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?

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

pls help

im not stuck im just lazy ;)

slader has an answer to c

(a) In order for X to satisfy the requirements for a binomial setting, the following conditions must be met:

1. The trials must be independent: Each child's situation of having a parent behind bars must be independent of other children's situations.
2. The number of trials must be fixed: In this case, 100 children are selected, so the number of trials is fixed.
3. Each trial must have two possible outcomes, success or failure: A child can either have a parent behind bars (success) or not (failure).
4. The probability of success must be the same for each trial: The probability of a child having a parent behind bars remains constant for each trial.

If X = B(n, p), where B represents a binomial distribution, n represents the number of trials, and p represents the probability of success, then n = 100 (as 100 children are randomly selected) and p = 2% or 0.02 (as approximately 2% of the nation's children have a parent behind bars).

(b) P(X = 0) represents the probability that none of the selected children have a parent behind bars. In words, this can be interpreted as the probability of selecting 100 children and none of them having an incarcerated parent. This can be calculated using the binomial probability formula:

P(X = 0) = nC0 * p^0 * (1 - p)^(n - 0)

where nC0 represents the combination formula for choosing 0 successes from n trials.

P(X = 0) = (100 C 0) * 0.02^0 * (1 - 0.02)^(100 - 0)
= (1) * 1 * (0.98)^100
= 0.1326 (rounded to four decimal places)

So, the probability that none of the selected children have a parent behind bars is approximately 0.1326.

Similarly, P(X = 1) represents the probability that exactly one child has a parent behind bars. This can be calculated using the same binomial probability formula:

P(X = 1) = (100 C 1) * 0.02^1 * (1 - 0.02)^(100 - 1)
= (100) * 0.02 * (0.98)^99
= 0.2707 (rounded to four decimal places)

Thus, the probability that exactly one of the selected children has a parent behind bars is approximately 0.2707.

(c) To find the probability that 2 or more of the 100 children have a parent behind bars, we need to find the complementary probability of none of them having a parent behind bars or exactly one of them having a parent behind bars.

P(2 or more) = 1 - P(X = 0) - P(X = 1)
= 1 - 0.1326 - 0.2707
≈ 0.5967 (rounded to four decimal places)

Therefore, the probability that 2 or more of the 100 randomly selected children have a parent behind bars is approximately 0.5967.