Question

The generalized negative binomial \({\rm{pmf}}\) is given by

\(\begin{aligned}{\rm{nb(x;r,p) &= k(r,x)}} \cdot {{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\\{\rm{x &= 0,1,2,}}...\end{aligned}\)

Let \({\rm{X}}\), the number of plants of a certain species found in a particular region, have this distribution with \({\rm{p = }}{\rm{.3}}\) and \({\rm{r = 2}}{\rm{.5}}\). What is \({\rm{P(X = 4)}}\)? What is the probability that at least one plant is found?

Short answer

The value of \({\rm{P(X = 4)}}\) is \({\rm{0}}{\rm{.1068}}\).The probability that at least one plant is found is \({\rm{0}}{\rm{.9507}}\).

Step-by-step solution

Concept Introduction

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

The amount of 'Success' in a series of n experiments, where each time a yes-no question is given, the Boolean-valued outcome is represented either with success/yes/true/one (probability \(\text{p}\)) or failure/no/false/zero (probability \(\text{q=1-p}\)) in a binomial probability distribution.

Probability of one plant being found

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

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

It is given that –

\({\rm{k(r,x) = }}\frac{{{\rm{(x + r - 1)(x + r - 2)}} \cdot {\rm{ \ldots }} \cdot {\rm{(x + r - x)}}}}{{{\rm{x!}}}}\)

From which, for\({\rm{p = 0}}{\rm{.3}}\)and\({\rm{r = 2}}{\rm{.5}}\), it is obtained –

\(\begin{array}{c}{\rm{P(X = 4) = nb(4;2}}{\rm{.5,0}}{\rm{.3)}}\\{\rm{ = }}\frac{{{\rm{(4 + 2}}{\rm{.5 - 1)(4 + 2}}{\rm{.5 - 2)}} \cdot {\rm{ \ldots }} \cdot {\rm{(4 + 2}}{\rm{.5 - 4)}}}}{{{\rm{4!}}}} \cdot {\rm{0}}{\rm{.}}{{\rm{3}}^{{\rm{2}}{\rm{.5}}}} \cdot {\rm{0}}{\rm{.}}{{\rm{7}}^{\rm{4}}}\\{\rm{ = 0}}{\rm{.1068}}\end{array}\)

For the second part, the following is true –

\(\begin{aligned}{\rm{P(X}} \ge 1) &= 1 - P(X = 0)\\ &= 1 - nb(0;2 {\rm{.5,0}}{\rm{.3)}}\\ &= 1 - 1 \cdot {\rm{0}}{\rm{.}}{{\rm{3}}^{{\rm{2}}{\rm{.5}}}} \cdot {\rm{1}}\\& = 0 {\rm{.9507}}\end{aligned}\)

Therefore, the values obtained are \({\rm{0}}{\rm{.1068}}\) and \({\rm{0}}{\rm{.9507}}\).