X}}\)be the number of failures that must occur before the \({\rm{r = 2}}\) success (a success is obtaining a prize).

The negative binomial distribution is a probability distribution for a variable that quantifies the number of failures required to achieve the \({\rm{r}}\)success (with independent trials and a constant probability of success), therefore \({\rm{X}}\) has a negative binomial distribution with \({\rm{r = 2}}\)and \({\rm{p = 0}}{\rm{.2}}\)

\({\rm{X\sim nb(x;2,0}}{\rm{.2)}}\)

Negative binomial probability formula:

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

If we had two successes in four boxes, we also had two failures. Calculate the negative binomial probability formula at \({\rm{x = 2}}\)

\({\rm{P(X = 2) = nb(2;2,0}}{\rm{.2) = }}\left( {\begin{array}{*{20}{c}}{{\rm{2 + 2 - 1}}}\\{{\rm{2 - 1}}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(1 - 0}}{\rm{.2)}}^{\rm{2}}}{\rm{ = 30}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(0}}{\rm{.8)}}^{\rm{2}}}{\rm{\gg 0}}{\rm{.0768 = 7}}{\rm{.68\% }}\)

Given: \({\rm{p = 0}}{\rm{.2}}\)

Let \({\rm{X}}\)be the number of failures that must occur before the \({\rm{r = 2}}\) success (a success is obtaining a prize).

The negative binomial distribution is a probability distribution for a variable that quantifies the number of failures required to achieve the \({\rm{r}}\)success (with independent trials and a constant probability of success), therefore \({\rm{X}}\) has a negative binomial distribution with \({\rm{r = 2}}\)and \({\rm{p = 0}}{\rm{.2}}\)

X~nb(x;2,0.2)

Negative binomial probability formula:

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

If we had two successes in four boxes, we also had two failures. Calculate the negative binomial probability formula at \({\rm{x = 2}}\):

\({\rm{P(X = 2) = nb(2;2,0}}{\rm{.2) = }}\left( {\begin{array}{*{20}{c}}{{\rm{2 + 2 - 1}}}\\{{\rm{2 - 1}}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(1 - 0}}{\rm{.2)}}^{\rm{2}}}{\rm{ = 30}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(0}}{\rm{.8)}}^{\rm{2}}}{\rm{\gg 0}}{\rm{.0768}}\)

If we had two successes in three boxes, we also had one failure. Calculate the negative binomial probability formula at \({\rm{x = 1}}\):

\({\rm{P(X = 1) = nb(1;2,0}}{\rm{.2) = }}\left( {\begin{array}{*{20}{c}}{{\rm{2 + 1 - 1}}}\\{{\rm{2 - 1}}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(1 - 0}}{\rm{.2)}}^{\rm{1}}}{\rm{ = 20}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(0}}{\rm{.8)}}^{\rm{1}}}{\rm{\gg 0}}{\rm{.064}}\)

If we had two successes in two boxes, we had no failures. Calculate the negative binomial probability formula at \({\rm{x = 0}}\)

\({\rm{P(X = 0) = nb(0;2,0}}{\rm{.2) = }}\left( {\begin{array}{*{20}{c}}{{\rm{2 + 0 - 1}}}\\{{\rm{2 - 1}}}\end{array}} \right){\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(1 - 0}}{\rm{.2)}}^{\rm{0}}}{\rm{ = 10}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{{\rm{(0}}{\rm{.8)}}^{\rm{0}}}{\rm{\gg 0}}{\rm{.04}}\)

Note that requiring fewer than two boxes is impossible because having fewer than two boxes means having fewer than two successes.

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)

Compile the following probabilities:

{P(X£ 2) = P(X = 0) + P(X = 1) + P(X = 2) = \({\text{0}}{\text{.0768 + 0}}{\text{.064 + 0}}{\text{.04 = 0}}{\text{.1808 = 18}}{\text{.08\% }}\)

The following formula calculates the mean (or anticipated value) of negative binomial variables:

\({\rm{\mu = }}\frac{{{\rm{r(1 - p)}}}}{{\rm{p}}}\)

Fill in the blanks and calculate:

\({\rm{\mu = }}\frac{{{\rm{2(1 - 0}}{\rm{.2)}}}}{{{\rm{0}}{\rm{.2}}}}{\rm{ = }}\frac{{{\rm{2(0}}{\rm{.8)}}}}{{{\rm{0}}{\rm{.2}}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.6}}}}{{{\rm{0}}{\rm{.2}}}}{\rm{ = 8}}\)

As a result, we expect to buy 8 boxes without knowing the price (failure) till we have the two quotes (success). We estimated that we would need to buy \({\rm{8 + 2 = 10}}\)cartons in total.