Answer:
The probability that 4 adults selected at random from the town all have health insurance is 0.1897
Step-by-step explanation:
Consider the provided information.
It is given that 66% of adults have health insurance.
Let p = 66% = 0.66
Therefore q=1-0.66=0.34
We need to find the probability that 4 adults selected at random from the town all have health insurance.
[tex]P(x)=^nC_r(p)^r(q)^{n-r}[/tex]
Here the value of r is 4.
Substitute the respective values in the above formula.
[tex]P(x)=^4C_4(0.66)^4(0.34)^{4-4}[/tex]
[tex]P(x)=(0.66)^4(0.34)^{0}\\P(x)=0.1897[/tex]
Hence, the probability that 4 adults selected at random from the town all have health insurance is 0.1897
Probabilities are used to determine the chances of an event.
The probability that 4 adults selected at random from the town all have health insurance is 0.190
The given parameters are:
[tex]\mathbf{p = 66\%}[/tex] --- the proportion of adults with health insurance
[tex]\mathbf{n = 4}[/tex] --- the number of selected adults
So, the probability that all selected adult have health insurance is:
[tex]\mathbf{Pr = p^n}[/tex]
So, we have:
[tex]\mathbf{Pr = (66\%)^4}[/tex]
Express percentage as decimal
[tex]\mathbf{Pr = (0.66)^4}[/tex]
[tex]\mathbf{Pr = 0.18974736}[/tex]
Approximate
[tex]\mathbf{Pr = 0.190}[/tex]
Hence, the required probability is 0.190
Read more about probabilities at:
https://brainly.com/question/11234923
