Binomial Distribution

Question

Sickle-cell disease is a condition common in persons of African ancestry where red blood cells contain an abnormal type of hemoglobin. Individuals whose cells produce both normal and abnormal hemoglobin are said to have "sickle-cell trait" and they are generally healthy. If both parents have sickle-cell trait, then each of their children has probability 0.25 of having the actual disease. Suppose two parents with sickle-cell trait have three children.

Identify the exact probabilities in this interim table of all possible outcomes and in the probability distribution table for X, where N denotes no sickle-cell disease and D denotes sickle-cell disease.

(Round your answers to six decimal places.)

Outcome	X	Probability
NNN	0
NND	1
NDN	1
DNN	1
NDD	2
DND	2
DDN	2
DDD	3

X	Probability
0
1
2
3

Answer 1

For each child, the probability they have the disease is 1/4 and the probability they don't is 3/4.

NNN = (3/4) * (3/4) * (3/4) = 27/64

NND = (3/4) * (3/4) * (1/4) = 9/64

NDN = (3/4) * (1/4) * (3/4) = 9/64

DNN = (1/4) * (3/4) * (3/4) = 9/64

DDN = (1/4) * (1/4) * (3/4) = 3/64

DND = (1/4) * (3/4) * (1/4) = 3/64

NDD = (3/4) * (1/4) * (1/4) = 3/64

DDD = (1/4) * (1/4) * (1/4) = 1/64.

(add 'em all up, you get 64/64)

X:

0 - 27/64

1 - 27/64

2 - 9/64

3 - 1/64