Suppose that in a region the chance that someone having a health insurance coverage is 80% and that a sample of five people is selected at random Suppose that in a region the chance that someone having a health insurance coverage is 80% and that a sample of five people is selected at random for a survey concerning health insurance. Let $X$ = number of people having health insurance in the sample. What is the probability that $a$) One person will not have health insurance ? $b$) At most three people will have health insurance ? $c$). What is the expected value for the number of people with insurance ? $d$). What is the variance and standard deviation for the number of people with insurance ?

The number of people with insurance follows a binomial distribution, with n=5 and p=.8.

Therefore the probability of $x$ people having insurance is $${n \choose x}p^x(1-p)^{n-x}$$

To calculate the probability of 3 or less people having insurance, you would independently solve the above expression for all desired values of $x$ (0,1,2,3) and add them together.

For a binomial distribution the mean is $np$ and the variance is $np(1-p)$. The standard deviation is the square root of the variance.