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A wildlife biologist examines frogs for a genetic trait he suspects may be linked to sensitivity to industrial toxins in the environment.

– 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 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:

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

\[ P ( x \leq 4  ) = 0.989 \]

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:

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

\[ P_5 = 0.0095 \]

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