Question

Black skin color is dominant to pink skin color in pigs. Two heterozygous black pigs are crossed. a. What is the probability that their offspring will have pink skin? b. What is the probability that the first and second off- spring will have black skin? c. If these pigs produce a total of three piglets, what is the probability that two will be pink and one will be black?

Answer 1

Answer:

$P(Pink) = \frac{1}{4}$

$P(Both\ Black) = \frac{9}{16}$

$Probability = \frac{3}{64}$ --- Two pink and One black

Step-by-step explanation:

Represent the pink skin with P and the black skin with B.

Since black skin is dominant to pink skin, there will be more occurrence of B than P

The punnet square for the breeding of the two pigs is then represented as:

$\begin{array}{cc} {B} & {B} \ \\ {B} & {P} \ \ \end{array}$

Solving (a): Probability of Pink

In the above square, there is only 1 occurrence of P out of a possible of 4.

So, the probability is:

$P(Pink) = \frac{n(P)}{4}$

$P(Pink) = \frac{1}{4}$

Solving (b): Probability that first and second are black

In the above square, there is only 3 occurrence of B out of a possible of 4.

$P(Black) = \frac{n(B)}{4}$

$P(Black) = \frac{3}{4}$

So, the probability that both are black is:

$P(Both\ Black) = P(Black) * P(Black)$

$P(Both\ Black) = \frac{3}{4} * \frac{3}{4}$

$P(Both\ Black) = \frac{9}{16}$

Solving (c): Probability of two pink and 1 black

This is calculated as:

$Probability = P(Pink) * P(Pink) * P(Black)$

$Probability = \frac{1}{4} * \frac{1}{4} * \frac{3}{4}$

$Probability = \frac{3}{64}$