Question

In a sample of 1000 u.s. adults, 180 dine out at a restaurant more than once per week. two u.s. adults are selected at random without replacement. (adapted from rasmussen reports
(a find the probability that both adults dine out more than once per week.
(b find the probability that neither adult dines out more than once per week.
(c find the probability that at least one of the two adults dines out more than once per week.
(d which of the events can be considered unusual? explain.

Answer 1

The probability that any one adults from the sample dines out more than once per week is $\dfrac{180}{1000}=0.18$. Taking two from the population without replacement sets the stage for an experiment involving a binomial distribution, with success probability $p=0.18$ and $n=2$ trials. So within this sample of 2 individuals, we have the probability mass function

$\mathbb P(X=x)=\dbinom2x0.18^x(1-0.18)^{2-x}$

a.
$\mathbb P(X=2)=\dbinom220.18^2(1-0.18)^{2-2}=0.18^2=0.0324$

b.
$\mathbb P(X=0)=\dbinom200.18^0(1-0.18)^2=0.82^2=0.6724$

c.
$\mathbb P(X\ge1)=\mathbb P(X=1)+\mathbb P(X=2)=\dbinom210.18^1+(1-0.18)^{2-1}+\mathbb P(X=2)=2(0.18)(0.82)+\mathbb P(X=2)=0.3276$

d. The event with the lowest probability can be considered the most unusual. Out of the three above, that would be the case that both adults dine out more than once per week.