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Eliza has a unique circumstance in that she has a 28% chance of having a child born with down syndrome for every pregnancy she has. She has always wanted to have two children. Using therandom variable X, find the probability distribution of Elizahaving two children with down syndrome.*P(X=0)=P(X=1)=P(X=2)=

Answer :

In this case, you are given the probability that if Eliza has a child, the child has the syndrome. X would be the number of children with the syndrome. 28% is equal to 28/100=14/50=7/25, which can be convenient to use to do the math.

P(X=0) means that Eliza has 2 children and none of them have the syndrome. The probability of Eliza having 1 child without the syndrome is 1-probability of having the syndrome, which equals 1-7/25=18/25

If she has 2 children, you must use the multiplication rule, because one case is independent of the other. Then:

[tex]P(X=0)=\frac{18}{25}\cdot\frac{18}{25}=\frac{324}{625}\cong51.84\text{ percent}[/tex]

Then, for P(X=1),

[tex]P(X=1)=\frac{18}{25}\cdot\frac{7}{25}=\frac{126}{625}\cong20.16\text{ percent}[/tex]

Finally, for P(X=2):

[tex]P(X=2)=\frac{7}{25}\cdot\frac{7}{25}=\frac{49}{625}\cong7.84\text{ percent}[/tex]

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