Question

Eliza has a unique circumstance in that she has a 25% chance of having a child born with a congenital heart defect for every pregnancy she has. She has always wanted to have 2 children. Using the random variable X, find the probability distribution of Eliza having two children with a congenital heart defects. *Input answer rounded to the nearest ten-thousandth for each answer, not a percent.* P(X=0)=____ P(x=1)=_____ P(x=2)=_____

Answer 1

P(x = 0) = 0.5625

P(x = 1) = 0.375

P(x = 2) = 0.0625

Step-by-step explanation:

She has a 25% chance of having a child born with a congenital heart defect, so it has 75% chance of having a child without the congenital heart defects.

Then, the probability to have 0 children with the heart defect can be calculated as

P(x = 0) = 0.75 x 0.75

P(x = 0) = 0.5625

Because each child has a chance of 75% to not have the heart defect.

In the same way, the probability to have 2 children with the heart defect is

P(x = 2) = 0.25 x 0.25

P(x = 2) = 0.0625

Since the total probability is equal to 1, we can calculate the probability to have 1 chid with the heart defect as

P(x = 1) = 1 - 0.5625 - 0.0625

P(x = 1) = 0.375

Therefore, the probability distribution is

P(x = 0) = 0.5625

P(x = 1) = 0.375

P(x = 2) = 0.0625