According to a 2000 study by the Bureau of Justice Statistics, approximately 2% of the nation's 72 million children had a parent behind bars - nearly 1.5 million minors. Let X be the number of children who had an incarcerated parent. Suppose that 100 children are randomly selected.

Question

According to a 2000 study by the Bureau of Justice Statistics, approximately 2% of the nation's 72 million children had a parent behind bars - nearly 1.5 million minors. Let X be the number of children who had an incarcerated parent. Suppose that 100 children  are randomly selected.

(a) Does X satisfy the requirements for a binomial setting? Explain. If X=B(n,p), what are n and p?

(b) Describe P(X=0) in words. Then find P(X=0) and P(X=1).

(c) What is the probability that 2 or more of the 100 children have a parent behind bars?

lol · Accepted Answer

damn 4 years and still no answer. now im stuck on this problem LOL

lauren · Answer

pls help

lol2 · Answer

im not stuck im just lazy ;) slader has an answer to c