Sickle-cell disease is a condition common in persons of African ancestry where red blood cells contain an abnormal type ofhemoglobin. Individuals whose cells produce both normal and abnormal hemoglobin are said to have "sickle-cell trait" and theyare generally healthy. If both parents have sickle-cell trait, then each of their children has probability 0.25 of having the actualdisease. 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 ProbabilityNNN 0012NND 13NDN 1DNN 1NDDDNDDDNX ProbabilityNN2DDD 3

Given,Probability of child having sickle-cell trait disease probability of child not having…

Answered: Identify the exact probabilities in…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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

Expert Solution

Step 1: we define the given terms then find out the problem solution

Given,

Probability of child having sickle-cell trait disease $bold left parenthesis bold italic D bold right parenthesis bold equals bold 0 bold. bold 25$

probability of child not having sickle-cell trait disease

$bold left parenthesis bold italic N bold right parenthesis bold equals bold 1 bold minus bold italic D bold space bold space bold space bold space bold space bold space bold equals bold 1 bold minus bold 0 bold. bold 25 bold space bold space bold space bold space bold space bold space bold equals bold 0 bold. bold 75$

Here having not having disease independent of each student

then we have $bold italic P bold left parenthesis bold italic A bold italic B bold right parenthesis bold equals bold italic P bold left parenthesis bold italic A bold right parenthesis bold asterisk times bold italic P bold left parenthesis bold italic B bold right parenthesis$

we have the selected children $bold left parenthesis bold italic n bold right parenthesis bold equals bold 3$

We identify the exact probabilities in this interim table of all possible outcomes and in the probability distribution table for X is,

outcome	X	Probability
NNN	0	0.750.750.75=0.421875
NND	1	0.750.750.25=0.140625
NDN	1	0.750.250.75=0.140625
DNN	1	0.250.750.75=0.140625
NDD	2	0.750.250.25=0.046875
DND	2	0.250.750.25=0.046875
DDN	2	0.250.250.75=0.046875
DDD	3	0.250.250.25=0.015625