In guinea pigs, long fur is dominant to short fur. if two heterozygous guinea pigs are mated, what is the probability that they will have (show your calculations for both questions):

a. three short hair babies in a row?
b. three long hair babies in a row?

Let the dominant long fur be represented by L and the recessive short fur be l. The parents are therefore:

Ll x Ll

This cross produces the following offspring:
LL, Ll, Ll and ll

From the above cross, we can see that there is a 3/4 chance of an offspring with long fur and a 1/4 chance of an offspring with short fur. Now, calculating:

a) Three short fur babies means the event has to happen in successions thrice, which implies multiplication in probability. This is:
P(3 short fur) = 1/4 * 1/4 * 1/4
P(3 short fur) = 1/64

b) the same method is used for babies with long fur:

P(3 long fur) = 3/4 * 3/4 * 3/4
P(3 long fur) = 27/64

16 ⋅ 3 = 48

16 ⋅ 16 ⋅ 16 = 4096