Chapter 16, Problem 4QC

Summary Introduction

To determine:

The probability that any two peas are green and yellow in a pea pod with exactly four seeds.

Conceptual introduction:

Multiplication rule is used when in two independent trials; the probability of obtaining a given outcome in both the trials is the product of their independent probabilities. Addition rule is applied if we want to know the probability of one outcome or the other, we add the probabilities together. The genetic inheritance principle follows the statistical approach of probability using multiplication and addition rule.

Explanation of Solution

$\text{The}\text{probability}\text{of}\text{the}\text{first}\text{pea}\text{is}\text{green}\text{=}\frac{\text{1}}{\begin{array}{l}\text{4}\\ \end{array}}$

$\text{The}\text{probability}\text{of}\text{the}\text{second}\text{pea}\text{is}\text{green}\text{=}\frac{\text{1}}{\begin{array}{l}\text{4}\\ \end{array}}$

$\text{The}\text{probability}\text{of}\text{the}\text{third}\text{pea}\text{is}\text{yellow}\text{=}\frac{\text{3}}{\begin{array}{l}\text{4}\\ \end{array}}$

$\text{The}\text{probability}\text{of}\text{the}\text{forth}\text{pea}\text{is}\text{yellow}\text{=}\frac{\text{3}}{\begin{array}{l}\text{4}\\ \end{array}}$

The probability of the pod with first and second pea as green and  that of the third and fourth pea is yellow will be calculated by using the multiplication rule as the given event are mutually exclusive in nature.

$\text{The}\text{probability}\text{of}\text{the}\text{given}\text{configuration}\text{=}\frac{\text{1}}{\begin{array}{l}\text{4}\\ \end{array}}\text{×}\frac{\text{1}}{\text{4}}\text{×}\frac{\text{3}}{\text{4}}\text{×}\frac{\text{3}}{\text{4}}$

$\text{The}\text{probability}\text{of}\text{the}\text{given}\text{configuration}\text{=}\frac{\text{9}}{\begin{array}{l}256\\ \end{array}}$

There are six different combinations that can be seen with a pod having two green and two yellow pea seeds. The green pea seed can be found on the positions $\left(\text{1}\text{and}\text{2}\right)$; $\left(\text{1}\text{and}3\right)$; $\left(\text{1}\text{and}4\right)$; $\left(2\text{and}3\right)$; $\left(2\text{and}4\right)$ or on $\left(3\text{and}4\right)$ positions.

$\text{The}\text{probability}\text{of}\text{the}\text{first}\text{pea}\text{is}\text{green}\text{=}\frac{1}{\text{4}}$

$\text{The}\text{probability}\text{of}\text{the}\text{second}\text{pea}\text{is}\text{yellow}\text{=}\frac{\text{3}}{\text{4}}$

$\text{The}\text{probability}\text{of}\text{the}\text{third}\text{pea}\text{is}\text{green}\text{=}\frac{\text{1}}{\text{4}}$

$\text{The}\text{probability}\text{of}\text{the}\text{fourth}\text{pea}\text{is}\text{yellow}\text{=}\frac{\text{3}}{\text{4}}$

$\text{The}\text{probability}\text{of}\text{the}\text{given}\text{configuration}\text{=}\frac{\text{1}}{\begin{array}{l}\text{4}\\ \end{array}}\text{×}\frac{3}{\text{4}}\text{×}\frac{1}{\text{4}}\text{×}\frac{\text{3}}{\text{4}}$

$\text{The}\text{probability}\text{of}\text{the}\text{given}\text{configuration}\text{=}\frac{\text{9}}{\begin{array}{l}256\\ \end{array}}$

As described in the two combinations of the position of green peas in a pod, the probability of every configuration will come out as $\frac{\text{9}}{256}$

$\begin{array}{l}\text{The}\text{probability}\text{of}\text{any}\text{of}\text{these}\\ \text{combination}\text{occuring}\text{is}\text{given}\text{by}\text{addition}\text{rule}\text{=}\frac{\text{9}}{\text{256}}\text{+}\frac{\text{9}}{\text{256}}\text{+}\frac{\text{9}}{\text{256}}\text{+}\frac{\text{9}}{\text{256}}\text{+}\frac{\text{9}}{\text{256}}\text{+}\frac{\text{9}}{\text{256}}\end{array}$

The probability that any two peas are green and yellow in a pea pod with exactly four seeds will be $\text{=}\frac{\text{9}}{\text{256}}\text{×}\text{6}$ $\text{=}\frac{\text{54}}{\text{256}}$ $\text{=}\text{21\%}\text{(approx}\text{.)}$

Conclusion

The probability that any two peas are green and two pea seeds are yellow in a pea pod with exactly four seeds present in the pod will be $\text{=}\text{21\%}\text{(approx}\text{.)}$.