Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go"), so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that $48\%$ of its customers order their food to go. If this proportion is correct, what is the probability that, in a random sample of $4$ customers at Anita's, exactly $3$ order their food to go?
Round your response to at least three decimal places.
$= 4C3(0.48)^3(0.52)=0.23003136=0.230$ (correct to at least three decimal places)
I got $0.230$. Am I right?
Yes you are right.
Generally, you can utilise the binomial distribution. $X$ ~ $B(n,p)$. Used to find the probability of $x$ successes (as opposed to failures) in $n$ trials where each trial has two possible outcomes (success or failure). $p$ is the probability of success of a single trial, and since there are only two outcomes of a single trial, $1-p$ is the probability of failure. The binomial formula below gives the probability of $x$ successes in $n$ trials.
$p(x)={n \choose x}p^x(1-p)^{n-x}$
In your problem, let "to go" be success and "eat in" be failure. Each customer is a trial. Then $p = 0.48$, $n = 4$ and $x = 3$. Then plug into the formula to find the probability of 3 "to go" (success) in 4 customers (trials).
$p(3)={4 \choose 3}0.48^3(1-0.48)^{4-3} = 0.230$ (which is what you have, except i think you just used logic instead of this known formula)