Question 1183306: Eliza has a unique circumstance in that she has a 26% chance of having a child born with spina bifida for every pregnancy she has. She has always wanted to have two children. Using the random variable X, find the probability distribution of Eliza having two children with spina bifida.
Find P(X=0) P(X=1) & P(X=2).
Thank you very much!!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p(x) = .26
p(not x) = 1 - .26 = .74
she wants to have two children.
the probabilties are assumed to be independent, i.e. the fact that she did or didn't have a child born with spina bifida has no impact on her probability of getting one on the second try.
in other words, the probability is the same, whether or not the first one was born with spina bifida.
p(x = 0) = .74 * .74 = .5476
p(x = 1) = .26 * .74 + .74 * .26 = 2 * .26 * .74 = .3848.
p(x = 2) = .26 * .26 = .0676
that middle probability is multiplied by 2 because the first child could be born with it and not the second, and the first child could be born without it and the second child born with it.
this is a binomial distribution type problem where p(x) = p^x * q^(n-x) * c(n,x)
p = probability of getting the disease.
q = probability of not getting the disease = 1 minus the probability of getting the disease.
c(n,x) is the probability of getting a set of x elements out of a set of n elements.
p(0) = .26^0 * .74 * 2 * c(2,0) = 1 * .74^2 * 1 = .74^2 = .5476.
p(1) = .26^1 * .74^1 * c(2,1) = .26 * .74 * 2 = .3848.
p(2) = .26^2 * 74^0 * c2,2) = .26^2 * 1 * 1 = .0676.
c(n,x) = n! / (x! * (n-x)!).
when n = 2 and x = 1, this becomes 2! / (1! * 1!) = 2/1 = 2.
when n = 2 and x = 0, this becomes 2! / (0! * 2!) = 2! / 2! = 1
0! is always equal to 1.
when n = 2 and x = 2, this becomes 2! / (2! * 0!) = 2! / 2! = 1.
the sum of all probabilities should always be equal to 1, as it does here.
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