Chapter 8.2, Problem 67SR

To determine

To calculate: The probability of the children having blue eyes.

Answer to Problem 67SR

The required probability is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The person with $BB\text{or}Bb$ genes has brown eyes and person with $bb$ genes has blue eyes.

Elisa has $Bb$ genes.

Bob has blue eyes.

Formula Used:

Probability formula is

$\text{Probability}=\frac{\text{number}\text{of}\text{favourable}\text{outcomes}}{\text{total}\text{outcomes}}$

Calculation:

Given that Elisa has $Bb$ genes and Bob has blue eyes.

Thus, Bob gene is $bb$ .

Elisa is a combination of brown gene and blue gene.

Thus, gene of Elisa can be written as $B+b$ .

Similarly gene of Bob can be written as $b+b$ .

Find the genes of their children’s, multiply both their genes.

  $\begin{array}{c}\left(B+b\right)\left(b+b\right)=B\left(b+b\right)+b\left(b+b\right)\\ =Bb+Bb+bb+bb\end{array}$

From the above expression the total posible case is 4 out of which 2 cases are there for the blue eyed children.

Now, find the probability. Use the formula for the probability.

Probability that their child would have blue eyes $\begin{array}{l}=\frac{2}{4}\\ =\frac{1}{2}\end{array}$

Hence, the required probability is $\frac{1}{2}$ .