170k views
4 votes
A study of immunizations among school‑age children in California found that some areas had rates of unvaccinated school‑age children as high as 13.5 % in 2014. Imagine a classroom of 20 children in one such area where 13.5 % of children are unvaccinated. If there are no siblings in the classroom, we are willing to consider the vaccination status of the 20 unrelated children to be independent. A) Obtain the probability that none of the 20 children in such classroom would be unvaccinated

B) Obtain the probability that at least one of the 20 children in such a classroom would be unvaccinated.

User Chele
by
6.0k points

1 Answer

7 votes

Answer:

a)
P(X=0)= (20C0) (0.135)^0 (1-0.135)^(20-0) = 0.0550

b)
P(X \geq 1)

And we can use the complement rule and we have this:


P(X \geq 1) = 1-P(X<1)= 1-P(X=0) = 1-0.0550= 0.945

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:


P(X)=(nCx)(p)^x (1-p)^(n-x)

Where (nCx) means combinatory and it's given by this formula:


nCx=(n!)/((n-x)! x!)

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by:
P(A)+P(A') =1

Solution to the problem

For this case we define the random variable X="number of children unvaccinated" and the distribution for X is given by:


X \sim Bin (n =20, p=0.135)

Part a

For this case we want this probability:


P(X=0)

Using the probability mass function we can replace and we got:


P(X=0)= (20C0) (0.135)^0 (1-0.135)^(20-0) = 0.0550

Part b

For this case we want this probability:


P(X \geq 1)

And we can use the complement rule and we have this:


P(X \geq 1) = 1-P(X<1)= 1-P(X=0) = 1-0.0550= 0.945

User Joshdcomp
by
6.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.