**– The genetic trait was previously found to be 1 in every 8 frogs.**

**– He collects 12 frogs and examines them for the genetic trait.**

**– What is the probability the wildlife biologist would find the trait in the following batches if the trait frequency is the same?**

**a) None of the frogs he examined.**

**b) At least 2 of the frogs he examined.**

**c) Either 3 frogs or 4 frogs.**

**d) No more than 4 frogs he examined.**

The question aims to find the **binomial probability** of **dozen frogs** with traits occurring **1** in every **8th** frog.

The question depends on the concepts of **binomial distribution probability, binompdf,** and **binomcdf.** The formula for a **binomial probability distribution** is given as:

\[ P_x = \begin {pmatrix} n \\ x \end {pmatrix} p^x (1 – p)^{n – x} \]

$P_x$ is **binomial probability.**

$n$ is the **number** of **trials.**

$p$ is the **probability** of **success** in a **single**Â **trial.**

$x$ is the **number** of **times** for specific outcomes for **n trials.**

## Expert Answer

The given information about the problem is given as:

\[ Number\ of\ Frogs\ n = 12 \]

\[ Success\ Rate\ is\ 1\ in\ every\ 8\ frogs\ have\ genetic\ trait\ p = \dfrac{ 1 }{ 8 } \]

\[ p = 0.125 \]

**a)** The **probability** that **none of the frogs** have any trait. Here:

\[ x = 0 \]

Substituting the values in the given formula for **binomial distribution probability,** we get:

\[ P_0 = \begin {pmatrix} 12 \\ 0 \end {pmatrix} \times 0.125^0 \times (1 – 0.125)^{12-0} \]

Solving the probability, we get:

\[ P_0 = 0.201 \]

**b)** The **probability** that **at least two of the frogs** will contain the genetic trait. Here:

\[ x \geq 2 \]

Substituting the values, we get:

\[ P_2 = \sum_{i=0}^2 \begin {pmatrix} 12 \\ i \end {pmatrix} \times 0.125^i \times (1 – 0.125)^{12-i} \]

\[ P_2 = 0.453 \]

**c)** The **probability** that **either 3 or 4 frogs** will contain the genetic traits. Now here, we will have to **add** the **probabilities.** Here:

\[ x = 3\ or\ 4 \]

\[ P (3\ or\ 4) = \begin {pmatrix} 12 \\ 3 \end {pmatrix} \times 0.125^3 \times (1 – 0.125)^{12-3} + \begin {pmatrix} 12 \\ 4 \end {pmatrix} \times 0.125^4 \times (1 – 0.125)^{12-4} \]

\[ P (3\ or\ 4) = 0.129 + 0.0415 \]

\[ P (3\ or\ 4) = 0.171 \]

**d)** The **probability** that **no more than 4 frogs** will have the genetic trait. Here:

\[ x \leq 4 \]

Substituting the values, we get:

\[ P ( x \leq 4) = \sum_{i=0}^4 \begin {pmatrix} 12 \\ i \end {pmatrix} \times 0.125^i \times (1 – 0.125)^{12-i} \]

\[ P ( x \leq 4Â ) = 0.989 \]

## Numerical Results

**a) P_0 = 0.201**

**b) P_2 = 0.453**

**c) P (3\ or\ 4) = 0.171**

**d) P (x \leq 4) = 0.989**

## Example

Considering the above problem, find the **probability** that the **5 frogs** will have the **genetic trait.**

\[ Number\ of\ Frogs\ n = 12 \]

\[ p = 0.125 \]

\[ x = 5 \]

Substituting the values, we get:

\[ P_5 = \begin {pmatrix} 12 \\ 5 \end {pmatrix} \times 0.125^5 \times (1 – 0.125)^{12-5} \]

\[ P_5 = 0.0095 \]